Fun Brain Teaser
Go to Dean’s World and try to answer this brain teaser.
Coincidentally, my father-in-law and I were just discussing this over the Christmas break. We tested the answer by actually performing the test. Using the principles mandated by the correct answer, I got it right five times in a row. Chance this was coincidence: 1 in 243.
I’m with Dean: if you disagree with the answer, let’s play the game — for money. I’ll take you to the cleaners.
Many people in Dean’s comments try to explain this in a way that is simple. I think my explanation is simpler and better than any other that I have seen. First go do the brainteaser, and then come back here and click on the extended entry for my explanation. My explanation makes more sense to me than any other one I’ve read:
[Extended entry]
Imagine that a ball could be in any one of 200 closed boxes, which are all on the floor of a large room. I draw a line down the center of the room. Now 100 of the boxes are on the left side of the room, and the other 100 are on the right side.
There’s a 50% chance that the ball is on the right side of the room, and a 50% chance that it’s on the left side. Remember this fact, because it will remain true throughout the entire explanation.
Now I open 99 of the 100 boxes on the left side of the room to show you no ball. There’s one closed box still there on the left side of the room. Meanwhile, there are still 100 closed boxes on the right side of the room.
Here’s the key fact: There’s still a 50% chance that the ball is somewhere on the left side of the room, and a 50% chance that it’s on the right side of the room. That fact was true before, and it’s still true. If I asked you which side of the room the ball was on, you would have a 50% chance of being right no matter which side of the room you picked.
But it’s very different if I tell you that, of the 101 boxes that are still closed, you can open only one box.
Which will you open?
There is a 50% chance that the ball is on the right side of the room. But that 50% chance is divided between 100 boxes. Each has 1/2% chance of holding the ball.
There is also a 50% chance that the ball is on the left side of the room. But that 50% chance all belongs to one box — because you know the other 99 don’t have it.
You can open one box of the 101 closed boxes in the room. Which box are you going to open? One on the right side of the room, where you have a 1/2% chance per closed box? Or the one on the left, where there’s a 50% chance that the ball is in the single closed box?
It’s a simple decision. You pick the single box on the left side of the room — the only box where your chances are still 50-50.
Does that help anyone? It helped me.
I have known about this “brain teaser” for years. I discovered it on my own, but the father-in-law mentioned above also happens to be my father. I know that Dean and “Patterico” and my father are right, but I find it very difficult to grasp intellectually.
My gut reaction is to say, if you’ve got 200 boxes, there is a 1/200 chance that the ball is in any particular box, and you can’t change that. It seems like a gambler’s fallacy to say that if you split the boxes into two sides and open some of them, you will have somehow changed those odds.
Since I know my gut reaction is wrong, the best way I can find to understand the correct answer, is to think of a little man on the left side of the room who is giving you information (allowing you to cheat by peeking into almost all of the boxes). No one gave you any information about the boxes on the right, so of course you’re going to going with the side where you peeked.
I recently bought a book called the curious incident of the dog in the night-time, which is what reopened the discussion for my family.
Comment by Mrs. Patterico — 1/7/2005 @ 10:48 pm
This is a GREAT brainteaser. Thanks for passing it along, Patterico.
I too had a hard time with it until something clicked. To me, the simple solution is this:
Given your initial choice, one out of three,
Comment by Tom — 1/8/2005 @ 12:54 am
Just a note: it’s important that the person who opens the 99 boxes knows where the ball is. Otherwise that line in the middle of the room becomes meaningless and the probabilities are still distributed evenly among all the boxes.
Comment by Doc Rampage — 1/8/2005 @ 1:01 am
This is a GREAT brainteaser. Thanks for passing it along, Patterico.
I too had a hard time with it until something clicked. To me, the simple solution is this:
Of the initial three doors, your chance that you picked a wrong door is 2/3. Monty then does you a favor by eliminating the other wrong door for you. Therefore, you jump to the remaining door, which has the best chance of being the cash. (Sure, you could have chosen the money door the first time, but your chances of getting it right were only 1/3, whereas your chances of getting it wrong were 2/3.) Thus, you’re betting that you were more likely to get it wrong initially, which you were, and then Monty eliminates the other wrong door for you–and suddenly, your chances increase dramatically.
Hope that helps–it’s the only way I can really understand it.
Comment by Tom — 1/8/2005 @ 1:04 am
Sorry for screwing up with an incomplete post. My browser is putting the “post” button in the middle of the form, for some unknown reason (I’m on my mom’s old computer–Mac OS 9. Eww.)
Comment by Tom — 1/8/2005 @ 1:06 am
Doc,
I think you’re wrong, but at least you’re thinking like me.
Comment by Mrs. Patterico — 1/8/2005 @ 1:23 am
Doc Rampage is entirely right — the whole Monty Hall problem (and your example) are dependent on the door-opener knowing where the good prize is.
Another way to explain: suppose the Monty Hall situation were 1000 boxes, with a “good prize” in one of them. You pick a box. I know where the good prize is. I pull up 998 boxes other than the one you picked, and all are empty.
So… do you want to keep the box you originally picked, or switch to the other one? You had a 1/1000 chance of being right with your first pick, which means….
(That explanation usually clues people into what’s going on)
More extreme case: I’m thinking of a real number between 0 and 1. Guess it.
(Listens to your guess)
Ok, it’s either that number you guessed, or it’s 0.918273312442
Would you like to change your guess?
Comment by meep — 1/8/2005 @ 2:01 am
Rather than offer yet another proof, I’ll show you why it works the way it does.
Here are all the possible ways the prize could be distributed. $ means the $1,000,000 prize, G means a goat, and B means a can of Bush’s baked beans. The doors are numbered 1, 2, and 3 (I have to use dots instead of spaces so the columns will more or less line up!):
Door:..1….2….3
a…….$….G….B
b…….$….B….G
c…….B….$….G
d…….B….G….$
e…….G….$….B
f…….G….B….$
There are six possible ways the prizes could be distributed: a, b, c, d, e, and f, as shown above.
Since everything is randomized except for Monty’s choice of a door to open, let’s just assume you always pick door 1, and let’s assume you always switch when allowed.
In case a, you pick 1 (which has the money); Monty opens either 2 or 3 (it doesn’t matter), you switch and you lose. You get bupkis.
Case b is exactly the same as case a: you lose.
In case c, you pick door 1 (which has the beans), Monty opens door 3, you switch, and you win.
In d, you pick door 1 (beans), Monty opens door 2, you switch, and you win.
In e, you pick door 1 (goat), Monty opens door 3, you switch, and you win.
In f, you pick door 1 (goat), Monty opens door 2, you switch, and you win.
Now go back and count the number of wins and the number of losses: four wins, two losses.
That’s why it works. It’s not completely random, because Monty will never open the door with the million dollars.
If it were random, then you would indeed have a 50-50 chance if the door Monty opened did not contain the money… but a third of the time, it would contain the money; and in those cases, you would have a zero chance: neither of the two doors you’re allowed to pick contains the money. That adds up to a 1/3 chance you get the money (it’s exactly the same as if no door were opened at all).
Dafydd
Comment by Dafydd — 1/8/2005 @ 3:57 am
Monty will always open an irrelevant door. The probability is 50/50 start to finish.
Comment by Stan — 1/8/2005 @ 7:59 am
Man we had fun with this one in grad school. You are right it is amazing the number of mathematics and physics wizzes who get fooled by it.
I think the key to understanding is recognizing that Monty’s intervention is giving you information. He *has* to choose an empty door.
Comment by Molon Labe — 1/8/2005 @ 8:40 am
The Monte Hall problem was popularized in a column by Marilyn vos Savant quite a few years ago. There was a huge amount of discussion at that time. The problem was originated by Stanford University Professor of Statistics Bradley Efron.
Comment by David — 1/8/2005 @ 8:51 am
What I meant when I said that Doc was wrong was that it doesn’t matter whether the person opening the boxes knew they were going to open only empty boxes or not — the fact of the matter is that they did open a bunch of empty boxes. We have been given new information whether or not the opener intentionally opened only empty boxes or he lucked into opening only empty boxes.
I first heard about this problem in the Marilyn vos Savant problem mentioned by David.
Comment by Mrs. Patterico — 1/8/2005 @ 9:04 am
Here’s the absolute simplest way of understanding it. Pretend that Monty DOESN’T open up one of the remaining doors, but instead offers you the chance to win the contents of BOTH un-chosen doors. (Think about it - it’s virtually the same thing.) You’ve just increased your odds from one out of three to two out of three. QED.
Comment by MrJimm — 1/8/2005 @ 9:23 am
I think you have a typo in your explanation. I believe that “The one on the left side of the room, where you have a 1/2% chance per closed box?” should read “The one on the right side of the room…” as I read it. otherwise a very good explanation.
Comment by Daver — 1/8/2005 @ 9:46 am
Thanks Daver. Mr. P isn’t here right now, but I think I fixed it.
I like MrJimm’s simple explanation.
Comment by Mrs. Patterico — 1/8/2005 @ 10:13 am
Daver, thanks for the tip about the typo. You’re right. Mrs. P. fixed it whlie I slept in.
Tom’s explanation is exactly right. My explanation is just another way of trying to explain what Tom explains.
Mr. Jimm’s explanation is also very, very good, and captures what’s going on here.
Doc Rampage: it’s only important that the person know where the ball is to the extent that he can’t open up a box with a ball in it. But if the ball is on the right side of the room, he’ll still open up 99 empty boxes on the left side — and choosing that last empty box on the left is still your best bet, statistically speaking. The line will not be meaningless. The probabilities will *not* still be evenly distributed throughout the boxes. You’re just wrong to say that.
(This also means that meep is wrong to say that Doc Rampage is “entirely right.”)
Dafydd is also right — you have a 2/3 chance of winning each time. (Mr. Jimm says the same thing a little more simply.)
Stan: If I understand you correctly, then you don’t believe me. Let’s get together and play the game — for money. We’ll do it with three doors and play it 100 times, each for a fixed sum of $1 to $10. I will be the chooser, and I will switch doors every time, and I will win most of the time. You up for it?
Comment by Patterico — 1/8/2005 @ 11:23 am
Mr.Jimm nails it good, real good.
It seems a key is that if you take one choice off the board, like in the contestant’s case and as you do with the whole right side of the room, the other side always starts to get better odds-wise as negative choices turn up. But it depends on the initial set-up to say for sure whether you should still take that side, I quess up to the 1/2-1/2 start, vs say the 1/3-2/3 start?
Comment by J. Peden — 1/8/2005 @ 1:02 pm
Brain Teaser
Dean Esmay has a doozy. Hat tip: Patterico.
Trackback by damnum absque injuria — 1/8/2005 @ 9:41 pm
Doc’s right: knowledge is essential. If you didn’t know which box the ball was in, the odds of the ball being in any given box are 1 in 200 before you open any of them, and 1 in (200 - n) when you have opened n of them. My only reason to bet on the lone box on one side of the room is because you told me in advance that you know where the ball is and you will open 99 boxes that do not contain it.
Similarly, the intuitive answer to the Monty Hall problem is that the odds are 50-50, since the prize had to be behind one of the three doors, and now there are only two. The reason it is wrong is because one of the two remaining doors was chosen at random, while the other was not. The only random selection was yours, so that’s the only one where you can fix any odds. At that stage, call the door you choose “A,” and there are three equally likely scenarios:
Based on those three possibilities, it’s easy to see why the odds are 2-1 in your favor if you always change your selection to whichever door Monty chose not to open.
Comment by Xrlq — 1/8/2005 @ 10:26 pm
Put simply: you can either see what’s in one of the three boxes, or you can see what’s in two of them. You can get two opened only if you opt to change your original selection.
Comment by L. Barnes — 1/9/2005 @ 12:11 am
The knowledge of the Gamemaster does not have anything to do with it. Monty can simply be told on the spot where one empty space is. Similarly, Patterico does not have to know where the ball is. Someone else could have set the whole thing up. Amazingly, the one box on the left has a 1/2 chance of having the ball while each of the others has only a 1/200 chance. P = 1 = 100/200 + 1/2. It is hard to believe, but has something to do with the extreme unliklihood of opening 99/100 boxes in a row without finding the ball, where the chance is 1/2 that it will be found on that side. Thus there is still a 1/2 chance that the ball is somewhere on the right side, but also with the left. The next [last] box on the left has now a 1/2 chance. If that box did not have the ball, then each box on the right would now have a 1/100 chance.
I’m making this up but I think it is something like: total remaining probabiliy = P = 1 = 1/2 + [1/2 - 1/99!] = pboxes on right + pboxes on left. 99factorial is a gigantic number, so 1/99! is almost 0. I might not be correct about it, but that’s what I think right now.
Comment by J. Peden — 1/9/2005 @ 1:18 am
Tom’s explanation made the most sense to me. By switching, you are betting that you were wrong to begin with, and the odds are that you were. Perhaps the reason the odds are not 2:1 with 2 doors left is that by opening the other “bad” door, Monty has provided more information. At first, we only knew that 1 door had the big bucks, and the other 2 did not. Now, we know what is behind one of the unchosen doors. Even though we are still operating with incomplete information, it is still more information than we had at the beginning.
But still, it all comes down to the fact that the odds are that the player picked the wrong door in the beginning.
Comment by The Calvinator — 1/9/2005 @ 1:48 am
Xrlq,
Doc is only right to the extent that the person who opens the boxes needs to make sure he doesn’t open the box containing the ball.
But even if he knows the ball is on the right side of the room, and opens up 99 boxes on the left side, your best statistical bet is to open that last box on the left side.
Comment by Patterico — 1/9/2005 @ 2:42 am
Xrlq:
Put another way, as long as the box opener opens up 99 empty boxes, it doesn’t matter whether he knew where the ball was, or whether he just lucked into finding only empty boxes.
His knowledge is not the key. The key is that you have one box on one side that has a 50% chance of having the ball, and 100 boxes on the other side, each of which has only a .5% chance. Obviously you will choose the box that has a 50% chance.
This is what Mrs. P said earlier on, and she was right.
Comment by Patterico — 1/9/2005 @ 2:52 am
Xrlq,
Here’s how I know you’re still not getting it, and that you’re letting the “knowledge” thing get in your way. On Dean Esmay’s site, you have a comment where you wrote the following, which is wrong:
Wrong. The odds still favor your making the switch.
If he has eliminated one of the two doors that you didn’t pick, it doesn’t matter whether he did it with or without knowledge. Your best bet is to switch. If you don’t understand this, then you don’t understand the problem.
Comment by Patterico — 1/9/2005 @ 3:11 am
Try this almost simular question:
All people in the world gets the first Monty question. Now he eliminates one of the false doors. If you were still in the game would you switch doors ?
(Why is it different ?)
Comment by Plys — 1/9/2005 @ 3:20 am
Sorry, Pattericos, but you’re both wrong. The only reason the odds were ever 50-50 that the ball was on one side of the room vs. the other was because at that point in time, there were the same number of boxes on each side of the room, and there was an equal probability that the ball could be in any one of them. As soon as those facts changed, so did the odds. If you had known from the start that the 99 boxes were empty, while knowing nothing about the other 101, you would have originally fixed the odds at 100-1, not 50-50. So if this result came about at random, that’s what you should be doing now.
The only exception is when the opened boxes were excluded systematically rather than at random. If you removed 99 boxes at random, I now have more information about all 101 unopened boxes than I had before, but I don’t have any more information about some of them than I have about any others. On the other hand, if you told me in advance that you were going to open 99 boxes, all of which would be on one side of the room, and none of which held the ball, then by doing so you would provide me with no new information about that side of the room, as I already knew there were at least 99 empty boxes there. You did provide me with significant information about the one box on that side that you didn’t open.
Or look at it this way. If you open 99 boxes without knowing where the ball is, and none of them reveal it, there are 101 equally likely possibilities. If you systematically removed 99 empty boxes from one side of the room, there are really only two possibilities:
Then, and only then, are the odds are 50-50 between the two sides of the room. And even that only works if your choice between sides of the room was made at random. Once you start rolling a loaded die, all bets are off.
Comment by Xrlq — 1/9/2005 @ 10:37 am
Another simple way to understand it — look at the expectation. Assume two doors with a payoff of 0 (worthless) and one door with a payoff of 1. If you don’t switch, there is a 2/3 chance of 0 payoff and a 1/3 chance of a 1 payoff, so your expectation is:
2/3*0+1/3*1 = 1/3
OTOH, if you switch and have chosen a worthless door, then you pick the worthwhile door, while if you initially picked the worthwhile door, you now have a worthless one, so your expectation is:
2/3*1+1/3*0 = 2/3
The switching strategy doubles your expected payoff, meaning it doubles the probability of choosing the worthwhile door.
Comment by jd watson — 1/9/2005 @ 12:17 pm
xrlq–
try this:
Each door has 1/3rd chance. Assume that Monty lets you pick TWO doors at the beginning, giving you a clear 2/3rds chance of winning.
Then he shows you one of the doors you picked was bogus and offers a switch. Do you switch?
If not, why is this any different than the original question? The doors and the choice are the exact same thing now.
Comment by Kevin Murphy — 1/9/2005 @ 1:13 pm
We really should see if we can find a way to have this argument on one thread only, without losing input from either thread. (A parallel version is going on here.)
If I understand you correctly, Xrlq, then your argument is very attractive and seems logical — but is dead wrong. Indeed, your fallacy is exactly what the brain teaser is designed to address.
Let me ask you a question to make it clear that we’re not just quibbling about semantics. In light of the language of yours I quoted above, I pretty much know the answer — if you are being consistent. But I want to absolutely nail you down on this. It’s the trial lawyer in me.
Let’s posit the following experiment: you actually perform the Monty Hall trick — say you have someone place a ball in one of three bags. Then you choose a bag.
To remove the factor of knowledge from the equation, you have a *separate* person (not the person who put the ball in the bag) choose a bag. This person is not allowed to know where the ball is before making their selection. Also, they are not allowed to know which bag you chose.
If they choose the one you picked, you ignore that result. Throw it out.
If they choose one you didn’t pick, then they look to see if the one they chose contains the ball.
If it does, you ignore that result too. Throw it out.
If they pick a bag you didn’t choose, and that bag doesn’t have the ball, this example counts. Now we are in your example that you used on Dean’s site:
My experiment replicates this “counter-example” precisely. Remember: the chooser *randomly* picked a bag, without knowledge of where the ball was (or even what your choice was).
Now: the ball could be in your bag, or in the other bag you didn’t pick. You don’t know. And the other chooser didn’t know at the time they made the pick.
True or false: in this example, the odds are not 50-50. You should switch your pick every time. It is no different from the example where the person did know where the ball was when they made the pick.
Based on your “counter-example,” I expect you to answer “false” to my assertions in the previous paragraph. If you answer “true,” then we’re in some semantic distinction that I don’t understand.
If you answer “false,” then you and I need to make a date to do this — for money. We’ll ignore the results that don’t count (person picks your bag, or the bag that you didn’t pick that has the ball). We’ll count the other ones. We’ll do 100 of those. I pick each time and I switch each time. Each time we bet a fixed sum — anywhere from $1 to $10, the same every time — that I get it right.
Expect to lose big.
What’s your answer?
True or false?
Comment by Patterico — 1/9/2005 @ 1:14 pm
Kevin:
Sssshhhhhhhh. I’m trying to take this guy’s money.
Comment by Patterico — 1/9/2005 @ 1:15 pm
JD: Right. But only because Monty’s method of choosing his door is one that tells you more about the unopened door he skipped than it tells you about the door you initially chose. If his selection process had given equal consideration to the two doors that didn’t have the prize (yours and the third), then his opening of Door #2 would have affected the odds of the two unchosen doors equally, i.e., it would have raised both to 1/2, rather than leaving one at 1/3 and boosting the other to 2/3.
Or, if Patterico manages to open 99 boxes on one side of the room without knowing where the ball is, one of two things just happened. One possiblity is that he chose the side of the room that the ball isn’t on. The other is that he picked the side of the room that the ball wasn’t on, defied almost impossible odds, and missed the box 99 times in a row. Which of these two do you think just happened?
Comment by Xrlq — 1/9/2005 @ 1:16 pm
Another way to look at it:
I have 3 playing cards. One is an ace, the others are jokers. I mix them up, you pick one. I look under the other 2. At least one of them is a joker. I turn it up and offer you a chance to switch.
Question: Considering that, until I looked under those other two, there was a 2/3rds chance that one of those two was the ace, how does my showing you that there was also a joker under there change the odds? You already knew that.
Comment by Kevin Murphy — 1/9/2005 @ 1:24 pm
Xrlq,
I assume you are answering “right” to Kevin, not me.
I require the answer to my question to be the form of a “true” or “false” answer, so it’s clear you’re answering me.
Since you’re a bright guy, I expect that you’ll see that “true” is the correct answer — but then you have to acknowledge that your “counter-example” is flawed. Monty’s previous knowledge doesn’t enter into it — at all.
Comment by Patterico — 1/9/2005 @ 1:25 pm
Kevin,
You can keep trying to explain it to him, but I get to be the one who takes his money if he remains obstinate.
(I don’t think he will.)
Comment by Patterico — 1/9/2005 @ 1:28 pm
It is interesting to note that Monty Hall changed the rules as the show matured.
In the end, TWO people chose from 3 doors, then one of them was shown they were zonked. The other person was then given the choice to switch.
Anyone want to suggest whether/how this is different?
Comment by Kevin Murphy — 1/9/2005 @ 1:56 pm
patterico–
ummm … I mistaked xrlq’s position on Monty’s problem. He’s got it right, of course. AND HE’S ALSO RIGHT about the 200 boxes.
The error you are making is with your 50/50 observation.
Then you open 99 boxes on one side of the original line. Now, one could say that this lone box has a 50% chance still, assuming your choice of which side to expose was random. But if you think it’s 50/50 you will have no problem with me picking ALL THE OTHER BOXES.
Comment by Kevin Murphy — 1/9/2005 @ 2:12 pm
Kevin - It’s a different problem when TWO people choose from three doors, and Monty shows that one of them ‘zonked’. The difference is that the percentage of time YOU zonked (and don’t get a chance to re-choose) makes up the ‘advantage’ of re-choosing. So there is NO advantage to re-choosing in THIS case - it’s 50-50. I have a marvelous proof of this, but it won’t fit in this tiny margin…
Comment by MrJimm — 1/9/2005 @ 2:30 pm
Mr Jimm–
I agree, and the thing to note is that one of the two contestants has picked the car 2/3rds of the time, and it is ALWAYS the remaining contestant. Monty’s choice of final contestant was constrained 2/3rds of the time. His choice of which door to open was also constrained in all cases, as he cannot open the unchosen door without altering the rules.
Note that who picks first is immaterial.
Comment by Kevin Murphy — 1/9/2005 @ 2:52 pm
All right, who wants to try a REALLY interesting & controversial version of this? (There is no right answer, and people seem to split 50-50 on it but vehemently defend their answers)
You meet someone who claims he can read your mind. You are shown 100 boxes, and are asked to choose either ONE of them, or ALL of them. The mind-reader tells you that, since he knows which choice you’re going to make, he’ll put a diamond ring in the box you choose (if you choose just one box), but if he ‘reads’ that you’re going to choose the contents of all the boxes, he’ll leave them all empty. He lets you do several trial runs to prove he can do it, and he’s always right. Whenever you choose one box, it’s got a ring of some kind (not diamond yet because it’s a trial run); whenever you choose all the boxes, they’re all empty. You know for sure he’s not using sleight-of-hand or anything, and if the ring was truely in a box you would get it if you chose all the boxes.
Now it’s time for the real thing. What do you choose?
Comment by MrJimm — 1/9/2005 @ 2:52 pm
No, Patterico, that wasn’t my example at all. My analysis applies equally to all rigged games, regardless of how such rigging is accomplished. Either Monty’s choice is made according to neutral criteria, or it’s not.
Here’s a bet I’ll gladly agree to. Fill up the room with 200 boxes, without keeping track of which box contains the ball. Flip a coin to decide which side of the room to open 99 boxes on. Once the boxes are open, I’ll place a bet as to which side of the room the ball is in. I probably shouldn’t be so generous as to let you take a mulligan each time you inadvertently open the box that contains the ball, but just to be a mensch, I will. Once we get to the point where there are 99 open boxes and one unopened one on one side of the room, and 100 unopened boxes on the other, you should have no objection to an even bet at that stage. Just to be really, really, really generous, though, I’ll let you have two to one odds that I guess the right side of the room every time.
Comment by Xrlq — 1/9/2005 @ 3:00 pm
Let’s try my randomized version of Monty Hall, only allow two people to bet instead of one. That won’t work for the rigged game, as it would leave Monty with only one door to open, but it should work fine in the randomized version, where he gets to pick from all three (either at random, or by randomly choosing between the two that don’t have the prize, or by any other criteria that give each of the three doors an equal shot at being picked). I randomly choose to bet on Door A. Patterico randomly chooses B. Monty randomly opens C, revealing no prize.
If Patterico is right, he and I can each boost our chances of winning to 2 out of 3, simply by trading doors.
Comment by Xrlq — 1/9/2005 @ 3:04 pm
Xrlq - Monty cannot “randomly open C, revealing no prize”. One-third of the time, there WILL be a prize there.
Comment by MrJimm — 1/9/2005 @ 3:08 pm
patterico–
Another way to look at the 200 boxes is to consider drawing additional lines, and noting, given your analysis, that the odds change as arbitraily as the lines. Which says that lines on the floor are meaningless.
Let’s say I draw another line, dividing the boxes into quadrants. You open the same 99 boxes on your randomly chosen side of the original line.
By your assumptions, the lone box now has a 25% chance, boxes in the two full qudrants have a 1/2% chnce, and there’s a divide-by-zero error in the empty quadrant.
Or, if you think that wrong, I’ll now erase the first line and one half of the remaining 101 boxes has a slight edge over the other set. (51/101 vs 50/101). Another absurd result.
The lines on the floor are a meaningless distraction.
The problem with your example is mainly that I don’t need to be present for your 99-box opening. I have no initial choice. If I did, I can’t tell from your procedure if my initial choice is a box or a side of the room. If the former, I will repick at random, as my odds, whatever they are, are better than 1/200 and I have no reason to prefer the “lone” box whether I picked it first or not. If the later, it’s always 50-50.
Comment by Kevin Murphy — 1/9/2005 @ 3:14 pm
Here’s why I will eventually own Patterico’s house if we play long enough. As long as the boxes are distributed randomly and he does not know where the prize is every time, for every 200 times we play, the prize will land roughly 100 times on the north side of the room, where he’ll open 99 boxes, and 100 times on the south side, where he won’t. Of the 100 instances in which the prize is on the north side, 99 will get thrown out because Patterico will stumble across the prize in the process. Among the remaining 101 games, I’ll bet on the south side every time, winning 100 times and losing once.
Comment by Xrlq — 1/9/2005 @ 3:30 pm
ummm … in my last post at the end, I’ll change that to picking the side with 100 boxes in it, just because it has more boxes. That side, whatever the odds, is not likely to be at a disadvantage.
Comment by Kevin Murphy — 1/9/2005 @ 3:36 pm
Of course he can randomly open C, revealing no prize. We’re talking about a single iteration of the game, not how it’s going to turn out each time. If we all play randomly every time, sometimes Patterico, Monty and I will all pick the same door. It doesn’t affect the odds of anything; it just makes the example more difficult to understand. That’s why I picked the clean result, where each of us just happened to land on a different door, and where Monty’s just happened not to contain the prize. Although I also noted that it’s OK if he does systematically exclude the prize, as long as he does so in a manner that is purely indepedent of my selection process and Patterico’s.
Comment by Xrlq — 1/9/2005 @ 3:57 pm
Quite a bit to respond to here. I’ll try to take them in order:
First, and this is the trial lawyer in me again: Xrlq, I am objecting to your answer as nonresponsive. My comment called for a clear answer of “true” or “false.” Unless you can explain how it was a “Do you still beat your wife” question — which it wasn’t — I’d appreciate a “true” or “false” so we know which road we’re traveling down here.
I’m reading your answer as “true, but that’s not what I was saying” — correct? But I am not understanding the difference between the “counter-example” you proposed and the real-life way of accomplishing it that I gave. I was just trying to find a real-life way to get to a situation where we had the brain teaser example perfectly replicated — but with the person revealing the empty bag doing so without foreknowledge.
How is my example not exactly that?
Kevin says:
But if you think it’s 50/50 you will have no problem with me picking ALL THE OTHER BOXES.
Correct. I have a 50/50 chance if I stick with the lone box. You also have a 50-50 chance if, rather than forcing you to open only one box, I allow you to pick a side — meaning you can open all 100 on the side with 100 unopened boxes. In that case, we each have a 50% chance of being right.
Xrlq, if I understand your bet correctly, I’ll take it in a heartbeat. It has to be with the “mulligan” you described. We each have a 50% chance of being right, but you’re giving me 2-to-1 odds. I think you’ll be surprised whose house gets owned if we do that all day long.
To make it practical, we’ll have to lower the number of boxes to something manageable.
I’m going to be kind to you and suggest that you run your game by a statistician first. But if you insist on being hard-headed, I’m totally up for taking your money. When do you want to do it?
Your explanation of why you think you’d win is nothing but an articulation of the famous Gambler’s Fallacy. You and I toss a coin ten times and it comes up heads every time, what are the chances it comes up heads next time? Assuming that it is a non-weighted coin, and that the chances were 50-50 for the first toss, the chances are still 50-50. It seems logical that the odds would be greater that it would come up tails, since it already came up heads so many times — but it’s 50-50 every time.
X, your “randomized” version of the Monty Hall game is a different example entirely — because when two people play, Monty can’t freely choose between two other doors for two separate people. In that example, he’s mandated to pick the third door — the one neither of us picked. That increases the chances for neither side — but it’s not the example used in the brain teaser.
I still want a “true” or “false” answer.
Comment by Patterico — 1/9/2005 @ 4:17 pm
Hmmmm. Hold on.
For all my bluster, I’m having second thoughts about your proposed bet.
I gotta think about this.
Comment by Patterico — 1/9/2005 @ 4:44 pm
True or false to what? I stipuled from the beginning that my version of the game - the only truly random one - requires that Random Monty not take into account my selection while making his. You can have him automatically exclude the door with the prize if you want, but he can’t exclude Door A simply because I chose it.
I’ve already explained why your box game will put you in the poorhouse in a heartbeat. On the average, 200 games equals 1 win for you, 99 mulligans, and 100 wins for me.
Comment by Xrlq — 1/9/2005 @ 4:47 pm
Talking to Mrs. P. about it, who says X may be right about the bet *he* proposed. I’m starting to think she (and X) may be right.
But I am still quite confident about the bet *I* proposed above, which I still claim gives the lie to the idea that knowledge is required:
http://patterico.com/archives/003286.php#12162
Comment by Patterico — 1/9/2005 @ 4:48 pm
You can have him automatically exclude the door with the prize if you want, but he can’t exclude Door A simply because I chose it.
But can *we* exclude it? In other words: *he* has no idea Door A is excluded, but if he chooses Door A, that example doesn’t count.
That is the original brain teaser, that is my example, and it doesn’t require knowledge on Monty’s part.
In other words, if Monty’s selection is truly at random — but it doesn’t count if he picks Door A, then it’s the same effect, but no knowledge on Monty’s part is required.
I think we’re getting closer to being on the same page here.
Re-read my comment where I asked for a “true” or “false” answer. I think your answer is “true.”
Comment by Patterico — 1/9/2005 @ 4:54 pm
I still say Monty’s knowledge is completely irrelevant — assuming that the rule is that it doesn’t count when he chooses the door you chose.
If you throw out those examples — and throw out examples where he chooses another door but it contains the prize — then his choice can be completely random, as far as his mental process goes.
I think Dean was right — we’re arguing about semantics.
Comment by Patterico — 1/9/2005 @ 5:02 pm
X:
Where we got off track was that Doc Rampage said: “it’s important that the person who
opens the 99 boxes knows where the ball is” and you said he was right. And he’s not.
Comment by Patterico — 1/9/2005 @ 5:05 pm
For example, you can have a person who *does* know where the ball is temporarily remove the box with the ball and one from the other side without the ball. Then the other person can open 99 boxes on either side. He doesn’t know where the ball is — but he does know to open 99 empty boxes.
Now you put back the two boxes, and there is one closed box on one side and 100 closed boxes on the other. And the chances that the ball is on one side or the other side is equal — 50-50. But if you’re allowed only one box to open, you’ll open the lone box. Every time.
You agree with that, yes? Because that was my original point.
Comment by Patterico — 1/9/2005 @ 5:09 pm
Hey, Mr. and Mrs. P. I’ll propose a more practical game than Xrlq’s. We’ll play “Let’s make a deal” The loser gives his money to the tsunami relief efforts and has to post an announcement on his blog that the winner is lot’s smarter than he is. I’ll actually just trust you to play for me and send in your check.
Here is the game: take three cards, a King and two Jacks. Shuffle and deal. Pick one at random (we’ll call that the choice card). Remove another one at random. Look at it, and it it’s the King, the hand is a push, no one makes money. Otherwise, if the choice card is a King, I win and if the other card is a King, you win.
According to you, you should win twice as often as me (2/3 vs. 1/3). According to me, we win equaly often. Let’s split the difference so each time you win I’ll owe you $1 and each time I win you owe me $1.50. Play as many times as you like, those poor people need the money.
Or, you could just sit back for a moment and think about those push hands and what they mean for your 2/3 to 1/3 advantage.
I’m going to post something on my blog about knowledge and probability. Hopefuly tonight.
Comment by Doc Rampage — 1/9/2005 @ 5:14 pm
But what I think your example (which I fell for, for a few blustery minutes) illustrates that *someone* has to know where the ball is, so that the 99 boxes are opened in a way that doesn’t reveal the ball.
Comment by Patterico — 1/9/2005 @ 5:15 pm
In other words, Doc Rampage is wrong, technically, the way he worded it — but the principle he was
trying to express (*someone* has to have that knowledge) was right.
Comment by Patterico — 1/9/2005 @ 5:19 pm
Doc,
I think you’re attributing to me an opinion that I don’t hold.
You said: “it’s important that the person who opens the 99 boxes knows where the ball is”
You see that this is not right, don’t you? You can argue that it’s a matter of semantics — because *someone* has to know — but it’s not right to say that the dude choosing the 99 boxes has to know. He just has to know he’s opening 99 empty boxes. He can be completely ignorant as to which side of the line the ball lies. (See my box-removal example above.)
To me that’s an important point.
But yes, *someone* must know, to be able to remove one box from each side (one of which holds the ball), leaving 99 empty boxes for the other guy to open. I’ll agree with you there.
When Xrlq agreed with you, on that point, that’s where we got lost. I think it’s a function of how you worded it.
See what I mean?
Comment by Patterico — 1/9/2005 @ 5:31 pm
Admit it, Patterico. Doc was right, period. At the point when he commented, there wasn’t one Monty who made a selection blindly and another who knew all and got to veto any selections that a more knowledgeable Monty would never have made in the first place. There was only one Monty involved, and the Full Monty either knew which doors to systematically exclude, or he didn’t.
Comment by Xrlq — 1/9/2005 @ 5:32 pm
Not at all. He was wrong only on a nitpicky technicality, but he was wrong.
He would have been right if he had simply said that the person had to have some way of knowing he was opening only empty boxes.
But he went further than that and said the person had to know where the ball is.
That is *not* the same thing. As my example illustrates, as long as there is a mechanism to assure that the ball with the box is not opened, it doesn’t matter that the person who opened the boxes knows which side of the line the ball is on.
Once the second person takes away a box with a ball and a box without a ball, the box opener can randomly choose which side on which to open 99 boxes, and it won’t change my example in the slightest. He can open them on the left side, and now the remaining box on the left side is your best bet. Or he can open them on the right side, and the remaining box on the right side is your best bet. All of this can happen without him knowing which side of the room had the ball.
As I said up front, this is all very nitpicky. Doc had the essential truth right.
Comment by Patterico — 1/9/2005 @ 5:50 pm
Sorry, I guess I misunderstood. I thought at one point that you were arguing that the box removal could be completely random.
I guess to be more precise I should have said that the box-removal must use a mechanism that guarantees the box with the ball will not be removed, otherwise you are essentially just throwing out all the cases where that unfortunate event happens and this screws up the probabilities.
Comment by Doc Rampage — 1/9/2005 @ 6:32 pm
Anyone who has made it this far (all two of you) presumably won’t mind if I put it one other way.
I take two cards from a deck: the A of spades (As) and another random card (x). I turn them face down and move them from hand to hand so that nobody, including me, knows which is As and which is x. I hold one in my right hand and one in my left.
I ask Doc Rampage to split the remaining 50 cards into two piles of 25, turn them face down, and remove one pack. I don’t care which.
He knows the As is in one of my hands but doesn’t know in which hand (neither do I).
I now put one card on the top of the remaining 25, and the other card face down on the table.
I shuffle the 26 cards.
There is a pile of 26 cards and a separate card face down. 27 cards total. Nobody knows where the As is.
You may pick one card of those 27.
Which one ya gonna pick? Obviously, the lone card. It’s not 1 in 27: it’s 1 in 2.
And Doc never had to know where the As was — as long as he knew I took it from the two piles of 25.
If Doc said: “it’s important that the person who removes one deck of 25 cards knows where the As is.” — that would not be entirely right. It would be right to the extent that it’s important that he know it’s not in either deck of 25. But beyond that, Doc never has to know where the As is — which of my hands it was in, whether I put it on the pile of 25 or whether I put it down on the table by itself.
Either way, selecting the lone card is the right way to go — and Doc need never have known the precise location of the As.
Does that help?
Comment by Patterico — 1/9/2005 @ 6:41 pm
I never meant to argue that the box with the ball could ever be removed. If I said something that sounded like that, that was wrong — but I don’t *think* I did. (Who knows? This is all so complex and abstract that misunderstandings are, well, understandable.)
Comment by Patterico — 1/9/2005 @ 6:44 pm
One more thing. When I made the first comment, I thought I was just making a nitpicky comment on your presentation :-).
Comment by Doc Rampage — 1/9/2005 @ 6:58 pm
Fair enough, but if drugs were legal, we never would have had to worry about this issue in the first place.
Comment by M. Simon — 1/9/2005 @ 7:25 pm
The answer is as simple as this. If I’m playing against you in a game of high card wins, and the deck consists of an ace and two threes, and I get one card and you get two, you can show me a three every time but it doesn’t change the odds that you’ll win most of the hands.
Comment by Boman — 1/9/2005 @ 9:33 pm
MrJimm had the right idea and then turned around and lost it when talking about “zonking”. When two people are playing “three doors” and Monty can zonk either one of two players, the remaining player is
playing odds of two out of three if he stays with the door he already possesses.
I understand that MrJimm is talking about continual play and the capricious choice that Monty could make but that is not the game we’re playing.
Comment by Boman — 1/9/2005 @ 10:27 pm
Pattericio’s example (of boxes divided into a right and a left section) forces all the action to the left section where I have one of two choices - has he fooled me or not? It is a wash.
But it does demonstrate how the odds should force you one direction or the other regardles of your intuition.
Comment by Boman — 1/9/2005 @ 11:30 pm
regardless - not regardles
Comment by Boman — 1/9/2005 @ 11:33 pm
Patterico–
After careful consideration, I think that your 200 box clarification of the long since disposed of Monty-problem was probably not as clear as you hoped.
Comment by Kevin Murphy — 1/10/2005 @ 12:51 am
probabilty and the universe
Probability isn’t about the outside universe at all, it is fundamentally about knowledge. A numeric probability is a concise summary of what is known (or believed) about a situation. It isn’t strictly about it’s apparent subject at all.
Trackback by Doc Rampage — 1/10/2005 @ 12:58 am
After careful consideration, I think that your 200 box clarification of the long since disposed of Monty-problem was probably not as clear as you hoped.
Heh. I think that, 71 comments later, you may be right.
I defer to Mr. Jimm’s initial explanation as the best.
Comment by Patterico — 1/10/2005 @ 8:39 am
Monty’s previous knowledge doesn’t enter into it — at all.
Xrlq is right; Patterico is not (at least with regards to the truth of the above proposition; as to technicalities of who asked which true-or-false question, or which statement of Doc Rampage one of them was agreeing with, I have no comment).
The mistake Patterico made was where he blithely “threw out” the results that led to Monty opening the door with the prize, or randomly selecting the same door that you chose (or, in Patterico’s example, the results where opening boxes led to finding the ball being found prematurely).
He called this: a mechanism to assure that the ball with the box is not opened.
But this process (of throwing out bad results) is the same thing as knowledge. Either it is actual, prior knowledge (by being the one who decided where the prize is), or it is constructive knowledge (obtained after the fact by process of elimination). The “process of elimination” in your example is the part where you are throwing out the results that you do not want.
Start with Dafyyd’s explanation, above. After you “throw out” the results you do not want (where Monty randomly chooses either the same door as the contestant, or randomly chooses one with the cash prize), you are left with only two possible scenarios: (a) or (b).
At that point, from that narrowed pool of possible scenarios, the odds of choosing the prize is 50-50, and switching your choice is irrelevant.
In other words, the fact that you are “throwing out” some of the results, you are narrowing the pool from which the results are taken.
And remember: statistics concerns the relationship between (a) the total array of possibilities, and (b) one possible scenario within that array. By altering the scope of (a), you alter the statistical probability of encountering (b).
Comment by George Gaskell — 1/10/2005 @ 10:55 am
George,
I completely agree with you that the mechanism performs the same function as knowledge. I think my point was just that you can have the mechanism without prior knowledge on the part of the person doing the choosing. As long as you set up the experiment so that something (prior knowledge by the chooser, or an after-the-fact mechanism) excludes what we want to exclude, the experiment will work. What I meant to say is that it is possible to accomplish this without regard to prior knowledge.
I said this in different language at Dean’s World, here.
Comment by Patterico — 1/10/2005 @ 11:07 am
you are left with only two possible scenarios: (a) or (b). At that point, from that narrowed pool of possible scenarios, the odds of choosing the prize is 50-50, and switching your choice is irrelevant.
I should correct myself. Dafyyd’s example randomizes the starting point, not the choosing, thereby making the explanation simpler.
What I should have said is that: (1) you always choose Door No. 1; Monty always opens Door No. 2; and you are throwing out the possibilities where Monty’s pick reveals the prize.
This process of elimination removes scenarios (c) and (e) from being possible. Thus, you have only FOUR possible scenarios, not two — a, b, d & f. My apologies.
In any event, by doing so, you are back to a 50-50 game, and switching is meaningless. My point is that throwing out some of the results narrows the array of starting conditions, such that it alters the odds of randomly encountering one of those that remain.
Comment by George Gaskell — 1/10/2005 @ 11:08 am
The only way an after the fact mechanism can make Patterico’s example work is if that mechanism automatically throws out a large number of otherwise valid selections. If Patterico tells you he’ll flip a coin to decide which side of the room to clean out, and that he will then open 99 boxes on that side of the room without revealing the ball, then odds end up 50-50 between the two sides of the room, just as they started. But if he does so by flipping a coin, opening 99 boxes on that side at random and throwing out the results every time he stumbles across the prize, the odds will be evenly divided among the remaining boxes, at 100 to 1. Or, if you prefer, at 100 to 100, but 99 of them don’t count.
Comment by Xrlq — 1/10/2005 @ 12:48 pm
In other words, to get Patterico’s 50-50 odds to hold, Patterico either has to know in advance which box to avoid, or his after-the-fact “correction” mechanism has to throw out just as many games (99 out of 100) where the ball ended up on the other side of the room as he had to throw out because the ball was on his side and he stumbled across it. Otherwise, you’ll sysematically exclude 99 of the 100 scenarios that put the ball on Side A, while leaving all 100 of Side B’s scenarios untouched, and the odds against Side A will be just as long as they appear (100:1).
Comment by Xrlq — 1/10/2005 @ 12:55 pm
The odds are never 50/50. The contestant either has odds of l of 3, or if allowed to switch, 2 of 3.
You make the mistake of “throwing out some of the results” when you should be adding the results.
Remember the old brain twister about the three men who paid $30.00 for a hotel room. Same thing - people add when they should substract, but here you are subtracting when you should be adding.
Comment by Boman — 1/10/2005 @ 1:14 pm
That last post is addressed to George Gaskell. Sorry for any confusion.
Comment by Boman — 1/10/2005 @ 1:19 pm
Here’s yet another angle on Monty’s original example. When originally picked A, you knew that the odds were 2/3 that either B or C had the prize, but you didn’t know which. By deliberately choosing the booby prize between them (and you knew all along that there was one), Monty has just told you which.
Comment by Xrlq — 1/10/2005 @ 1:20 pm
Boman,
I know that. In the original question, the odds are never 50-50. What we have been discussing is an alternative scenario, designed around the question of whether Monty’s knowledge (of which door holds the prize) is a relevant factor.
To do that, we set up the possibility that Monty does NOT know which door holds the prize, but merely guesses a door, just as the contestant did.
But that presents some problems: first, Monty might choose the same door as the contestant. Second, he might choose the door with the prize. Only if you throw out these two scenarios from consideration do you get a situation in which: you have selected a door, and Monty has revealed another door that doesn’t hold the prize.
Only under these more limited set of circumstances would there be a 50-50 probability that switching wins the prize (i.e., switching would be meaningless).
The only reason that this alternative situation presents a 50-50 probability is that you have selectively REMOVED several scenarios from the set of possible starting points, thus altering the calculation of probabilities.
If we go back to the original question, the reason that switching is beneficial is precisely BECAUSE Monty’s reveal of one of the wrong choices is NOT made at random. It is done with knowledge of: (a) which door you already chose and (b) which of the other two is a wrong choice.
Another alternative scenario which gives you an even, 50-50 chance of improving your chances by switching would be for the prize location to be re-randomized after Monty’s reveal — i.e., you pick a door, then Monty picks one of the wrong doors, then the prize is either moved or not, at random, to either your door or the other. In this case, you’d be starting over with only two doors.
I think this is what most people think is happening, which is why the 50-50 theory seems “intuitive” to many people. The reason that this situation (Alterntative Scenario #2) presents a 50-50 chance of improving your odds (whereas in the original example it is better to switch) is that the prize is NOT re-randomized after you select a door. It is assumed to be fixed. They do NOT move the prize (and cannot do so). Thus, the odds remain constant from start to finish, and the only thing that changes is your level of knowledge when Monty reveals a wrong choice.
Some people seem to assume that because, after “Monty’s reveal” you still have imperfect knowledge, that the situation is random to the same degree as if the reveal never occurred. It is not; the prize is fixed from the original starting point, but after the reveal, your level of knowledge has greatly improved.
Comment by George Gaskell — 1/10/2005 @ 1:53 pm
I think this is what most people think is happening, which is why the 50-50 theory seems “intuitive” to many people. The reason that this situation (Alterntative Scenario #2) presents a 50-50 chance of improving your odds (whereas in the original example it is better to switch) is that the prize is NOT re-randomized after you select a door. It is assumed to be fixed. They do NOT move the prize (and cannot do so). Thus, the odds remain constant from start to finish, and the only thing that changes is your level of knowledge when Monty reveals a wrong choice.
I am having a bad day! (This is what you get for trying to solve brain teasers while also working on requests for production of documents.)
What I meant to say is that the reason that the re-randomizing scenario (Alternative Scenario #2) makes switching meaningless is because you are, in effect, starting over. The randomizing happens twice, thus making the original guess meaningless. It is no different that if you just started out with two choices.
I think this is the situation that most people think of when they (wrongly) conclude that switching your guess has only a 50-50 chance of improving your situation. They imagine the situation after the reveal, and they see that the contestant still has imperfect knowledge, and they assume that the situation is still random to the same degree as when they started.
But it is not. When compared to the original starting conditions, Monty’s reveal introduces important information precisely BECAUSE (a) the reveal is not random, and (b) they do not move the prize once the game starts.
What I want to know is how this discussion went from statistics to talk of the “full Monty” and “Monty’s reveal.”
Comment by George Gaskell — 1/10/2005 @ 2:07 pm
George - tell me what is the difference between Monty with knowledge, able to reveal the door with no prize, and a random pick by Monty where you throw out the times when Monty picks the same door as the contestant, or when he picks the door with the prize. There is no difference. Knowledge that reveals information and a random choice that reveals the same information is the same.
The odds are not changed to 50/50 with the random choice as constrained by your example. They remain at 1/3, or 2/3.
Comment by Boman — 1/10/2005 @ 2:29 pm
There is a difference, but maybe not in the way you think I mean.
In the Monty with Knowledge scenario, the odds of improving by switching are 1/3 - 2/3, for reasons which we both understand.
In the Monty at Random scenario, the odds are also 1/3/ - 2/3 in any particular game (for the same reason that flipping a coin is 50-50 every time you do it, and previous and subsequent flips do not affect it in any way).
However, you calculate odds based on the total universe of starting sets. If you “throw out” a portion of the possible starting sets (not selecting the ones you throw out at random but doing so precisely because those sets fit a certain pattern), then the overall odds can be said to have changed, because you are calculating the anticipated chance of getting one particular outcome from a smaller, selectively reduced universe of starting conditions.
In the Monty at Random scenario, a portion of the set that would have fallen into the 1/3 side of the line have been “thrown out” because Monty picked the door with the prize. If you base your odds on the whole universe of possible starts, then the odds are 1/3 - 2/3. If you base your odds on only the subset of games that you don’t throw out, the odds are 50-50.
Comment by George Gaskell — 1/10/2005 @ 2:51 pm
There is no “Monty at Random.” He does not exist. Therefore he cannot play in any of your games.
When “Marty at Random” has three doors to choose from and you disallow two of those choices, there is no randomness no matter how many or how few games you play.
Your “Monty at Random” is a chimera.
Comment by Boman — 1/10/2005 @ 3:30 pm
George is right. When you select Door 1 and the Full Monty selects Door 2, he communicates valuable information about Door 3, which cannot be communicated simply by censoring Random Monty after the fact. Here’s why. Between your random three-way choice, Random Monty’s, and the prize, there are 27 possible combinations, 15 of which will be systematically excluded. Among the remaining 12, your odds are no better or worse if you switch.
To illustrate, let’s limit the 27 choices to 9, by stipulating that you play Door 1 every time. Also stipulate that you, like Bornan, expect the Random Monty + Correction model to work just like the Full Monty, so you change your bet every time the game isn’t canceled.
Scenario 1: Door One holds the prize.
Game 1: RM selects Door 1 (yours, and the prize). Game canceled.
Game 2: RM selects Door 2, you switch to 3. You lose.
Game 3: RM selects Door 3, you switch to 2. You lose.
Scenario 2: Door Two holds the prize.
Game 1: RM selects Door 1 (yours). Game canceled.
Game 2: RM selects Door 2 (prize). Game canceled.
Game 3: RM selects Door 3, you switch to 2. You win.
Scenario 3: Door Three holds the prize
Game 1: RM selects Door 1 (yours). Game canceled.
Game 2: RM selects Door 2, you switch to 3. You win.
Game 3: RM selects Door 3 (prize). Game canceled.
I count five scratches, two wins, and two losses. Multiply everything by three to allow the same combinations of events when you choose Doors 1 and 2, and now you’re up to fifteen scratches, six wins, and six losses. No overall advantage or disadvantage to keeping vs. switching your bet.
Comment by Xrlq — 1/10/2005 @ 4:17 pm
In other words, I should have taken Patterico up on his bet.
Comment by Xrlq — 1/10/2005 @ 4:53 pm
When “Marty [sic] at Random” has three doors to choose from and you disallow two of those choices, there is no randomness no matter how many or how few games you play. Your “Monty at Random” is a chimera.
“Monty at Random” is (a) not mine, and (b) no less real than Contestant at Random. You are not disallowing two of the choices — it is only two out of three if you picked the wrong door. If you picked the right door, then Monty could pick two of the three, at random. See Xrlq’s example, Scenario 1, Game 1.
Of course, you do not know until it’s too late if you are in this scenario or not.
Comment by George Gaskell — 1/10/2005 @ 5:12 pm
What I MEANT to say was:
Compare Xrlq’s example, Scenario 1, Game 1 with Id., Games 2, 3.
And, in total, you disallow 5 of 9 possible outcomes (or 15 out of 27), not 2 out of 3.
Sheesh! I’m not my usual totally-error-free self today.
Comment by George Gaskell — 1/10/2005 @ 5:16 pm
You’ve got questions. We’ve got answers.
Dean Esmay started the discussion by posing a classic brain teaser, then Patterico picked up on it, and before long, Xrlq had, too, betting the deed to his house on his confidence in his conclusions. The puzzle is simple. Monty…
Trackback by George Gaskell — 1/10/2005 @ 5:42 pm
You’re right. When the contestant has the door with the prize, Monty can be random because he has two choices. It is only when the contestant does not have the door with the prize, is it not possible for Monty to be random. (other choices might seem to be random but since the rules do not allow them, they must be disregarded as random choices.)
Now, I’ve got to start all over.
Comment by Boman — 1/10/2005 @ 9:27 pm
I fall into the switching makes no difference camp. My explanation (yes! yet another attempt to clarify) is that although the odds of correctly picking the correct door are originally only 1 in 3, by removing one door, Monty brings the odds up to 1 in 2 (as choices are removed, the odds necessarily change; they do not remain fixed at 1 in 3. This is key).
By offering the opportunity to switch, Monty is allowing you to choose between two doors, so your odds now become 1 in 2, rather than the original 1 in 3.
Not switching is the same as choosing anew; this time with 1 in 2 odds. Therefore, switching or not switching should produce the same result when enough games are played to be statistically significant (anyone care to do the appropriate power analysis?).
I did a small test of 20 tries. My results came out close enough to CBS work to convince me.
The original explanation for the brain teaser is seductive, but sometimes common sense pervails.
Apologies to anyone who already provided this explanation. Which seems, now that I read more posts, to be everybody who agrees with me and is far more articulte than me.
Comment by Pigilito — 1/11/2005 @ 1:12 am
Random Monty does not exist. Yes, I’m back to that. It appears that RM has a choice when the contestant chooses the correct door but it is a similarity and does not count as two choices but one. Whether RM chooses door 2 or door 3 the results are the same and because you can only play one game with the contestant choosing door one(containing the prize) the contestant loses once - not twice. Random Monty is once again a chimera. You have been misled by the idea that the contestant can lose twice in one game. That is an impossibility.
Comment by Boman — 1/11/2005 @ 6:43 am
Sorry Pigilito, you cannot remove one door. That is your mistake. You must include that door in your calculations as information, information that you can use to your advantage. The odds will always be either 1/3, or 2/3, determined by whether you switch or not. We are playing 3 doors, not 2 doors. You can remove one of the doors in your mind but you cannot remove the door in real calculations of the odds.
Comment by Boman — 1/11/2005 @ 7:15 am
Boman, you seem to be a bit unclear on the concept. No one ever claimed that “Random Monty” existed. He’s a hypothetical counter-example, offered to rebut Patterico’s earlier (now-discarded?) theory that the real Monty’s knowledge of the prize was irrelevant to how he played - or relevant only in the sense that there had to be some method of discarding the illegal games in which he lands on your selection or the prize. To demonstrate that this was not true, we posited an alternative Random Monty who truly acts at random, then compared and contrasted the outcome of Random Monty’s games, vs. the outcomes associated with the real game as played by the real Monty (a.k.a. The Full Monty).
Now do you get it? Finally? I’d think that if you had really applied yourself, you might have figured it out on your own in less time than it took you to pull the word chimera out of your pocket thesaurus.
Comment by Xrlq — 1/11/2005 @ 8:55 am
Boman,
I see it that Monty removes a door, not literally, of course, but by demonstrating that no prize lies behind it. Doing so, in conjuntion with offering the contestant the chance to switch brings the odds back up. This has the effect of Monty offering you a choice between two doors; thus upping your odds. If the contestant had not had the opportunity to switch, the odds would remain 1 in 3.
Odds change as doors are eliminated. For example, I choose door A, Monty opens door B, offering me the opportunity to stitch. My odds are up to 1 in 2. Monty opens door C; it is empty. My odds now are 1 in 1.
OK pretty simplistic. Here’s another: 100 doors are offered. I choose door 1. As each empty door is opened, my odds of having picked the right door go up. The odds that I picked the right door do not remain at 1 in 100. When we are down to two doors my odds of having picked the right door have increased from 1 in 100 to 1 in 2, notwithstanding that no door was actually removed.
If you disagree, sell me your door when 98 doors have been elimanated. I imagine you will want more than 1/100th the value of the prize by that point.
Comment by Pigilito — 1/11/2005 @ 9:30 am
Pigilito: your theory works for the Random Monty counter-example, but not in the real game. Since Random Monty shoots blindly, his selections don’t give you new information about the prize, except in the sense that they reveal it isn’t behind the door he just opened. Real Monty doesn’t shoot blind. He knows where the prize is. One time out of three, it’s out of his reach. Two times out of three, he finds the prize behind one of the two doors available to him, and invites you to pick that door by eliminating the other.
Comment by Xrlq — 1/11/2005 @ 9:45 am
Xriq-
Just so there is no misunderstanding, I quote you, “(There is)no overall advantage or disadvantage to keeping vs switching your bet.”
Is this your position?
Glad I was able to introduce a new word to your vocabulary. It was already in mine. It probably irritated you that you had to look it up. Sorry about that.
Comment by Boman — 1/11/2005 @ 10:02 am
Pigilito-
In your game of l00 doors, if random chance has eliminated 98 doors with yours still standing, the chance that you have chosen the right door go up dramatically. I would stay with your door.
If Monty (with knowledge of the right door) opens 98 doors, your door does not increase in value. It retains the original l/100 odds. I would switch to the door that Monty has not chosen. The odds have moved to that door.
Comment by Boman — 1/11/2005 @ 10:33 am
While I’m way too broke to gamble with Patterico, Dean, XLRQ, or anyone else… I have to say that I disagree with them all.
I completely understand the explanations in the threads on their web pages & the more scientific proofs (http://www.panix.com/~mshaw/3door.txt) posted… all given by people with much bigger brains than I. But IMO, there is a serious flaw in the logic presented.
Let me try to explain.
In order for there to be a 1/3 or 2/3 probability in choosing the correct door after Monty opened the losing door (Door C)… Door C would still have to be a possible choice for the contestant. Since it’s not… there in fact is never a 1/3 choice to make… only a 1/2. The game would have essentially been the EXACTLY same if Monty only asked the contestant to pick one of two doors at the start. The first selection is a fraud… smoke & mirrors… because as soon as Door C is opened and eliminated, you are left with only two… one which is definitely the winner, and the other definitely the loser. The only possible probability at this point is 1/2… 50-50.
Check out the Mathematical proof:
This is correct… P(omega) = P’(Door A) + P’(Door B) = 1
The reason this proof works out algebraically, is because after Monty opens and eliminates a losing door, the equation continues as:
P(Door A) = P’(Door A) = 1/3 and P(omega) = 1/3 + P’(Door B).
Of course continuing with this logic, however flawed, will yield a 2/3 chance of success in switching your choice.
But the equation SHOULD have continued as:
P(Door A) = P’(Door A) = 1/2 and P(omega) = 1/2 + P’(Door B).
Door C was eliminated… no longer part of the equation… yet was still included (incorrectly) as a possible choice. But in fact, it was not. In the riddle, Monty said, “Okay, now, would you like to stay with your original guess, or would you like to switch to the other door that’s still closed?”
Door C was not an option. Only Doors A & B were. The choice is 1/2… not 1/3.
The same reasoning applies to the Set Theory, The Step-by-Step Theory, The Empirical Model, and all the rest (I’ll have to exclude the Computer Simulation Proof, cuz I’ve no idea what’s going on there.) At the point where one door is eliminated, and the original equation for probability also eliminated… the elimination should be reflected in the proofs. It should be a clean slate… 1/2… 50-50. The only time the contestant has an actual chance of picking the winning door is in the final selection. The first choice, 1/3, is just smoke & mirrors… a setup, like in the old shell game. No matter what door was originally selected, Monty was going to open AND ELIMINATE one losing door. No matter what door was originally selected, the contestant was going to be left with two doors… one absolutely a winner, and the other absolutely a loser. Picking the correct one is 50-50. The door originally selected is irrelevant. Switching your choice after the known losing door was opened is also irrelevant. No matter what, the game always comes down to a choice of one in two. Period.
So all of us who first believed we were correct in stating that a switch wouldn’t matter, but changed our minds once reading the proofs… well, now we know why, and can switch back. If you’re having a hard time admitting you were wrong… Dean has an excellent post in dealing with that dilemma on his page. Check it out. :o)
-Amerikan
Comment by Amerikan — 1/11/2005 @ 11:18 am
Boman and Xrlq (damn that’s hard to type): Random Monty? Yes, after reading a few more posts I see that this has been discussed already. Sorry to have rehashed that bit.
Nevertheless, Boman, even if all knowing Monty opens 98 doors, my odds have improved to 1 in 2. If I understand your argument, the other door has vastly improved odds, whereas my door is for losers.
Can you quantify the improvement? And what if 1000 doors were originally offered? At what point do the odds approach 1 to 1? It seems from your argument that the more choices originally offered, the more of a sure thing it would be to eventually switch. I can’t buy that.
Please, someone test this idea. Some 500 times ought to do it for my peace of mind.
As it seems that there are just a few of us left, and I seem to be doing the least to uphold the proper point of view, I think I’ll sit out for a while unless something really really important occurs to me.
Comment by Pigilito — 1/11/2005 @ 11:19 am
All right American and Pigilito, let’s get practical.Forget the theories. Get out the cards - ten of them. Make one of them an ace - that’s the prize. Make all the rest something other than an ace.
Do a blind shuffle so you don’t know where the ace is. When you’re done, take a card, any card, without seeing it and set it off to the left. That’s your card. Set all the rest of the cards to the right. Now, pretend you’re Monty and see if the ace is in the right hand deck. If it isn’t then the left side wins. If it is in the right side, the right side wins. Got it? That’s all there is to it. Don’t be fooled by the elimination of cards from the right hand deck. It
either has the card or it doesn’t and since its got more cards, the chances are that’s where the ace is.
It’s the same with 3 doors,or 3 cards or whatever. The odds are still with the side that had the most doors, that is, the side you didn’t start with.
Comment by Boman — 1/11/2005 @ 12:32 pm
Amerikan, Pigolito:
Look at it this way: there are three places where the prize could be located — Door 1, 2, or 3. These locations are selected at random. The three scenarios are A, B and C. The probability matrix looks like this:
….1…2…3
A…$…x…x
B…x…$…x
C…x…x…$
Each of these three scenarios (A, B and C) occurs 1/3 of the time, at random. Look what happens as you play:
Switching strategy (where you pick Door 1):
A. Monty opens 2 (half of these times) or 3 (half of these times), you switch, you lose.
B. Monty opens 3, you switch, you win.
C. Monty opens 2, you switch you win.
Keeping strategy (again, where you pick Door 1):
A. Monty opens 2 or 3 (each 1/2 of this 1/3 scenario), you keep your selection, you win.
B. Monty opens 3, you keep, you lose.
C. Monty opens 2, you keep you lose.
Switching: Two wins, one loss.
Keeping: One wins, two losses.
Scenario A (where you picked the right door the first time), will happen 1/3 of the time. Of those situations, Monty will open Door 2 half of the time, and Door 3 half of the time (i.e., out of those 1/3 occurrences of Scenario A). Scenario A will still only happen 1/3 of the time, because the initial starting conditions are chosen at random.
Multiply this result over the selection of other doors (to account for your selection of Door 1, 2 or 3, which is also at random, at 1/3):
Switching strategy (Door 2):
A. Win
B. Lose
C. Win
Keeping strategy (Door 2):
A. Lose
B. Win
C. Lose
Switching strategy (Door 3):
A. Win
B. Win
C. Lose
Keeping strategy (Door 3):
A. Lose
B. Lose
C. Win
Total: Switching wins 6 of 9. Keeping wins 3 of 9.
Comment by George Gaskell — 1/11/2005 @ 12:46 pm
Boman:
As to the dumb-luck alternative version I was discussing, yes. As to the real game, where Monty purposely avoids both your choice and the prize, no.
Amerikan: as pretentious and annoying as Boman may be, he is right about this. Opening Door 2 at random and revealing a booby prize would have re-set the odds for the other two doors, but Monty (the real one, not his random, bumbling hypothetical counterpart) usually does not select his door at random. For each game, the odds are only one in three that he chose the door at random, and 2-3 that he selected of the two doors he consider specifically because the other door held the prize. In effect, Monty is throwing a loaded die, where all six numbers are possible but not equally probable. Meanwhile, you are ignoring the lead in his die, and placing your bet as if he were throwing a regular one.
If it’s still not clear why, go back to the beginning of the game, select a door (the only truly random move), assume you end up on Door 1. Now calculate the odds of every possible outcome from this point forward:
Now Monty opens Door 2. Options 2 and 3 both called for Monty to open Door 3, so they’re off the table now, leaving only 1 and 4. Either the scenario that originally had a 1 in 6 chance has materialized, or the one that originally had a 1 in 3 chance did. Where do you place your bet?
Comment by Xrlq — 1/11/2005 @ 1:37 pm
Xriq,
In a three door game it is impossible for Random Monty to act any differently or make different choices than Knowledgeable Monty (excluding the choices that Random Monty cannot make because they are against the rules).
You have made a mistake in your outline of Random Monty.
You have caused a contestant to lose twice in one game, which is impossible. You did this when you gave Random Monty two choices when there is in reality only one. The “two choices” are a similarity. Therefore, there is but one choice.
If you go back and study your Random Monty, I’m sure you will see the mistake.
Comment by Boman — 1/11/2005 @ 2:06 pm
When I speak of “two choices” being one choice, it would probably be more intellegent and easier to understand if I said that the “two choices” should be treated as one choice because the effect is the same. There are not two effects from the two choices. There is but one outcome - not two.
Comment by Boman — 1/11/2005 @ 2:17 pm
make that intelligent - not intellegent
Comment by Boman — 1/11/2005 @ 2:27 pm
Boman, you’re wrong. In the randomized version of the game, Random Monty ends up voiding five games out of nine, leaving the odds split evenly between the four remaining games, two of which you can win by keeping your bet, and two of which you win by switching. That’s because when your selection hits the prize, there’s only 1 in 3 chance Random Monty will void the game, while there’s a 2 in 3 chance when your door and teh prize door are not the same.
Note, however, that limiting Random Monty to the two doors you didn’t select will not cause him to replicate the results of the real Monty, either. Once again, RM screws up two games, leaving the same four legal (and equally probable) possiblities I demonstrated above:
Strike out the two illegal games (3 and 6), and you’re left with the same four legal games I identified before, the only difference being that you had three less void games to throw out in the first place (the three scenarios in which RM would have picked your door). Once again, Random Monty fails to replicate the results of the real, purposeful, non-random Monty.
The reason why Random Monty can never replicate the results of the real Monty is simple: if Real Monty chooses Door No. 2, there’s a 2:1 chance he did so precisely because the prize was behind Door No. 3. Random Monty, by contrast, will never do that. If Door 3 has the prize, RM is just as likely to screw up and void the game as he is to point you to Door No. 3 by opening No. 2.
Comment by Xrlq — 1/11/2005 @ 3:01 pm
If you don’t think switching makes a difference, just do it. Play the game yourself. You’ll quickly change your mind.
Comment by Patterico — 1/11/2005 @ 4:43 pm
Which game? The original game (Monty intentionally picks 2 when the prize is behind 3, and vice-versa), or the randomized version (Monty picks at random between either 2 or 3 doors, and every illegal game is thrown out afterward)? Big difference.
Comment by Xrlq — 1/11/2005 @ 4:51 pm
Original game. I was responding to someone further up the thread. I forget who. But the comment applies to anyone who doesn’t believe it.
Comment by Patterico — 1/11/2005 @ 4:56 pm
There’s a simple, non-numerical way of looking at this puzzle, which operates on logic rather than symbolic (mathematical) notation:
1. At the beginning of the game, when you select a door, it is more likely that you will guess a wrong door than the right one. (Specifically, you will be wrong 2 out of 3 times, but “more likely to be wrong” is close enough for a logic-based non-numerical explanation.)
2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. In this case, the door that neither you or Monty picked MUST be the right door.
3. The only way in which switching your guess would cause you to lose is if your first guess had been correct.
4. However, as stated in point 1, you were more likely to be wrong in your first choice than correct.
5. Since it is more likely that your original guess was wrong, then changing your guess to the only other available door MUST improve your chances of winning.
QED
Comment by George Gaskell — 1/11/2005 @ 6:07 pm
Since you did not respond to my proposition and simply repeated what you have said before, let me show you how the chart for Random Monty should be laid out (call him - RM).
Prize behind door one. Contestant chooses door one.
RM can reveal door 2 or door 3 - it makes no difference. (all other reveals are illegal and are not to be considered).
Contestant switches and loses. This is game one.
Prize behind door two. Contestant is still on door one.
RM can reveal only door 3. (all other reveals are illegal and are not to be considered).
Contestant switches and wins. This is game two.
Prize behind door three. Contestant is still on door one.
RM can reveal only door 2. (all other reveals are illegal and are not to be considered).
Contestant switches and wins. This is game three.
Xriq - I believe you are confused about what is a game. It appears that because RM in game one, can
reveal either door 2 or door 3, you believe this should be counted as two games. It does not. It is one game.
You have not set up RM with the right program, therefore your RM fails. Mine does not.
Comment by Boman — 1/11/2005 @ 6:44 pm
Sorry, the previous post should be addressed to Xriq.
Comment by Boman — 1/11/2005 @ 6:46 pm
George - you’re absolutely right.
Comment by Boman — 1/11/2005 @ 6:59 pm
Sorry, Boman, but you are the one who is confused here. It doesn’t matter how you define “game.” What matters is that you throwing out three possible games/scenarios as though each were equally likely to occur in the form of a valid game. They’re not. Random Monty is equally likely to encounter the three scenarios, but he won’t handle them equally well. When the contestant chooses the door that holds the prize, RM faces only one illegal door and two legal ones, and as a result, two-thirds of the resulting games will end up being valid. When the contestant chooses a different door, the odds are reversed. Now two doors are illegal, and two-thirds of RM’s attempted games will have to be discarded as as result.
Once you understand that 2/3 of the Game 1 scenarios produce valid games at random, while only 1/3 of the Game 2 and 1/3 of the Game 3 scenarios do, it’s not hard to see the odds are now evenly split between the contestant’s original choice and the remaining choice in a valid, random game.
In a nutshell, your “Random” Monty only works because he isn’t really random after all.
Comment by Xrlq — 1/11/2005 @ 7:59 pm
Xriq,
What we’re trying to do here is see how RM plays the 3 door game, with the prize distributed among the 3 doors. We’re not trying to see how many variations RM can find. That is not playing the game. It is something else. What you’re trying to do is play golf with me and at the end of the round say, “I found more golf balls than you did. I win.” You may have won something but that doesn’t mean you won at golf.
If you’re looking for variations, RM can find four.
If you’re looking to see how RM plays the 3 games of 3 door, then see my chart above.
Your RM is random in an area we are not investigating in this thread.
Comment by Boman — 1/11/2005 @ 8:35 pm
I misspoke in the above - I should have said that
Xriq’s RM finds one variation - not four.
Comment by Boman — 1/11/2005 @ 9:11 pm
You’re right, Boman, your chart does illustrate the three possible outcomes. Congratulations on mastering the obvious. Now, perhaps you can finally rejoin the class, where the rest of us have been discussing the non-obvious for several days now. Rather than simply tell us what the three potential outcomes are (as you’ve chosen to define them), tell us how often each of your three scenarios is to materialize, and why.
Without that last part, your chart is as useless as a chart informing me I will either get lung cancer, or I won’t.
Comment by Xrlq — 1/11/2005 @ 9:24 pm
Admist that you were wrong to find two games in the variation in game one and we can continue the conversation. I wrote out the obvious because you evidently could not see it.
Comment by Boman — 1/11/2005 @ 10:24 pm
change admist to admit.
Comment by Boman — 1/11/2005 @ 10:26 pm
Boman, I have nothing to “admit” to you, and you have nothing to “explain” to me. I understand the concept fully, and you don’t. I’ve proven that my model is correct by laying out all the possibilities, while you’ve skipped a crucial step and assumed that your three non-random “games” (as defined by you) occur at random, long after I had proved that they don’t.
And get off this silly semantic kick already. You can’t change the odds in your favor simply by redefining the word “game” to suit your fancy. You can define it any way you want; you just have to apply that definition consistently, and apply your logic right, and you’ll end up proving me right: if RM is the game master, there is no advantage to changing your bet over sticking with your original choice; the odds are 50-50 either way.
First, to recap, let’s try proving you wrong by using my sensible definition of the word “game,” which includes each of the 27 possible combinations of prize placements, contestant choices, and doors Random Monty might have opened along the way. We can simplify the problem by making one (and only one) of these three random choices a constant, e.g., by stipulating that the contestant always chooses Door 1, which a real-life contestant might as well do anyway if he trusts that everyone else is behaving randomly. So now we’ve pared it down to nine possible “games,” only four of which are legal since the other five result in RM opening the contestant’s door, the prize door or, in one instance, both. So while nine “games” (as defined by me) get played, only four actually count, and those are the only four worth worrying about now. Among those four, you’ll win two and lose two, if you switch your bet every time - or if you don’t. 50-50 odds either way.
Now let’s run the same little experiment using your funky definition of “game.” Based on all the above facts, the four equally probable outcomes are now defined as consisting of only three “Games.” Two of the above random patterns lead to “Game” 1, while only one leads to each of the other two “Games.” Therefore, “Game” 1 pops up twice as often as either of the other two “Games.” Thus, each time you play, there is a 50% chance you are playing “Game” 1, and a 50% chance you are either playing “Game” 2 or 3 (based on a 25% chance for each). If you happen to be playing “Game 1″ (50% probability), you can only win by sticking with your original bet. If you happen to be playing either “Game” 2 or “Game” 3 (50% odds bewteen them), you can only win by switching your bet. Since you never know which “Game” you are playing, though, the only thing you do know is that each time you play, there’s a 50% chance you are playing the one “Game” that causes you to win if you keep your bet, and a 50% chance you are playing one of the two “Games” that causes you to win if you switch. Therefore, your odds are 50-50 whether you keep your original bet, or switch.
A third option, which is similar to yours but a little less retarded, would be to define “Games” 2 and 3 as a single “Game.” I mean, hey, if you are going to count two separate situations as a single “Game” just because the strategy rules are the same, why not do the same for the other two scenarios, for the same reason? Now it’s simple. With only two “Games,” you can win Game 1 by keeping your original bet, or you can win Game 2 by switching it. Each game represents two of the original random scenarios, so each game as a 50-50 chance of occuring every time. Once again, there’s a 50-50 chance you are playing a “Game” you can win by sticking with your original chance, and there’s a 50-50 chance you are playing the “Game” you can win by switching it.
Comment by Xrlq — 1/12/2005 @ 12:42 am
Evidently you believe that repeating yourself will improve your argument. Face it - your model is flawed. Random Monty, with the proper instructions, acts no differently than Intelligent Monty in the three door game. But you insist upon an unrestricted Random Monty and think you have discovered something. GIGO. This conversation is over.
Comment by Boman — 1/12/2005 @ 7:41 am
As opposed to what? A restricted, non-random Random Monty?
Moron.
Comment by Xrlq — 1/12/2005 @ 8:21 am
Games have restrictions. Therefore Random Monty must follow the rules of the game, randomness within the rules of the game. You want to play a game where the rules say three strikes and you insist on four strikes before you’re out.
If you can’t see that you have broken the rules of the game with your RM then you are hopeless.
And why don’t you can the insults. I once had a teacher who would say to certain students, “I see you have run out of arguments and now must retreat to vituperation.”
This really is the end of the conversation, as far
as I’m concerned.
Comment by Boman — 1/12/2005 @ 9:17 am
Boman, if your version of “Random” Monty knows where the prize is, determines his moves according to its location (and only acts randomly in the one situation where regular Monty does, too) then you have completely missed the point of this exercise. The reason for the hypothetical, randomized version of the game was to test (and ultimately prove) my hypothesis that the gamemaster’s knowledge of the prize is essential to the 2-1 odds that favor switching over staying. Throw in a gamemaster who doesn’t know where the prize is, and you’ll end up with a bunch of illegal games that have to be thrown out, and 50-50 odds among the ones that don’t.
Your version of “Random” Monty, by contrast, is a RINO (random in name only). He never has to be corrected, because he never makes the illegal moves in the first place. You call this “randomness within the rules of the game.” I call it an admission that your version of “Random” Monty knows where the prize is.
Comment by Xrlq — 1/12/2005 @ 1:01 pm
I’m going to respond because you are now responding to my arguments, instead of just repeating yourself.
You are quite accurate in describing the situation.
Now, respond to this. Your Random Monty breaks the rule that once a game is played, it’s over. That’s why your results are skewed. To play the same game twice is an illegal move. One of the games has to be thrown away. Reprogram your RM so that it does not make this mistake and your results will be identical to my RM.
You might say that is not true randomness. I say that you have set up rules for which games must be declared illegal. Now, set up one more rule, that the same game played twice is illegal and the “second game” will be thrown out of the equation. Then do all the random playing you want and you will find that your RM plays the game exactly like my RM, without knowledge of where the prize is, and also, plays it exactly like the real Monty who does have knowledge of where the prize is.
Comment by Boman — 1/12/2005 @ 1:52 pm
The real game doesn’t have any rules prohibiting any given combination of prize placements and contestant choices from happening twice, so the simulated, randomized version shouldn’t, either. Canceling out every other instance of Scenario 1 because … well, just because … is the statistical equivalent of affirmative action. It has no basis in how the game is played; it’s just an artificial mechanism for skewing the results in some way that just happens to match the odds that occur in the real game.
You seem to be resisting the notion that the game can be truly random when Scenarios 1, 2 and 3 fail to produce equal numbers of valid, legal games. Not so. Invalid games, as well as legal ones, are an important part of the model. Three of the five illegals occurred because Random Monty landed on the contestant’s door. Those three affect the three scenarios equally, so you can ignore them. The other two represent the 2-1 edge that all knowing Monty would have used to your advantage, but which Random Monty randomly pissed away instead on invalid games (or, if you prefer, valid games, which could be counted as either automatic wins or automatic losses, but in either case, games where you know where the prize is so the switching vs. staying dilemma does not arise).
Last attempt to explain. Choosing a door at random and not opening it tells you nothing about what’s behind any of the doors. Opening one of the three doors at random tells you something about the other two, by process of elimination, but it doesn’t tell you more about one of the two unopened doors than it tells you about the other. The only way to open one door and leave the odds at anything other than 50-50 for the remaining two, is to play by rules where the item behind one door influences the likelihood that the game master will decide to open another. That, in turn, requires the game master to know where the prize is, and play accordingly. RM will stumble across that result every now and again, but no more often than he stumbles across any other random action, ergo, not often enough to help (or hurt) your overall strategy.
Comment by Xrlq — 1/12/2005 @ 4:04 pm
Face it - you didn’t see the anomaly and you didn’t allow for it. Give it up. Goodby.
Comment by Boman — 1/12/2005 @ 11:08 pm
Face it - you’re full of crap. I proved that the odds were 50-50, and rather than be man enough to admit it, you tried to throw in a hack rule of the form “if P, then Q, unless P and Q are happening more often than I hoped they would, in which case P and Q should sit this one out.” If you were allowed to throw in rules like that, you could make the odds anything you want.
I apologize for going through two whole comments without calling you a moron, moron.
Comment by Xrlq — 1/13/2005 @ 6:11 am
Now, now . . .
Comment by Patterico — 1/13/2005 @ 6:18 am
OK, OK, I’ll try to be a little more gentle with this idio…distinguished individual. The bottom line is that there is no “anomaly” in my model. Once you exclude the door chosen by the contestant, Game/Scenario 1 becomes foolproof. Of the two possible doors for Monty to pick, both are equally valid, so it’s no surprise that Random Monty performs just as well as the real Monty (who is himself choosing randomly) under those circumstances. It’s only in Games/Scenarios 2 and 3 where the real Monty gives you any valuable information about one of the doors by opening the other. And the only reason he’s giving you any information is because he had a 50-50 chance of screwing up that game by revealing the prize, but he made a conscious choice not to. In short, it’s not an anomaly for a random idiot to perform better on the idiot-proof version of the game than he does on the versions that require him to actually know something. That’s what SHOULD happen.
I do like Boman’s “if P then Q but not twice in a row” rule, though. It can come in handy in plenty of other situations. Maybe I can persuade MLB to adopt a rule making it illegal to beat the Angels in two back-to-back games. Then again, now they’ve made themselves the laughingstock of the country by working two cities into their name, perhaps they deserve to lose.
Comment by Xrlq — 1/13/2005 @ 6:56 am
If you ever play chess with Xriq beware of this scenario - “Checkmate. That’s one game for me. Oh,
by the way, I also could have beaten you this way. Oh, you see that. Then you recognize that that’s another game for me. I’m ahead - two games to none.”
Comment by Boman — 1/13/2005 @ 7:47 am
What’s that you say Xriq? Talk a little louder, I can’t hear you. O.K. I got it. Xriq just said “I can’t believe I’m losing to this moron.”
Comment by Boman — 1/13/2005 @ 11:25 am
If the only to “win” against a moron is to convince him that 2 + 2 really does equal 4, then I guess I am “losing” to this moron after all. Although moron is probably the wrong word. Religionist is more like it. You started with an idea of what the “right” answer was supposed to be, then devised whatever tortured rules you felt you needed to defend that answer rather than concede your original gut feeling was wrong.
I, by contrast, have no problem admitting I’m wrong, when I am. Follow the link to Dean’s thread, and you’ll have an example of that, where I initially fell for a different brain teaser.
On the off chance any non-morons are still reading this thread, it occured to me that the a similar game can be played from a different perspective. Suppose you are a distant observer, who can see the prize but can’t see which door the original contestant selected. Rather than betting on the prize (that’s too easy), you place your bets on which door the original contestant will bet on. If either version of Monty uncovers the prize, the choice between the other two doors is a toss-up. But if regular Monty opens a door other than the prize, you can now be certain that the contestant bet on the third.
It all boils down to the fact that when Monty chooses one door to open, he also chooses two doors NOT to open. If you don’t know how he reached the decision not to open either of the doors, then the best you can do is to guess randomly between them. But if I do know why, you can sleuth your way back to the answer. In the regular game, you know that one door was selected not to be opened because of a reason having nothing to do with the prize (Monty left it alone simply because you chose it yourself, at random). The other door may have been selected not to be opened for a reason that is very relevant to the prize - because it had it.
Also, while waiting on jury duty this morning, I figured out that that you can’t randomize Monty’s knowledge of your door, either. If he didn’t see you picking a door but does know where the prize is, that’s not good enough. He has to know where the prize is, and also know which door you selected, and deliberately avoid opening either. Unless, of course, some genius like Boman is standing nearby, ready to cancel one-quarter of the otherwise valid games just to fix the odds in his favor.
Comment by Xrlq — 1/13/2005 @ 1:02 pm
Xriq
I don’t have time today to do more than scan your posting, but I did notice a few things that need commentary.
First, thank you for testifying that you are able to admit when you are wrong. I had you pegged for the scum that scum eats. Now I know I’m dealing with an honorable person, who nevertheless has a curious blind spot in his thinking.
I too have admitted when I was wrong. It was in a post on this page when I said I was wrong and you were right. But that was a careless and cursory reading of your position, taking the obvious conclusion of your progressions without questioning them. However, upon closer examination I soon realized that you had made a mistake, that you had taken one situation to be two and therefore your conclusion was skewed. I then laid out the correct chart, using the rules that you and others had declared to be proper and you announced that it was correct but obvious. Then with great scorn you declared that the conversation was about that which was not obvious.
I declared that you could talk about that which was not obvious, but you obviously did not see your problem.
You made no comment about my response to your screed about a ball team playing back to back games. Why? Is it because you still refuse to face the fact that you have a problem with your model?
Comment by Boman — 1/13/2005 @ 2:55 pm
OK, this semantic thing is making things harder than they need to be. If we can’t agree on anything else, we should at least be able to agree on the terminology, without which no rational debate is possible. Here’s the terminology I propose:
So far, so good? OK. Now to the assumptions that I think we share:
These are the assumptions I’m working under. Where am I wrong?
Comment by Xrlq — 1/13/2005 @ 5:40 pm
Xriq.
Number l will not work. Every combination of contestant choices results in more choices than games. By your definition you will have more “scenarios” than games. Somewhere, to balance the equation, contestant choices must be combined so they equal the number of games.
In other words, a contestant may have two ways to lose, but it counts as one loss.
If you have scenarios on one side of the equation and games on the other, and they don’t balance, are you telling me you should just leave it that way?
If you don’t mean that, then what do you mean?
And if you shouldn’t balance scenarios and games, tell me why?
Comment by Boman — 1/13/2005 @ 7:54 pm
And Random Monty does not make the mistake of trying to play the same game twice, as it appears you are trying to do. Once the game is played, we go on to the next combination.
Comment by Boman — 1/13/2005 @ 8:00 pm
Crap, you’re right. I over-edited my comment and ended up screwing up the definition. Here is what I meant to write:
Under that definition, assuming you agree that it’s OK to super-glue the contestant (C) to Door 1, as I’ve been doing all along, there are only three potential scenarios: (P-1 + C-1), (P-2 + C-1), and (P-3 + C-1). This part of the original game really is determined at random, so regardless of which Monty is involved, nine games can be expected to mean that the relevant Monty gets to play each of the three scenarios thrice. Random Monty will play it a different way each time (first M-1, then M-2, then M-3). Smart Monty, by contrast, won’t do that. Unlike RM, SM can’t choose the contestant’s door or the prize, so depending on which scenario he’s playing three times, he’ll choose the same door at least twice, or all three times. Depending on which scenario you’ve dealt him, he will play at least two identical games, if not three. But no game is being repeated. Like watching your favorite team beat its main rival twice in a row, by the same score; the two games look a lot alike, but they’re still two different games.
Correct. However, to allow all nine situations to occur only once, we have play the game nine times, meaning that either Monty faces each of the three initial scenarios three times. I’m not giving Random Monty more opportunities to play Scenario 1. I’m giving him the same number of opportunities to play each of the three. He just does a better job on Scenario 1, producing a valid game two times out of three, rather than three times out of three. Even there, he trails beind the Real Monty, who bats 1.000 under all three scenarios. Smart Monty knows what he’s doing, so as long as you give him at least one way to produce a legal game, he’ll find it. Random Monty produces only as many legal games as the odds dictate. The more ways he has to screw up a given scenario, the more often he will.
I wonder if you may be hitting the same mental snag that Dean did before I finally won him over in the other thread. His initial reaction to me invalidating the illegal games was to say I was “pulling a fast one” by canceling certain games and leaving others standing. As I explained then, we don’t actually have to “cancel” anything, as far as the rules of the game are concerned. We can call them automatic wins (Monty revealed the prize, so now you know which door to switch to), automatic losses (you only get to switch between your unopened door and the other unopened door, and both contain booby prizes), or whatever. My preference is to discard them since they represent combinations of C/P/M that will never occur in the real game, but it doesn’t really matter what you do with the illegal games. What does matter is that you ignore them while devising an overall strategy of switching between unopened doors vs. holding on to your original bet. Because in this situation, whether it counts as an automatic win, loss or draw, you can’t improve or worsen your chances of winning by switching or staying.
Comment by Xrlq — 1/13/2005 @ 10:47 pm
Did you answer the question? If you did, I missed it.
Comment by Boman — 1/14/2005 @ 8:42 am
I’ll make it clearer to you. Every combination of contestant choices results in more than 27 possible games. Putting “before” in rule one does not answer the question.
Comment by Boman — 1/14/2005 @ 11:03 am
I just have one thing to say: Boman has the best explanation of the original Monty Hall paradox I have ever seen, and I have read hundreds of them. It’s so simple the way he explains it but I’m going to restate it anyway.
There are three doors, W,L,L. You don’t see the letters on the back but You get to put them into two piles, one with 2 doors and one with one door. Monty always gets the pile with two doors. Now Monty gets to look at his pile and flip over one door that is a loser. The rule he uses to flip over the door is irrelevant, whether it is “flip a coin until the choice is a loser” or “go top to bottom until you find a loser.” Now he says “Do you want to switch piles?” If you do you go from 1/3 chance to 2/3 chance of winning. Monty doesn’t have to know where the winner door is in advance. He doesn’t even need to know where the winner is. If he looks at just one of his two doors and it is a loser and he flips it over and then offers you the choice it is still the same thing. He gives you the choice of his pile of two doors or your pile of one door.
This is only true when the game is with three doors. With 4 or more doors Monty won’t give you his whole pile of three doors when he flips over one.
Also, technically Monty doesn’t give you the loser door he showed you, but you don’t want that one anyway. What you want is his original pile with 2/3 odds.
Comment by RoseDude — 1/14/2005 @ 1:16 pm
That was a nice compliment, Rose. Isn’t it fun to use your own words to explain the puzzle! Everyone should try it. Writing it out always clarifies the mind.
Well, not always. See Xriq’s explanation of his position on Jan l0th, 4:17p.m. Take a good look at “Scenario 1″. Notice how he has divided scenario 1 into 3 games. Do you see what he has done? It a bit of sleight of hand. Scenario 1 is not 3 games. It is one game with 3 variations. So, what happens is that sometimes Xriq calls this 3 games and sometimes it’s 1 game and he gets the whole thing confused to the point that he has confused himself.
Xriq refuses to see his problem and I doubt that he even wants to see it.
Comment by Boman — 1/14/2005 @ 3:10 pm
Rose -
Which explanation are you referring to? I’ve given several in this thread.
Comment by Boman — 1/14/2005 @ 3:12 pm
I didn’t answer it because I assumed it was based on the terminology I had previously botched. I figured that once you were clear on what I meant by “scenario” (combination of prize + contestant, NOT Monty) you’d recognize that the question didn’t make sense anymore. You can’t have more scenarios than games, as there are only three possible scenarios, and either 9 or 27 games (depending on whether you play out all 27 theoretical possibilities, or stipulate that the contestant always selects the same door).
Absolutely. Three scenarios occur at random, with equal odds, so if you play nine times, you shoudl expect to play each game TYPE three times. That much parity arises by chance. The rest does not, as RM is twice as likely to void any given Scenario 2 game as he is to void a Scenario 1. There’s nothing “right” or “wrong” about that result; it just is.
Because doing so would spoil the randomness of the model. Only the pre-game scenarios occur at random, with equal odds, so that’s the only thing you should expect to occur in equal numbers (i.e., nine plays should translate into three instances of each of the three scenarios, not four instances of one, three of another ,etc.). The ultimate outcomes of the legal games, the illegal games, etc. is not a factor to be “balanced,” just data to be observed.
Comment by Xrlq — 1/14/2005 @ 4:17 pm
Xriq - you say you can’t have more scenarios than games and then you turn around and say that scenarios don’t have to balance with games.
You continually contradict yourself (see above post written to Rose).
I give up. But I do encourage you to continue work on that Perpetual Motion Machine you’ve got in the basement. I have absolute confidence that you’ll make it work.
And if you want to reply to this posting, fine, but don’t bother writing it to me. I won’t waste any time reading it, as I’m sure you will find the proper invective with which to chastise me.
Comment by Boman — 1/14/2005 @ 4:41 pm
God, you’re dumb. Three scenarios, played out three times each, equals nine games. Of course that means more games than scenarios. If you’d rather play only one game on each scenario, fine, but then you can expect to have 2/3 of a valid game the first time, and 1/3 of a valid game each of the next two times. If that makes sense to you, try explaining it to your 1.3 children.
Damn, you really are a moron, aren’t you? I was beginning to think you might be a troll, but real trolls rarely do as consistent a job as you’ve done at not getting it. You’re probably a Democrat, too, I presume?
Comment by Xrlq — 1/14/2005 @ 5:42 pm
The shortest and simplest explanation is best (Occamss Razor)
Boman said “The answer is as simple as this. If I’m playing against you in a game of high card wins, and the deck consists of an ace and two threes, and I get one card and you get two, you can show me a three every time but it doesn’t change the odds that you’ll win most of the hands.”
It took mne twice as many words to restate it.
Comment by RoseDude — 1/15/2005 @ 12:50 pm
Boman, xrlq is quite correct in what (s)he states, I’m unsure as to what you’re arguing so I’m loathe to say you’re wrong.
A Random Monty will accidentally mess games up; if he doesn’t he can’t be truly random.
Put as simply as I can, he can’t mess up games where you pick the prize first time, which is 1/3 of the time. He will randomly pick from the two remaining boxes, will reveal an empty box, and you’ll win if you stick, but lose if you switch.
2/3 of the time, you won’t pick the prize first time, and he now can potentially mess up. Half of the time he will open one of the two remaining boxes to reveal an empty one, and all is well - in this instance, stick you lose, switch you win. However the remaining half of the time, he will accidentally reveal the prize, and the game is screwed.
So one-third you’ll win if you stick, and one-third you’ll win if you switch, and the remaining third will be ruined by Monty accidentally revealing the prize. This is what xrlq is correctly asserting, that you have a fifty-fifty chance whether you stick or switch, provided Monty hasn’t already messed the game up.
Comment by JC — 1/17/2005 @ 7:07 am
Exactly. Either Monty selects his door according to criteria that destroy randomness vis a vis the prize, or he doesn’t. If he ONLY excludes the prize, but considers both non prize doors equally, then he gives you no more infomration about one remaining door than the other. Similarly, if he systematically excludes your door, but chooses between the other two according to criteria that do not take the location of the prize into account, then again, it’s no different than if you had opened the first door, revealing no prize, and were then given a random choice between the remaining two.
Boman is an idiot, but he’s probably not a big enough idiot to play me for real money. I could be wrong about that, though. Maybe I can prove my theorem and the time-honored “fool and his money are soon parted” theory in one feel swoop?
Comment by Xrlq — 1/17/2005 @ 10:32 am
I don’t know why I’m doing this because I got bored with this three or four days ago. But here we go again.
JC - you’re right. We have no quarrel. But that is not what Xriq has been saying according to my understanding.
Xriq - let’s get some things basic.
Do you still say:
“If Smart Monty is the game master, nine scenarios will result in nine legal games.
If Random Monty is the game master, each of the nine scenarios will occur exactly once.”
I believe we both agree that the illegal games of RM are thrown out. Am I right>
Comment by Boman — 1/17/2005 @ 10:55 pm
That’s basically right, depending on what you mean by “scenario.” As I used the term, it applied to the proverbial hand Monty was dealt, i.e., the combination of prize placement + contestant selection. By that definition, the distribution of the scenarios will be the same, regardless; the difference is what Monty does with those scenarios. Every scenario offers either one or two possible legal moves by Monty, so in the real, non-random game, every scenario (or, if you prefer, every potential game) will result in a legal play. When Monty is random, he’ll churn out a legal game twice as often while playing the scenario that has two legal moves as he does while playing the one that allows only one.
Bottom line: if Monty knows which door you selected and knows where the prize is, and purposefully avoids both, then the odds are 2:1 in your favor if you switch. Otherwise, they’re 1:1, just as they would be if you yourself had randomly uncovered one of the two booby prizes (or, for that matter, if there had only been two games to begin with).
Your view, as I understand it, is that as long as Monty plays randomly between the two doors you did not select, and as long as all illegal games are excluded from the mix, your odds remain 2:1 . If that is not your view, then our entire debate was pointless. If it is, then I’d like to challenge you to test it empirically, for money. But if you are that foolish, be forewarned that I just tested both theories using Excel, playing 1,000 games at a time, then hitting F9 65 times to play 1,000 more each time. It soon became clear they were hovering around 1-1, and didn’t reach the 2-1 numbers a single time.
Comment by Xrlq — 1/18/2005 @ 12:43 am
Boman, correct me if I’m wrong here (it is a little confusing to determine your exact stance due to the size of the thread now) - you are saying that in the Random Monty game you should STILL switch where possible, yeah? You’re saying that REGARDLESS of how you arrived at the situation, once you’ve picked a box and you’ve seen another empty one opened, you should always switch - regardless of how that empty box came to be opened, be it through knowledge of the prize location, or by chance?
In the original game, you are ALWAYS in this position. As you seem to have no trouble grasping (which makes your stance on the random game all the more confusing), the chance of you being in this position because you have picked the prize box is 1/3, and the chance of you being in this position because you have picked an empty box is 2/3, and that sticking/switching gives a 1:2 ratio of wins.
In the random game, you are only ever in this position because:
a) you picked the prize box, with probability 1/3 (ergo whichever box Random Monty opened revealed an empty box).
b) you picked an empty box (with probability 2/3) and Random Monty picked the other empty box (which he had a 1/2 chance of doing from the remaining two), which again equals probability 1/3 (2/3 x 1/2).
Once you are have opened a box and then seen another empty one opened IN THE RANDOM MONTY GAME, the chance you have done so because you picked the prize box IS THE SAME as the chance you have done so because you picked an empty box.
If you’re gonna say “it’s not a 50:50 chance because they’re both only a 1/3, we’ve already established that we’re in this position so how can you say we’re here with only probability 2/3!”, it’s akin to saying we’ll play a game rolling a dice, which you’ll win if a 1 is thrown, and I’ll win if a 6 is thrown. We’ll force ourselves into a position (much like we’re doing with the Random Monty game) where somebody has won. Clearly the die shows a 1 or a 6 with probability 1/6 each, but since we’re forcing ourselves into the position where one is thrown, we conclude that since the probabilities are the same (1/6), it is a 50:50 chance that either has happened.
One of the best ways of explaining the original (”Full” Monty) game is by extending the game to, say, 100 boxes. Most people soon catch on at this point. I’d recommend to you Boman that you give this some thought with respect to the random game. With 100 boxes, I’ll maintain that in a random game, once I’ve picked mine and then seen 98 other empty boxes opened, I gain no advantage by sticking or switching. Would you still say switching gives a 99% chance of success?
Comment by JC — 1/18/2005 @ 3:02 am
Xriq-
We start with your words - “When Random Monty is a game master, each of the nine scenarios will occur exactly once” and will result in l2 games.
Do you agree with this statement?
JC - I’m afraid I haven’t read your posting. Sorry.
I am trying to clear things with Xriq first. I have spent too much time on this and was going to drop it and probably am silly to waste much more time on it.
Comment by Boman — 1/18/2005 @ 5:34 am
Fair enough Boman, I understand it gets confusing to have more than one concurrent discussion ongoing.
Could you please just clarify one thing for me though please?
Do you still maintain, as you appeared to earlier, that if you have chosen a box, and you are then shown one other empty box AT RANDOM, that you still improve your chances of winning if you switch?
Cos if you don’t, then you don’t need to bother reading my posting anyway.
Comment by JC — 1/18/2005 @ 5:59 am
JC - if I am shown an empty box by chance, by random choice, is that not information? And if it is information (the box is empty), does it not convey the same information that a knowledgeable Monty would give by opening the same box? If you are saying that a knowledgeable Monty gives more information by opening an empty box, just how does he convey that?
The exception,as far as winning, is that if you have chosen the prize box, then opening another box at random, and switching, will cause you to lose. But knowledgeable Monty also can choose at random, in this scenario. And if you switch you will lose. So, Random Monty can randomly choose one of two boxes, and so can Knowledgeable Monty.
Comment by Boman — 1/18/2005 @ 6:42 am
I may have used the word that way once in error, but I can’t find it. In any event, when Monty plays randomly, all of the nine outcomes I identified here will occur with equal frequency. However, playing all nine out results in nine games (four valid, five invalid), not 12. I don’t know where that number came from.
As to your question to JC, if an empty box is shown by chance, or if a door is opened and random, it does indeed convey some information about the other two. However, it does not convey any more information about the unopened door you didn’t previously choose than it does about the one you did.
When in doubt, try playing the game on Excel.
Comment by Xrlq — 1/18/2005 @ 7:22 am
Boman, the whole point is that YES it does convey more information when knowledgeable Monty reveals an empty door because of the rules he is bound by - it’s the crux of the original brainteaser. It baffles me that you have apparently given such an eloquent explanation of the solution elsewhere yet have really failed to grasp this point. When Random Monty shows me an empty box it just tells me that either I picked the prize in the first instance (which will happen 1/3 of the time), or that I didn’t and he then failed his 50/50 chance to then pick it (which will also happen 1/3 of the time).
What you’re unsurprisingly having trouble with is that yes, two games could be identical - I could pick box A and then be shown that box B is empty. If knowledgeable Monty has shown me this, the chance my box (box A) contains the prize is 1/3; if Random Monty has shown me this, the chance my box contains the prize is 1/2. Sounds crazy, but I’m afraid it’s true.
My earlier post (which you didn’t deign to read) suggested that you try with 100 boxes, which is a common way to help the non-believer of the original brainteaser, as they soon see that playing the game with knowledgeable Monty gives a 99% chance of success with an “always switch” policy.
I’m assuming you’ll stick by your assertion that you maintain your advantage when Random Monty just happens to present you with an identical situation to Knowledgeable Monty, which in this instance would mean 98 empty boxes.
We’ll label the box you choose number 1, and all the other boxes between 2 and 100. The prize is in a random box, neither of us knows which. I’ll open all the boxes between 2 and 99. A lot of the time I’ll stumble across the prize, and we’ll start over again. The times I don’t are clearly when the prize was originally in box 1 or box 100. If our ball happens to be randomly placed in box 100 ninety-nine times more often than in box 1, I’d suggest we double-check our random number generator.
If (as has been said countless times about the original brainteaser) you accept this is the case with 100 boxes, why not 3?
Comment by JC — 1/18/2005 @ 8:23 am
JC - I think if you and I had had this conversation in the beginning we would have solved this a long time ago. I congratulate you on your clarity of presentation.
As originally stated Random Monty was not allowed to play illegal games. An illegal game would be, for example, revealing the prize door,or revealing the door you had chosen. This meant that Random Monty would play exactly like the real Monty. However, Xriq insisted that because Random Monty has two choices in one of the situations that this be counted as two games. I have insisted ever since that this is playing the same game twice and should not be counted as two games. I suggested that Xriq’s model be changed and he has insisted ever since that I can’t add. Thus the stalemate and why I’m bored.
Comment by Boman — 1/18/2005 @ 8:45 am
In other words we have two different scenarios. I said that Random Monty plays exactly as Intelligent Monty does, if you throw out the illegal games. Xriq says this is nonsense and then proceeds to talk about the statistics of randomness. I, in turn say, that if Random Monty plays the same games that Intelligent Monty plays he will play them the same, with the proviso, that illegal games are tossed out.
Xriq wants to include extra games because Random Monty has more choices and therefore an extra game.
I in turn say that Intelligent Monty has the same choices but does not insist on the extra game. Now, who is right?
It depends upon whether you want to talk about the statistics of randomness, or whether you insist on playing the game the way the game is played.
Comment by Boman — 1/18/2005 @ 9:05 am
One more thing - and then I’m done.
I have not disagreed with Xriq’s statistics. I have stated so previously. I have simply disagreed with his use of them.
Statistics can leave out many relevant contingencies. That’s why the models for global warming are so screwed up.
And I am a libertarian-conservative. You insult me,Xriq, when you suggest I am a democrat. I think you should know a person before you demean him. That way you won’t look like an idiot when you are wrong.
Comment by Boman — 1/18/2005 @ 9:12 am
Which I disproved early on, by illustrating the nine random combinations that can occur for any given contestant choice. Unfortunately, at that point in the thread you were too fixated on disproving the existence of Random Monty to even grasp what the issue was. Now that you presumably do know what the issue is, you might want to give it another read, along with earlier comments on the topic by Doc Rampage, Dafydd and others.
Nope, I never insisted any such thing. You are the one who made up that “playing the same game twice” bit, not me. All I did was point out that Random Monty was twice as likely to produce a valid game under Scenario 1 than he was under either of the other two, and therefore could be expected to produce twice as many valid games (and half as many invalid ones) under that scenario over the long haul. I didn’t give RM more opportunities to produce a valid game under Scen. 1 than I did under 2 or 3; he just did better under 1 than the other two on his own. If he had fared equally well/badly among all three scenarios, e.g., voided one-third of the games of each type, or voided two-thirds of each type, or whatever, then remaining games would indeed be indicative of how Smart Monty plays. But he doesn’t, so they’re not.
Judging by your last comment to JC, it sounds like you are finally starting to grasp the problem. That’s great. It would be greater still if you would own up to your own failure rather than attempting to blame others for it, such as me or my supposed lack of clarity. My words were clear enough to persuade Patterico, Mrs. Patterico, Dean Esmay and probably more than a few others a full week ago. Just not you.
Comment by Xrlq — 1/18/2005 @ 9:47 am
Xriq -
In the paragraph above I said “if Random Monty plays the same games that Intelligent Monty plays, he will play them the same, with the proviso, that illegal games are tossed out.”
You chose to disprove the less clear statment. Why didn’t you chose the clarified statement?
When Random Monty plays the same games there is no difference between RM and Intelligent Monty. When RM goes off to play by himself, he plays an absurd version of the game, not getting much of anything right. This has been my position from my first posting. That is why, under the controlled experiment, Random Monty does not even exist. It’s because he plays the game the same as Intelligent Monty.
Now, if you want to prove there is indeed an unintelligent Random Monty that does not know how to play the game, who would argue with that. It would be silly to argue that.
No, the problem has been that you wanted to argue with a straw man. I have been arguing for a controlled experiment and you have demanded an uncontrolled experiment.
Comment by Bo Newman — 1/18/2005 @ 3:23 pm
I disproved both statements. Allowing Monty to play randomly and then throwing out the illegal games will leave you with the same four combinations that can occur under Smart Monty, but not in the same proportions.
I’m not sure what kind of a “controlled” experiment you have in mind. Either Random Monty knows which doors to avoid or he doesn’t. If he does, he’s not playing “randomly” at all; he’s playing purposefully, under exactly the same rules as Smart Monty. If he doesn’t know where the prize is (or if he does know, but doesn’t know which door the contestant chose), then once the illegal games are thrown out, your odds for switching vs. staying among the remaining legal games will be 50-50, not 2-1.
Comment by Xrlq — 1/18/2005 @ 5:02 pm
O.K. I’ll try to show you again.
“If R.M. plays the same games that Intelligent Monty plays, he will play them the same, with the proviso, that illegal games are tossed out.”
What this means is that we have Smart Monty play a game of three door. Then we have Rm play that same game. What will be the difference? Nothing. This is the controlled experiment and not the unrestricted RM
that plays the same game twice without knowing that it has already played that game.
Play the game of actually comparing individual games of Smart Monty with the same game played by R.M. There is no difference except that illegal games are thrown out.
And if we’re going to control the experiment and say that illegal games are thrown out, why not say that repeating a game is illegal and also to be thrown out.
If you’re not going to control the experiment you have an experiment run amuck, producing figures that distort the reality of the situation.
Comment by Boman — 1/18/2005 @ 5:35 pm
In other words, your RM isn’t acting randomly at all. He’s playing by exactly the same rules as Smart Monty: knowing which door the contestant chose, knowing where the prize is, and purposefully avoiding both. That’s not a randomized experiment; it’s a series of rules to describe the real, non-random, knowledge-based game.
We throw out the illegal games because they can never occur in the real game, and thus make for lousy comparisons. On the other hand, throwing in a new rule against “repeating a game” (i.e., playing two games that happen to pan out in essentially the same way) has no basis in the real game, or anything else. Each game is effectively played in a vacuum as far as the others are concerned. Even Smart Monty would never deliberately avoid opening any particular door just because he opened that same door the last time he played, when the prize just happened to be in the same location and the contestant just happened to pick the same door. If even Smart Monty won’t constrain his game that way, why on earth should his allegedly random counterpart do so?
Here’s what I think is being distorted. Your insistence that the 2-1 ratio is a product of the structure of the game rather than of Smart Monty’s knowledge and subtle communication to the contestant is causing you to distort the bejeezus out of the meaning of the word “random.” There’s a reason why the intuitive, albeit wrong, answer to the original problem is 1-1. Those are the odds one should expect when the two unopened doors are selected at random, or according to any non-random criteria that are irrelevant to the placement of the prize. The trick here is that the rules are relevant to Monty’s knowledge of the prize, they’re just subtle enough that the average reader fails to understand them, and thus ignores them and proceeds as though they were random (either purely random, determined by criteria unrelated to the location of the prize). Arbitrarily forcing Random Monty to never open the door with the prize he’s not supposed to know about, or arbitrarily voiding half of the otherwise valid Scenario 1 game just to keep things even, makes no sense.
Comment by Xrlq — 1/18/2005 @ 7:34 pm
Xriq,
Why did you leave out the most important part of the sentence - “the unrestricted RM that plays the same game twice without knowing that it has already played that game.”
Isn’t it true that you insist on seeing two games where there is but one to protect you own version of RM that finds four games where there are but three.
You say “arbitarily voiding half of the otherwise valid Scenario 1 game”. There is nothing arbitrary about it. We recognize that the lst game has been played. One might even say that both of the options have been used up in that one game. We’re not doing it to keep things even - we’re doing it because anything else is a farce. To make two games out of one just because there are two ways to lose is nonsense. It cannot be allowed. You want this to be two games because it preserves your RM that finds four games where there are but three.
Your whole dissertation rests on this one thin reed - two options is two games.
I think it is time to agree to disagree. We have whipped this particular horse long enough. If you wish to have the last word - be my guest. And may peace be with you and your new baby.
I give you a new question - if an iceberg melts in the sea, does the water level go up, stay the same, or go down. For laboratory purposes, a large chunk of ice in a tub of water will do quite well.
Comment by Boman — 1/18/2005 @ 9:30 pm
Disregard the different densities of fresh and salt water.
Comment by Boman — 1/18/2005 @ 9:34 pm
I guess we’ll never agree if we can’t agree on what is or isn’t a “game.” To me, every time you play, that’s a game. If you play twice and get similar or identical results, that’s two similar/identical games, not one, and if certain scenarios result in higher proportions of valid than invalid games, that’s just the way the cookie crumbles. Whatever, I agree we’ve beaten this horse enough times over, and I appreciate your kind words about my son, who is bouncing on my knee as I type. And I apologize for the occasional personal attacks; I was frustrated but rather than lashing out I should have just bitten my lip instead.
As to the iceberg question, I don’t know. The intuitive answer - and given this thread, I presume, the wrong one - is that of course it will go up. The less intuitive answer is that ice contracts when it melts, so the part that is already under water will displace less water than it does already. What I don’t know, and would need to, is the respective ratios: what percentage of the iceberg starts out above vs. below the surface, and what percentage of its original size the entire iceberg loses when it melts.
Comment by Xrlq — 1/18/2005 @ 10:22 pm
Archimedes has done most of the work for you with his most famous “Principle.”
Comment by Boman — 1/18/2005 @ 10:50 pm
Sorry to keep this going, but I’m still unclear as to where Boman stands on the issue of a TRULY random Monty, rather than his/her utterly-non-random Random Monty. I haven’t as yet seen him/her retract the statement that it doesn’t matter how the knowledge is gained, as long as it is gained.
So could Boman *please* answer the following question:
You’re actually on the show. You have just selected your door. Monty is about to do his bit when suddenly there’s a crash - one of the other doors, by a complete accident, has come off it’s hinges. It wasn’t your door, and it’s empty. The cameras stop rolling while the stage crew attend to the broken door. The door is quickly patched up, and filming restarts with Monty walking over and opening the same door, obviously showing you again that it is empty.
What do you do?
Comment by JC — 1/19/2005 @ 2:18 am
JC, I know this question isn’t for me, so I wno’t answer it, but I would like to ask for clarification on one point. Are we to assume that Monty’s choice was influenced by the unfortunate event that preceded it, or did we just chance upon an early peek at the door Monty was about to open anyway?
Comment by Xrlq — 1/19/2005 @ 11:35 am
JC-
Go ahead Xriq. I have faith you will answer the question as I would.
Comment by Boman — 1/19/2005 @ 12:32 pm
Sorry JC - reading back on one of your posts I’m afraid Xriq is going to beat you up a bit on this.
Go ahead Guys.
Comment by Boman — 1/19/2005 @ 12:55 pm
To JC
I will answer one of your questions from an earlier posting and then I’m afraid I’m going to quit posting. Frankly, I’m bored with it. If I choose box 1 and there are 99 other boxes, the odds are that the prize is in the 99 boxes. I’m sure we can agree on that.
If a random number generator is used to open boxes, as the boxes are opened without finding the prize the odds that I have chosen the correct box go up. If we get down to 2 boxes without finding the box, the odds are more than 50/50 that I have the correct box. If knowledgeable Monty is opening boxes, my original box only contains odds of 100/1.
Comment by Boman — 1/19/2005 @ 1:34 pm
As Boman has declined to answer, I’ll answer as I think he would, and then I’ll answer as I would. Both examples assume your door is #1 and the inadvertently opened door is #2.
Boman (I think): Your odds are 2:1 in favor of switching to Door 3, just as if the accident never happened.
Me: It all depends on whether the accident influenced Monty’s behavior. If he can assure you that he already decided on Door 2 before the mishap, or that he was otherwise able to ignore it while deciding which door to open, then you can pretend the accident never happened, and the odds are still 2:1 in favor of switching. If, however, he points to the inadvertently opened Door 2 and says “Oh, crap, I guess I’ll have to open that one now,” then you know that both Doors 1 and 3 were left unopened for reasons unrelated to the location of the prize, and that the only new information imparted by opening the third door was to eliminate one booby prize from each. Your odds are 50-50.
Comment by Xrlq — 1/19/2005 @ 3:24 pm
I was just going to sneak a peak at Xriq’s answer. I was just curious. And it was fun to see how Xriq answered. I’ve been trying to not post but something has gotten the best of me. Here it is.
I believe the better answer is that if you believe that Monty is telling the truth then you have to stay with door 1. It is a l00% winner if you’re right in your judgment about Monty.
This is true because if Monty had been planning to choose a different door it would have had to be door 3.
If it made no difference whether it was door 2 or door 3, then the prize has to be behind door 1.
If you don’t believe Monty then you switch from door 1 to door 3. If you don’t know whether or not to believe Monty, you still switch from door 1.
O.K. That’s it. I’m not going to peak at your answer and so I won’t be tempted to post. I’ve got other things I’ve got to do.
Comment by Boman — 1/19/2005 @ 9:00 pm
I blew it. I meant to say that if you believe Monty when he said “Oh crap etc”
Comment by Boman — 1/19/2005 @ 9:06 pm
Guys,
After I posted that I thought I should have clarified a few points, but I refrained in the hope I wouldn’t need to and that you’d realise that I was trying to do was make Boman answer the “Random Monty” question that the whole debate blew up out of, but phrasing it in a different way so it was easier to visualize. Just assume you’re on the show, obviously everybody can see all the doors so it’s no secret after the accident, it was a great big glittery star-covered door that crashed down, you didn’t “sneak a quick peek”. I did think about shoving in a line explaining Monty’s decision, let’s assume he says something like:
“Well, at least we don’t have to set the whole game up again, it would’ve been a pain for us to take you out while we shifted the car around”, so you can be sure that the prize doesn’t move.
I also thought it was safe to assume that the accident DID influence Monty’s decision - after all, he couldn’t POSSIBLY open any other door now that you’ve seen that one’s empty, otherwise he really will give it away. He just *HAS* to open that one, so Monty really becomes an irrelevance once the accident has happened.
If I were in that position, I would insist that the whole competition be restarted because my slender advantage of 2/3 has been lost down to 1/2.
Boman - read your post above regarding the 100 boxes - why is the other box any less likely to have the prize in than yours? Sure, you’ve defied some pretty big odds to get down to the last two, but it’s 50/50 once you’re there. What kind of magical forces do you think are at work here?
Imagine the same scenario as above with 100 doors - Monty’s having a really bad door day, and 98 come crashing down, all empty, after you’ve made your choice. By luck, yours is left and one other. This has told you NOTHING other than that the prize is in one of these two doors, nobody has influenced which doors were left like they do in the original game. I find it incredible that you can’t see this. You stating your choice has made NO DIFFERENCE whatsoever, which is the absolute key of the original brainteaser!
Comment by JC — 1/20/2005 @ 1:03 am
Boman wrote re the 100 box puzzle:
“If a random number generator is used to open boxes, as the boxes are opened without finding the prize the odds that I have chosen the correct box go up. If we get down to 2 boxes without finding the box, the odds are more than 50/50 that I have the correct box. If knowledgeable Monty is opening boxes, my original box only contains odds of 100/1.”
Pardon? PARDON?!! I think you’ll find your chances are EXACTLY 50/50!! Suppose you had picked 37 and I had picked box 45 (an important difference is we CAN play the “random” variant of the game like this, we can both pick boxes initially because the others aren’t being removed as bound by the rules of the original game, which only work when one contestant is playing). After 98 boxes have been randomly removed, your 37 is left and so is my 45. Please explain to me how your odds of winning are better than mine !!
Comment by JC — 1/20/2005 @ 6:13 am
I’m also curious as to what Boman thought Xlrq was going to “beat me up a bit” over. As far as I can tell we are in complete agreement with each other on absolutely everything. (And also, incidentally, in complete agreement with all relevant probability laws :oP)
His question regarding my scenario was a fair one, he was just trying to eliminate *any* former knowledge that Monty was going to choose that door anyway, and also confirm the fact that it subsequently influenced his decision. I gave no reason whatsoever to assume Monty was going to open door 2, and I don’t really think the second point needed clarification, because unless the whole game was restarted from scratch, opening door 3 (if it were indeed available for opening in the first place; we don’t know) would undoubtedly guide the contestant to the prize.
Comment by JC — 1/20/2005 @ 6:55 am
Actually Boman, as well as being wrong, you have also started arguing against yourself - on the one hand you say that you should still switch when presented with the a valid Monty game scenario in the random environment, yet recently claimed in the 100 box game that your *original* choice had the greater chance of being correct.
You really are a crazy, crazy person.
Comment by JC — 1/23/2005 @ 3:36 am
Well, I’m back. I got some of my stuff done and thought I’d see how this thread is doing.
Well, what shall I do? I see I’ve been called a few names. JC., you really shouldn’t do that. It might come back to bite you.
But, JC. is partially right. I really screwed up on one point - I got it assbackwards. But unfortunately, JC., so did you. You got it wrong but in a different way.
Now, that I’ve come to my senses, am no longer crazy, (see what happens when you rush things), and seen the error of my ways, we’ll see if JC. will do the same.
So, here’s the scenario. JC. wanted to talk about 100 boxes, with one box containing a prize. If I understand JC. correctly, he stated that if I chose one of the boxes out of the 100 boxes, and something happened to cause 98 of the remaining boxes to open, revealing all empty boxes, the remaining two boxes have a 50/50 chance of containing the prize.
Did I state your position correctly, JC.? Because if I didn’t I apologize. But I believe I have stated it just the way you wanted it, right?
I say the odds of 50/50 between the two boxes is incorrect.
First, of all, it is important to recognize that the box I chose is not eligible to be opened. I believe that was your intention, wasn’t it JC.?
With that understanding, to make it clear, let’s put all the boxes in a circle. Now, let’s place one of the boxes outside the circle - that’s my box (unopened). It has a 1/100 chance of containing the prize. O.K. so far?
Now, we can see that all the boxes in the circle (99 of them) have a 99/100 chance of containing the prize. Something opens (at random) 98 of the boxes in the circle. They’re all empty. One box is left and it’s unopened. There is one box in the circle and one outside the circle. JC says the box inside the circle has the same chance to contain the prize as the box outside the circle. Does it? Aren’t the circumstances different?
What we have is a box outside the circle with a 1/100 chance of containing the prize. Nothing that happens inside the circle can change those odds. It is separate - the odds cannot change.
But the box inside the circle - that is a different matter. Since 99/100 of the odds remain in the circle, when 98 are opened and are empty, the odds shift to the unopened box inside the circle. Now, the odds that the box inside the circle contains the prize, are 99/100.
It’s not 50/50, it’s 1/100 and 99/100. It makes not difference either, if the boxes are opened at random (in the above situation), or by an intelligent “Monty”.
Where did I screw up? I gave the odds to the box outside the circle. What can I say - assbackwards.
Tomorrow, I’ll show you (prove?) where Xriq screwed up, why we disagreed and what is the correct outline for the dispute between us.
Question to JC - is your “box 45″ inside or outside the circle? You intended it to be inside the circle, didn’t you? If you meant for it to be outside the circle, we have a different question.
Comment by Boman — 1/24/2005 @ 10:32 pm
Oh, goody. I was afraid we’d heard the last of this.
Comment by Patterico — 1/24/2005 @ 10:42 pm
Oh goody is right, Patterico. Boman’s sticking to the very same theory that very nearly won me your house. Here’s hoping he’ll stick to it long enough for me to collect.
Comment by Xrlq — 1/24/2005 @ 11:04 pm
Welcome back Boman, I missed you
“You are a crazy, crazy person” was meant kinda affectionately, not as a crushing insult. But anyway, yes, you stated my position correctly.
First things first, my examples didn’t make the chosen box ineligible for opening, there were random forces, “acts of nature” at work which had no idea which box was chosen or where the prize was. They *could* have made the door fall off the chosen box, but they didn’t. But actually that doesn’t even matter, let’s make the chosen box ineligible if you wish. It doesn’t affect the conclusion. What is of paramount importance is that we don’t also then make the prize box ineligible - if we make both ineligible then we are playing the original “Smart Monty” game, rather than analysing the “Random Monty” situation which has occurred by chance. For the random game, either the chosen or the prize box (or both) must remain eligible to be opened, but good fortune just so happens not to open them. This distinction is crucial. I reckon there’s a good chance that this is where you are tripping up, not appreciating the difference between arriving at a situation by chance and arriving at the same position purposefully.
The whole crux of the original brainteaser is that Monty is bound by rules to perform exactly as he does each game. The only reason you should switch is because you know he always plays the game that way. This is why you have to be careful about how you word the puzzle when you give it to people. If you just offer it as a single iteration of the game, it doesn’t work:
“A friend hides a dollar under a cup and invites you to try and find it. You make your selection. He then lifts up one of the other cups, and it’s empty. He now asks if you want to stick with your original choice, or choose the other cup. What should you do?”
We don’t know whether he was always going to lift one cup, or that he was doing so intentionally to reveal an empty one (might he have just said “bad luck” upon lifting this cup to reveal the prize?). If you can’t psychoanalyze your friend sufficiently to work out his motives etc (we don’t even know he knew where the prize was), the best you can do is flip a coin and take your 50/50 chance. It is only because we can be so absolutely specific about (Smart) Monty’s actions, and therefore every possible scenario, that the argument to switch can be made.
What we’re effectively doing is choosing TWO boxes from the 100 AT RANDOM (yours, and one other which will remain at the end). It doesn’t matter a jot where all the boxes are positioned, put them in circles, draw lines, arrange them in groups, remove a chosen box if you like, it has no bearing on anything. Just identify two that are to remain unopened. Luckily, the other 98 all turn out to be empty. Doesn’t it sound crazy that either of the remaining boxes should have more or less chance of containing the prize than the other?
If I’ve read both our stances correctly, we should each consider this a good bet:
We put 100 boxes in a circle. You & I leave the room. A third party, we’ll say Xrlq, randomly conceals a ball in one box. We return. You then select one, and Xrlq removes it from the circle. I then randomly open 98 of the remaining boxes in the circle - if I accidentally uncover the ball, we leave the room again, and start over. If I don’t, we are now in the position in question. I’ll put $100 in the middle, and you $500 (since you should win 99 times as many games than me this is more than generous!). You can pick whichever of the remaining two boxes you like (I assume you’ll want to switch, but I’ll leave it entirely up to you), and if you get the ball, you get the $600 pot, otherwise I win it. We’ll agree to play *at least* 20 games so that any flukey results do not distort the overall trend. Fair?
(I don’t need anywhere near as many as 20 games to be well enough assured of being ahead, but why take any chances?)
Comment by JC — 1/25/2005 @ 3:14 am
JC - maybe if you answer my question about “box 45″ I’ll be able to answer your question.
Comment by Boman — 1/25/2005 @ 7:39 am
Sorry Boman, I thought I did cover it when I said I didn’t care what you did with the boxes. You can make a circle and remove your “box 37″ and leave my “box 45″ in it. You can leave both in it if you like, or you can remove both and then open all the other 98, I couldn’t care less. Do what you like with it to try and prove your point, and I’ll counter-argue or concede defeat, whichever is appropriate.
Comment by JC — 1/25/2005 @ 7:58 am
Xriq -
I quote Xriq:
To illustrate, let’s limit the 27 choices to 9, by stipulating that you play Door 1 everytime. Also stipulate that….you change your bet everytime the game isn’t canceled. (that is, canceled because it’s and illegal game) (RM - Random Monty)
Scenario 1: Door One holds the prize.
Game 1: RM selects Door 1 (yours, and the prize). game canceled.
Game 2: RM selects Door 2, you switch to 3. You lose.
Game 3: RM selects Door 3, you switch to 2. You lose.
Scenario 2: Door Two holds the prize.
Game 1: RM selects Door 1 (yours). Game canceled.
Game 2: RM selects Door 2 (prize). Game canceled.
Game 3: RM selects Door 3, you switch to 2. You win.
Scenario 3: Door Three holds the prize.
Game 1: RM selects Door 1 (yours). Game canceled.
Game 2: RM selects Door 2, you switch to 3. You win.
Game 3: RM selects Door 3 (prize). Game canceled.
I count 5 scratches, two wins, and two losses…No overall advantage or disadvantage to keeping vs. switching your bet. (end of quote. )
In other words, Xriq is claiming that RM is so stupid that he screws up and can’t get to the 1/3-2/3 ratio that an Intelligent Monty obtains.
We can simplify the above outline by showing the appearance possibilities of the above games.
To keep it simple only legal games are considered.
The prize is on door one.
There are four ways the game can be played.
Here’s how Xriq would do the outline, using the outline above as our guide.
If the contestant chooses:
l- Door 1, host can open door 2. 1/4 possibility.
2- Door 1, host can open door 3. 1/4 possibility.
3- Door 2, host can open door 3. 1/4 possibility.
4- Door 3, host can open door 2. 1/4 possibility.
total 1
I believe this is an accurate summary of Xriq’s position from his outline above. To follow these possibilities to their logical conclusion will result in 2 wins and 2 losses for RM.
However, there is a problem - we are playing 3 doors, not four doors. In this scenario, the contestant picks one of three doors, resulting in 3 possible games. So, where did Xriq get four games?
He did it by giving the wrong values to the possibilites.
Here’s how the outline should look. (Same pre-outline directions)
If the contestant chooses:
1-Door 1, host can open door 2 - 1/6 possibility.
2-Door 1, host can open door 3 - 1/6 possibility.
3-Door 2, host can open door 3 - 1/3 possibility.
4-Door 3, host can open door 2 - 1/3 possibility.
total 1
Door 1 will occur 1/3 of the time.
Door 2 will occur 1/3 of the time.
Door 3 will occur 1/3 of the time.
total 1
Using these possibilities Xriq should have said, “I count 5 scratches, two wins, and one loss.”
He has based his assertions upon an incorrect understanding of the possible occurance of the variations of the games.
When illegal games are thrown out we find that RM plays the 3 door game exactly like Intelligent Monty. It is still 1/3 and 2/3.
Comment by Boman — 1/25/2005 @ 8:26 am
Only in the real game, where Intelligent Monty plays. Random Monty doesn’t know (or care) where the prize is, remember? So when the prize is behind Door 2, RM is just as likely to open Door 2 as he is to open Door 3. Thus, in the randomized version of the game, the last two outcomes only have a 1/6 probability each, with voided games accounting for the other 1/6.
But I’m bored of Monty, and I’m really looking forward to owning your house, so let’s return to JC’s boxes. Do you really think that by drawing a circle around a randomly chosen box, drawing happy faces on a randomly chosen box, peeing on a randomly chosen box, or doing anything else to a randomly chosen box - short of having a knowledgeable game master systematically avoiding opening both it and the box containing the prize - can affect its odds of holding a randomly placed ball?
Comment by Xrlq — 1/25/2005 @ 9:07 am
No, Xrlq did nothing of the sort. He correctly outlined the possibilities which arise from picking doors randomly, showed which resulted in illegal games, and which resulted in you being presented with the same choice that you have in the original game, and from here showed that half the time you won if you switched. It is *you* that has assigned each of the variances an incorrect probability. They all occur with probability 1/6. You have 3 choices. Monty then has 2 between the remaining boxes (he doesn’t know the location of the prize). Ergo 6 variances occurring with probability 1/6 each. 2 of these result in illegal games and are removed from the “universe” of possible results (since we’re forcing ourselves into the position of picking a door and seeing another empty one), so we’re now looking at one of 4 possible variances, all with the same original probability (1/6) of occurring. Half of which you lose if you switch.
It seemed clear at that time that you didn’t understand the difference between the proposed “Random” Monty (which everyone else was talking about) and his intelligent counterpart, your concept of him appeared to be randomly opening doors that weren’t the contestants and also weren’t the prize, i.e. identical to normal Smart Monty.
So I then tried to remove him altogether and make it clearer, by speaking of doors falling off without any human intervention. You still maintained that it didn’t matter, and as long as you were presented with the same situation, i.e. you’ve picked a door and another has been shown (through whatever means) to be empty, you should switch. Expanding it to 100 boxes still didn’t make you change your mind.
So: is the bet I proposed earlier agreeable to you? According to your odds, you have a 99-1 chance of winning $100 and a 1-99 chance of losing $500, an average gain to you of $94 per game.
Please try to concentrate on answering that one question. I think I outlined all the conditions clearly enough to avoid any ambiguity.
Finally, Xrlq, bug off - this is my bet =P
I think you hesitated slightly when having the same argument with Patterico and missed out on your fortune. I don’t intending making that same mistake!
Comment by JC — 1/25/2005 @ 9:32 am
Nah, I didn’t hesitate at all. Patterico and I talked past each other a few times, but by the time we were both absolutely positive of what the other meant, he saw the error of his ways.
Boman, by contrast, appears to be impervious to learning, so what say we take turns betting him? Don’t worry, he’ll play long enough to leave plenty for both of us, or even a third player if he wants in. He may end up having to pay you in U.S. dollars, however, so be forewarned that your huge jackpot will buy you f— all in Britain. Might as well save yourself the trouble and leave all the money to me.
Comment by Xrlq — 1/25/2005 @ 10:40 am
Okay, sounds reasonable. I did kind of butt in on your discussion in the first place, to be fair. Tell you what, we’ll get 1000 boxes and I’ll put our winnings in one of them. Then we’ll arrange them all in a big circle, and you’ll be shut in the room with them and you have to open them at a rate of one per minute. All of which should have given me ample time to flee back to the UK with the booty.
I’m under the impression that Boman is unable to comprehend reducing the universe of possible results due to the fact that we’re enforcing certain conditions, and thus failing to adjust his probabilities accordingly.
Comment by JC — 1/25/2005 @ 12:09 pm
Oops, I didn’t finish that last comment. I meant to go on to say:
What he’s doing is saying his original box maintains it’s 1/100 chance all along (which is kinda true depending on how you want to phrase it), the mistake he’s then making is saying that the other remaining box HAS to have 99/100 chance (because he feels the probabilities should add up to 1) - it doesn’t, it also had a 1/100 chance and your universe of possible results has reduced to just 2/100 of the original set, and each remaining box represents a 50% share of this.
But couldn’t you also say this of the original Monty puzzle when a door has been removed? No. To start with, the box could be in door 1, 2 or 3, with equal probabilities for each 1/3 + 1/3 + 1/3, giving us our complete universe of possible results = 1. When we choose box 1 and see box 2 is empty, our universe of possible results has indeed reduced, but the ratio of box 1:box 3 is the key thing. The chance that this happened because the prize is in box 1 is 1/3 x 1/2 (1/3 for the prize originally being there, 1/2 because box 2 was chosen out of the two remaining ones) = 1/6. The chance that this happened because the prize is in box 3 is 1/3 x 1 (1/3 for the prize originally being there, and 1 because box 2 HAD NO CHOICE but to then be removed) = 1/3. So the ratio of the probability of box 1 containing the prize to box 3 is 1/6:1/3, or 1:2.
It’s interesting how in the second example (the original Monty puzzle) removing one box has actually reduced the universe of possible results down to 1/2, rather than the 2/3 you might instinctively think - of course 2/3 is what you get by RANDOMLY removing an empty box!
Comment by JC — 1/25/2005 @ 12:52 pm
I agree with everything except your proposal for divvying up the jackpot. I propose this game instead. First, whichever one of us has the loot puts it in escrow, to be paid out to whoever wins the ball/box game. Then, I look away just long enough to allow you to hide the ball in one box and to mix them around a bit, keeping all of them inside the circle. I choose one box, carry it to an area of the room well outside the circle, and set it down. You open 98 of the other 99 boxes without revealing the prize. [It's up to you whether to do so purposefully or randomly, canceling and restarting the game each time you inadvertently reveal the prize.] Once we’re down to two unopened boxes, I mentally recall carrying the box. If it seemed a bit heavy, or if I heard a ball bouncing around inside while carrying it, I stick with my original choice. If I don’t, I switch.
Comment by Xrlq — 1/25/2005 @ 2:30 pm
Ha ha, do you know, that’s exactly why I had YOU remove the box from the circle in the outline of my bet with Boman earlier! I had originally written that he picked a box and removed it from the circle, and then realised that of course it would probably be fairly obvious whether or not that box contained a ball….
Here’s a better game: Each of the following sequences of coin tosses occurs equally frequently, agreed?
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
You tell me which sequence you want to pick, then I tell you which one (of those that remain) I want. Then we toss a coin continually, writing down the result. The winner of the loot is the person whose sequence we hit first. Sound fair?
Comment by JC — 1/26/2005 @ 1:08 am
It certainly sounds fair, but judging by the fact that you suggested it, I assume it is not. What’s the catch?
Comment by Xrlq — 1/26/2005 @ 9:15 am
Oh, so you’re calling me a cheat now are you? It’s completely fair. When have I ever offered a loaded bet? ;o) I’m even being nice by giving you first choice, so even if there were a “best” sequence, you could grab it first…
I just noticed earlier that I called you Xlrq in an earlier message, so I apologise for that error. I have a friend who has a son called Xlrq you see, so you can understand the confusion.
Comment by JC — 1/26/2005 @ 9:25 am
Take one of those “earlier”s out of that last bit. I look stupid otherwise.
Comment by JC — 1/26/2005 @ 9:26 am
I know, I just assumed it was the British spelling.
Comment by Xrlq — 1/26/2005 @ 9:59 am
Still, probably better than Xriq, eh?
Are you gonna take me on at the coin-tossing game then? Your sequence has a 1/8 chance of coming up for any given 3 tosses, as does mine. What could be fairer?
Comment by JC — 1/26/2005 @ 2:40 pm
Fairer would be for each of us to make our choice separately, then pick something else if we happened to land on the same combination. Your proposal allows you to choose a sequence that ends as mine starts, but doesn’t start as mine ends. That, in turn, will allow you to win some of the games I was in the process of winning, but prevent me from doing the same to you.
That said, I’ll agree to your bet anyway. We’re playing for Boman’s losses, and he appears to have buggered off for good, so why should I care that you have an unfair advantage in winning a $0.00 jackpot?
Comment by Xrlq — 1/26/2005 @ 3:50 pm
Pretty much spot on. My precise formula is to take your first two and make them my last two, and then make my first the opposite of my last (so if you choose THT, I choose TTH). My chance of winning is either 2/3, 3/4 or 7/8, depending on which sequence you chose. The game is called Penney-Ante if you want to research it further…
Actually, something just occurred to me. What if a third player joins as well? If B is more likely to appear before A, surely we can pick C based on B using the above formula, so that it is more likely to appear before that, and therefore also more likely to appear before A? Eep. I need to give that some thought. Unfortunately the maths are a little messy for some of the sequences (for HHH it’s a simple case that if they aren’t the first three tosses, my THH will occur first every time)
Actually, there’s a dice game that’s similar, where you have a red one with sides 2,2,6,6,8,8 a green one with 1,1,7,7,7,7 etc (these aren’t the right numbers, I just plucked them out of nowehere, but you get the idea). Each one rolls the same AVERAGE value, yet for any one you pick you can select another which will have a GREATER chance, on a single roll, of showing a higher value. The maths are a little easier for this, and it’s a similar idea, so hopefully the conclusions will also be similar. Now all I need to do is find out the specifics of the game….
As for Boman, the silence may finally suggest he’s seen the error of his ways, although it would be nice if he were man enough to admit it. That’s probably wishful thinking though, and he’ll return in a day or two to tell me again that I’m wrong because I wasn’t clear about which boxes were inside the magic circle.
Comment by JC — 1/27/2005 @ 2:22 am
Over 200 comments on this post now. I think that’s a record.
Comment by Patterico — 1/27/2005 @ 8:18 pm