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 anyt