The Answers to the Fun Statistics Question
Yesterday I posed the following problem:
50 standard coins are thrown onto the floor. Before the toss, each had an equal chance of coming up heads or tails. Before you see how they came up, all 50 coins are covered up.
Tom goes and uncovers 25 of them. All are heads.
Joe offers to give you $10 for every remaining coin that came up tails, if you will give him $12 for every remaining coin that came up heads.
1) Do you accept Joes offer?
2) Is there any question you would like to ask Tom before you decide?
Here are my answers:
1) It depends on the answer to #2.
2) Yes. I would ask Tom if he knew the coins he uncovered were heads before he uncovered them.
If Tom knew the coins would be heads, then we’re in a “Monty Hall” situation. He has uncovered 25 coins on a non-random basis, which gives us extra (and non-random) knowledge about the remaining coins.
If he didn’t, we’re in a “Gambler’s Fallacy” situation. If he removes coins on a random basis, then he has told us nothing new about the ones that remain.
It’s an interesting problem, and there’s no shame in getting it wrong. It fooled me a while back.
If we didn’t know which situation we had — and if we knew the game wasn’t rigged by dishonest people — we would have to assume we’re in the Monty Hall situation, since the odds against the first 25 coins randomly coming up heads are astronomical.
P.S. Here is a fun follow-up question. What if Tom didn’t know whether he would be uncovering heads, but he decided in advance to throw out every situation in which he didn’t uncover 25 heads?
(That’s a stipulation that wasn’t given in the original problem.)
In such a case, is there still prior knowledge, just arrived at in a different way, making it a Monty Hall situation?
Or is it a Gambler’s Fallacy situation because of the lack of knowledge?
Don’t ask me. Xrlq tells me that my intuitive answer is wrong, and I’m not inclined to argue with him. So debate it out.
It’s not a Monty Haul problem.
If Tom flipped 50 coins, and selected 25 heads to show to you, it doesn’t affect the results of the remaining coins.
If the Monty Haul problem looked like this, there would be fifty doors and at least 25 goats that he’d show you. There would be an unknown number of cars and goats behind the remaining doors. This is not a Monty Haul problem.
CliveStaples (49fc6e) — 8/23/2008 @ 7:02 pmI got it.
As for your follow-up question, I’m not sure what this means. Are there multiple trials of 50?
Cyrus Sanai (4df861) — 8/23/2008 @ 7:11 pmAs far as the follow-up question goes, it’s a Gambler’s fallacy as you don’t have any knowledge about the remaining coins. It’s just like tossing 25 coins.
For the original problem, I still don’t take the bet whether he knew or not. It would only affect whether I turn the bet around.
Stashiu3 (460dc1) — 8/23/2008 @ 7:14 pmPatterico:
Those odds don’t affect whether the 26th coin was heads. Yes, it’s exceedingly unlikely that the first 25 coins were all heads–1/(2^25)–but the odds of getting 25 heads in a row followed by 1 tails is no greater than the odds of getting 25 heads in a row followed by 1 heads.
CliveStaples (49fc6e) — 8/23/2008 @ 7:16 pmIf he removes coins on a random basis, then he has told us nothing new about the ones that remain.
This is where I disagree (with no knowledge to back it up). Since he has removed all heads, then the removal consists of coins outside a random distribution, which should, to me, affect the probability of the distribution of the remaining coins in aggregate. The closer the coins removed are to random distribution, the closer the remaining would be to a random distribution, and the reverse should also be true.
To say that all coins have a 50% chance is true per toss but to apply that to the entire 50 coin toss in aggregate would necessitate that the toss should end up 50H as often as 25H/25T.
I’m saying that the Gambler’s fallacy applies to each specific toss, not the complete toss in aggregate. You cannot tell which coin specifically will be H or T, as that is a 50% chance, but you can tell how many out of a group should be H or T according to probability.
It is true that there’s an equal 50% chance of the toss of a coin being T, even if there’s 49H before, but it’s not true that there’s an equal chance of 49H/1T and 25H/25T.
Which is why I see no difference between this new question and the old one.
But I’m probably wrong, as I’m just BS’ing here and someone will use math to prove it.
Apogee (366e8b) — 8/23/2008 @ 7:21 pmI think that pretty much defines the Gambler’s Fallacy Apogee. If the coins were uncovered at random, the remaining coins are just as random. It’s the appearance that something just “has to start coming up because it’s due” that gamblers fall for. But the dice (or coins, or cards, or wheel, etc..) do not have a memory unless the game is fixed.
Stashiu3 (460dc1) — 8/23/2008 @ 7:27 pmIt’s a Gambler’s Fallacy only if Joe allows it to be. Whether Tom knows the distribution of the remaining coins is only good for Tom’s own peace of mind. In either case it’s a sucker bet. You’re betting more than your opponent on even odds.
From “Guys and Dolls”:
nk (3c7a86) — 8/23/2008 @ 7:31 pmMost of the math classes I had to take I just barely managed to pass. I pick answer 3.) Tell Tom to let me think about his offer, then find a math geek to answer the question and act accordingly.
doubleplusundead (caccb2) — 8/23/2008 @ 7:34 pmAnd yes, I’d give said math geek a cut of the profits, assuming there were any.
doubleplusundead (caccb2) — 8/23/2008 @ 7:35 pmdpud, do not take the bet. Cider is sticky.
Stashiu3 (460dc1) — 8/23/2008 @ 7:38 pm😉
Stashiu3 – Then why is counting cards illegal?
Apogee (366e8b) — 8/23/2008 @ 7:41 pmBecause they are not shuffled after each hand.
Stashiu3 (460dc1) — 8/23/2008 @ 7:42 pmAnd neither are the coins re-thrown. That’s what seems off for me.
Apogee (366e8b) — 8/23/2008 @ 7:45 pmOnce a card is taken from the deck, it is not possible for it to come back into play until the deck is reshuffled. On the other hand, each coin always has a 50/50 chance of heads or tails, no matter what the coins around it do.
One action changes the possibility of the remaining results, and therefore the odds… the other doesn’t change the odds at all, the results are already there waiting to be uncovered.
For example, if I lay out every spade in a single deck face down, you have a 1 in 13 chance of picking the Ace. If you miss, the odds change to 1 in 12. After you find the Ace, the odds of finding another are zero.
Same with counting cards. In a six-deck shoe there are 24 Aces and 60 ten-point cards. If too many of the Aces and/or face cards come out, you have a much smaller chance of getting one. If you reshuffled after each hand, your chances are the same each hand (not very good btw).
Stashiu3 (460dc1) — 8/23/2008 @ 7:55 pmWell, each head revealed changes the probability of the available number of tails left, since it cuts down the number of possibilities; for instance, if four coins are flipped and he shows you two heads, that removes the possibility of T-T-T-T.
I suppose that the selection method actually does matter. If he has simply revealed the result of the first 25 coins he flipped, the remaining 25 coins will most likely follow a standard distribution–somewhere close to 50% heads, 50% tails.
If he has revealed 25 heads of his choosing, that accounts for many more possibilities of various results. Imagine he only flipped 4 coins instead of 50; if he reveals the first 2 coins to be heads, that means the possible results are thus:
HHHH
HHHT
HHTH
HHTT
That’s only a 25% chance that you’ll net a profit–exactly the same probabilty if he had only flipped two coins.
Whereas if he simply chose to reveal two heads, the possible results are thus:
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
THHT
HTTH
THTH
TTHH
That’s ~54.5% chance of netting a profit.
Of course, whether you’d take that risk depends on your degree of risk aversion.
CliveStaples (49fc6e) — 8/23/2008 @ 8:12 pmDo coins know statistics?
nk (3c7a86) — 8/23/2008 @ 8:16 pmProbability theory is an attempt to turn ignorance into a science. Nothing more.
nk (3c7a86) — 8/23/2008 @ 8:21 pmSo, thinking of it as just red or black cards, the difference is that each card can only be red or black, and each coin has the equal possibility of being H and T. Thereby the distribution is set for the cards at the beginning of each game, and increased information allows you to increasingly identify each leftover card.
I’m sure you’re right about this, and I appreciate you responding. I just can’t shake that distribution question, as the counter-intuitive part for me is the probability of the leftover percentages from the aggregate. It’s hard to believe that you have an equal chance of finding 11H and 12T in the leftover coins as you do 5H and 20T, once you remove 25H from a single toss of 50 coins.
Apogee (366e8b) — 8/23/2008 @ 8:25 pmharumph! I’m caught in the filter I guess… no links even. What’d I say? 🙁
Stashiu3 (460dc1) — 8/23/2008 @ 8:29 pmAhh… bet it was the word “c a s i n o”. I think someone else had that problem before.
Little help?
Stashiu3 (460dc1) — 8/23/2008 @ 8:31 pmnk – Do coins know statistics? No, and neither do I. But coins have a hell of a time typing.
Apogee (366e8b) — 8/23/2008 @ 8:36 pmI can’t get the spam filter to release the comments so I’m reprinting them.
Comment from Stashiu3:
That’s why it’s the Gambler’s Fallacy Clive. Sure, the possible number of tails decreases with each head revealed, but so does the possible number of heads remaining.
If two coins are remaining, the possible distribution is:
HH
HT
TH
TT
The previous results do not factor in because you’re only looking at two coins. You’re taking a sucker-bet if you do it. Even odds for greater risk than the person you’re betting against.
By your reasoning, if I toss a thousand coins, reveal 998 of them and find 700 of them are heads, the last two just have to be tails because they’re due. Not so, it’s still 50/50 for each coin and you’re a sucker if you take the bet. Now, betting $10 on heads against $10 on tails would be even odds (still not good enough… ask the casino owners).
Aug 23, 8:23 PM
DRJ (a5243f) — 8/23/2008 @ 8:41 pmNot quite the same thing. You have the same chance of finding 5H and 20T as you would no matter what the first 25 were. In other words, whether the first 25 were evenly distributed, all heads, or all tails has no bearing on the odds of the remaining coins. It’s still even odds for greater risk ($10 vs $12).
Stashiu3 (460dc1) — 8/23/2008 @ 8:42 pmThank you DRJ. 🙂
Stashiu3 (460dc1) — 8/23/2008 @ 8:42 pmYes. Thank you DRJ, and Stashiu3.
Apogee (366e8b) — 8/23/2008 @ 9:17 pmContrary to my original intuition, it’s a Monty Hall situation, only this time it favors heads instead of favoring tails. Yes, it’s possible that someone will pitch 25 heads and 25 tails, and Tom will just “happen” to randomly uncover all 25 of the heads and none of the tails. Possible, yes, but not bloody likely. Realistically, the only games that will ever survive a screening process like that are the ones where nearly all of the coins landed on heads to begin with, leaving heads with a solid majority even after the first 25 were removed.
Xrlq (62cad4) — 8/23/2008 @ 9:17 pmXrlq 9:17 – Realistically
That’s cheating.
Apogee (366e8b) — 8/23/2008 @ 9:25 pmThe second question is highly relevant in explaining why I don’t want to play his game.
If Tom didn’t know in advance which coins were Heads and which were Tails, then his turning over 25 coins in a row, all of which are Heads, has a 1 in 2^25 or 1 in 33,554,432 chance of occurring. Which is comparable to the odds in winning the jackpot in one of the powerball type of lotteries.
Which means that either Tom is cheating or incredibly lucky. In either case, I don’t think I would bet against him.
Now, if Tom asserts that he did know which coins were Heads before turning 25 of them over, then he isn’t necessarily cheating or incredibly lucky, but we have the Monty Haul problem.
Of the 1.126×10^15 possible coin toss sequences, we can immediately eliminate 4,997×10^14 of them because they wouldn’t have 25 Heads to turn over.
Of the remaining 6.262×10^14 cases, slightly over 20% (1.265×10^14) only had 25 Heads, all of which have already been turned over by Tom, so I lose $10 per coin or $250.
Another 19.4% of the remaining cases have 24 Tails and 1 Heads, so I lose $228.
My expected loss is $194.48.
It is a sucker game, and today isn’t my turn to be the sucker.
GaryC (4c03da) — 8/23/2008 @ 9:29 pmThat’s if he uncovers them randomly. That would also make it even more of a sucker bet, but I would say that’s also Gambler’s Fallacy if you try to apply it to the remaining coins. If the odds of randomly uncovering 25 heads in a row out of 50 tossed coins quoted in the earlier thread were accurate (about 1 in 16 million per toss of 50 coins), it still doesn’t change the odds on the remaining coins. Statistically, it’s still a 50/50 shot on the remaining 25, despite beating 1/16,000,000 odds with the first 25. Real world however, no bet either way.
Stashiu3 (460dc1) — 8/23/2008 @ 9:35 pmLook at it this way. No matter how many coins are tossed, what if 25 are removed for the bet? Those results are fixed already. They won’t change no matter if the other 25, 250, or 25 millions coins that were tossed are looked at or not. Anyone who tells you differently is someone you should be making bets with.
Stashiu3 (460dc1) — 8/23/2008 @ 9:42 pmXrlq:
What’s the selection method, then? Does Tom just randomly reveal the results of 25 coins? Does he choose which coins to reveal?
If it’s a Monty Haul problem, then he does choose which coin to reveal; he’ll reveal heads.
Odds are that out of fifty coin tosses, near to 25 of them will be heads. Probably not the first 25 tosses, but out of the entire fifty tosses. So if he shows you 25 heads that he’s selected from the pool of tossed coins, odds are that there are more tails left than heads–since there are probably around 25 tails, too.
It’s basically a bell curve distribution. The break-even point (out of the remaining 25 unrevealed coins) is at 12 tails, 10 heads. At $10 per tail and `$12 per head, you’d come out exactly even.
So what are the odds that out of fifty coin tosses, more than 12 are tails? It’s actually quite good.
This presumes, of course, that Tom hasn’t simply revealed the results of the first 25 coin tosses.
CliveStaples (49fc6e) — 8/23/2008 @ 10:29 pmStashiu3 #30:
That’s true only if the revealed coins are picked at random.
As has been said many times, if 25 consecutive heads are revealed, that’s very strong evidence the reveal is not random. The reveal is independent of the flip, so the fact that the flip is honestly random is irrelevant.
There is nothing in the problem setup that specifies the reveal is random, and I don’t know why people are insisting it must be.
===
The revised statement of the problem, where Tom only reveals in the case where the first 25 coins he looks at happen to be heads, simply means that, on average, he and we all die of old age waiting for that to happen.
DJMoore (3c32ad) — 8/23/2008 @ 10:34 pmExcuse me, I wrote the proper ratio of tails to heads rather than the actual number out of 25.
The correct break-even point is at ~13.64 tails and ~11.4 heads. So as long as there are at least 14 tails, you’ll come out ahead.
CliveStaples (49fc6e) — 8/23/2008 @ 10:37 pmOk. How about:
The coins are digital, and the single toss of 50 is displayed on two different screens. The results are the same, but on the first screen Joe knows which ones are Heads and uncovers 25 of them.
On the second screen, Tom, by chance, happens to do what is improbable, but possible, and uncover, in random order, 25 Heads.
Remember, it’s two different actions on the same throw.
Is one the Monty Hall situation and the other not?
Apogee (366e8b) — 8/23/2008 @ 10:46 pmIf you end up with 13 tails you get $130, but that leaves 12 heads and you pay $132… so you lose money. Besides, you’re tossing all 50 coins at the same time. The results are already fixed.
Xrlq specified that the overturned coins were done randomly, but that doesn’t affect what has already happened. If the overturned coins were selected because they were heads, it’s a sucker bet. If they were random, it’s an even chance on the remaining coins… no matter how many were already looked at (or even if they weren’t). 50/50 odds on a 6/5 wager is a sucker bet as well. You have to beat the 50/50 odds to come out ahead (actually more because even getting 13 out of 25 puts you $2 in the hole).
Stashiu3 (460dc1) — 8/23/2008 @ 10:46 pmStashiu3 – did you just answer me before I commented?
Apogee (366e8b) — 8/23/2008 @ 10:48 pmI posted a response to Clive just as you posted, but the same thing applies. If the toss is truly random, it’s a sucker bet no matter what. Don’t do it. The only variable is how big a sucker you are if you do.
That said, you can’t win if you don’t play. If you’re not averse to risk, you might beat the odds even on a sucker bet. Sometimes the ring does fall onto the Coke bottle. Long term though… the house is going to clean up.
Stashiu3 (460dc1) — 8/23/2008 @ 10:51 pmxrlq–
I’m not sure what it is, but I don’t think it’s Monty Hall OR “Gambler’s Fallacy.”
There is the extreme situation where Tom tosses 50 heads and then (certainly) picks 25 of them. As the number of heads tossed decreases, the chance of Tom not missing decreases. The toss distribution at which there is a 50% expectancy of a miss would be the key to making or not making the bet, adjusted by the odds.
I am unconvinced that this set of odds matches either of the two original choices (50-50 or highly asymmetric). I’d also expect a fairly large standard deviation. But I don’t really feel like doing the math.
Kevin Murphy (0b2493) — 8/23/2008 @ 10:54 pmBTW, Tom is going to be tossing and randomly choosing coins for quite a while before he gets it all set up.
Kevin Murphy (0b2493) — 8/23/2008 @ 10:59 pmComments are still getting caught up. I just saw DJMoore’s comment, which wasn’t there before. Nor was Clive’s 10:37 comment. Strange. I have another one in the filter, so I’m going to hold off on more comments for a bit until it clears.
Kevin, all 50 coins are tossed at the same time… you see that, right?
Stashiu3 (460dc1) — 8/23/2008 @ 11:09 pmHere’s another one.
Stashiu3 @ Aug 23, 11:01 PM:
LOL! That was rather amusing when I hit post and the answer actually applied to Clive and you. Rule of thumb… if there isn’t any skill involved, you’re eventually going to get screwed by the odds. Doesn’t matter if they’re long or short, chance will catch up with you. It’s always better to play poker than Blackjack (assuming you know how to play both well), and better to play Blackjack than roulette. Even the best slot machines only pay out 98% of what they take in.
DRJ (a5243f) — 8/23/2008 @ 11:21 pmStashiu3:
Yes, that’s true. So if at least 14 of the 50 tosses came up tails, you net a profit. Which is a probably–although not guaranteed–outcome.
If the overturned coins were selected because they were heads, then you should play. Odds are that the total number of heads is ~25; eliminating 25 heads will probably leave a good number of tails.
If the overturned coins were revealed as the results of the first 25 tosses, then you’re actually less likely to win money. You’d need at least 14/25 of the tosses to be tails, rather than 14/50.
CliveStaples (49fc6e) — 8/23/2008 @ 11:22 pmShash–
This thread has mutated from the blog post, I think. As I understand it, the question was: OK, let’s assume that Tom’s toss AND coin’s revealed are both random AND that Tom does both over until he manages to reveal exactly 25 heads.
Such a task will take a while.
Xrlq’s point that success most likely requires a toss that contains MANY heads is well taken. But how many? It is probably not 50-50, but it might be close to 50-50, making it neither “gambler’s fallacy” nor an obvious one-sided bet (maybe 80-20) like the original “Monty Hall” scenario..
So, yes, I do get it. But most of the commenters seem to be responding to some other question.
Kevin Murphy (0b2493) — 8/23/2008 @ 11:23 pmKevin Murphy:
All that matters is whether Tom can only reveal the results of the first 25 tosses, or if he can reveal the results of any of the 50 tosses.
I don’t really see the point of stipulating that he ‘randomly’ selected 25 coins to reveal, but they just happened to all be heads. This is statistically indistinguishable from the options above.
The question is whether Tom is eliminating “heads” (i.e., “goats”/losses) from the available pool or not. If he just reveals the results of the first 25 tosses, that’s like having a fourth door on Monty Haul that is revealed to be a goat before you ever make a decision. The results are exactly the same.
CliveStaples (49fc6e) — 8/23/2008 @ 11:30 pmBut it’s not 14 out of 50… it’s 14 out of the 25 remaining. The 25 that were already revealed are irrelevant no matter what they turned out to be.
Kevin,
The original question did not assume the reveal was random, but the follow-up did… with multiple 50-coin tosses until the first 25 randomly revealed coins were all heads. Somebody else figured the odds at 1 in 16 million, but whatever they are, it’s extremely unlikely. The fact remains that this has no effect whatsoever on the remaining coins. They might all be tails, all heads, or any combination thereof. The answer is already fixed and doesn’t change just because someone beat the odds with the first 25 coins revealed.
So yes, it would probably take a long time in that sense. Or it might happen with the first 50-coin toss.
Stashiu3 (460dc1) — 8/23/2008 @ 11:32 pmYou’re right, as long as Tom was only revealing the results of the first 25 tosses. If some of the heads he revealed were from later tosses, then the statistics are a bit different.
Here’s the two possible scenarios:
1) Tom flips 50 coins, and reveals the results of the first 25. All of which are heads. This means that of all the possible outcomes for the fifty tosses, you eliminate all that have a “tails” result in the first 25 tosses.
2) Tom flips 50 coins, and reveals exactly 25 “heads” results. This means that of all the possible outcomes for the fifty tosses, you eliminate all that have fewer than 25 “heads” results.
Do you see the difference?
CliveStaples (49fc6e) — 8/23/2008 @ 11:37 pmScenario (2) has quite a few more options that include tails, since you can take into account the possibility of any of the first 25 tosses being tails. Thus, out of the 50 tosses, you only need 14 tails.
Scenario (1) has quite fewer options that include tails, since it only concerns the results of the second 25 tosses. Thus, out of the last 25 tosses, you need 14 tails.
CliveStaples (4c1930) — 8/23/2008 @ 11:44 pmCliveStaples – 11:30pm
From your 11:22pm If the overturned coins were selected because they were heads, then you should play.
Which is the clarification I was seeking in my #34 (if it’s still #34 -10:46pm)
In one scenario Joe is eliminating heads from the available pool, and the other version of the same toss Tom reveals the results of the first 25 tosses to be all heads.
Since it’s the same toss, 25 heads are chosen out of the 50 available coins and 25 heads are revealed from the first 25 tosses. They are, however, the same heads. How is one a different set of odds than the other? (or not)
Apogee (366e8b) — 8/23/2008 @ 11:47 pmI answered and it’s caught up again, probably because I linked to my answer in #3 where I said (essentially) if he cherry-picked the results to get the first 25 heads, he’ll know what the rest are and wouldn’t offer the bet unless he was sure to come out ahead.
The better option would be to turn the bet around and shade the odds in your own favor, but the best option is not to make the bet at all.
Stashiu3 (460dc1) — 8/23/2008 @ 11:50 pmApogee:
If the 25 heads came from the first 25 tosses, then the odds don’t favor you. You’re essentially playing the results of the last 25 tosses, which are generally going to be 50% heads and 50% tails, making it quite difficult to make your rubber of 14 tails.
If the 25 heads were selected from any of the fifty tosses that resulted in heads, then the odds favor you. You’ve eliminated far fewer possibilities that include tails.
CliveStaples (4c1930) — 8/23/2008 @ 11:52 pmApogee, we keep cross-posting and I am answering you before I even see your post, lol.
Don’t take the bet. If he can pick out 25 heads, he knows how many heads/tails there are left and wouldn’t offer the bet unless he was sure to win. Even if you stipulate that he grabbed them one at a time and stopped once he reached 25 heads, he still has more information than you do since he’ll know how many tails he saw (no pun intended) before he reached 25 heads. It’s a sucker bet I tellz ya! 😉
Stashiu3 (460dc1) — 8/23/2008 @ 11:53 pmStashiu3:
You’re going beyond what was stipulated. Patterico said that Tom only discarded the results that did not result in the first 25 tosses being “heads”. Of course, in this scenario the odds favor Tom anyway.
That’s true only if the 25 heads were the results of the first 25 tosses.
The set “Results that include 25 heads” contains far more tails than the set “Results that include the first 25 tosses being ‘heads'”.
CliveStaples (4c1930) — 8/23/2008 @ 11:58 pmStashiu3:
Where was it stipulated that Tom would only offer bets that he was sure to win?
CliveStaples (4c1930) — 8/23/2008 @ 11:59 pmIt wasn’t, but at best you’re wagering 6/5 on 50/50 odds for 25 coins. That’s the best-case scenario and it is not in your favor. You continue to think of it as being out of 50 coins total, but it’s not… if the reveal is random it’s only the last 25 that matter. If it’s not random, Tom still has more information than you do and if you trust someone to make a bet that they’re more likely to lose than win… not wise. Hope you like cider. 😉
Stashiu3 (460dc1) — 8/24/2008 @ 12:06 amStashiu3:
If the reveal is randomly selected from among the 50 coin tosses, then you should play.
It isn’t the “last” 25 that matter, in that case. It’s the 25 that weren’t the revealed tosses that matter.
The only time the 25 that weren’t revealed are the last 25 tosses is when the first 25 tosses were all heads (an infinitesimally small chance) and Tom just happened to choose exactly those 25 to reveal (another infinitesimally small chance).
If he chose the heads randomly from the pool of 50 results, then most likely he didn’t choose the first 25 results, since they were very unlikely to all be heads.
The question is which possibilities get eliminated. If he reveals the first 25, and they’re all heads, then he’s eliminated all results that aren’t HHHHHHHHHHHHHHHHHHHHHHHHHX…. That is, he eliminates half of the possibilities off the bat.
If he reveals any set of 25 heads, then he’s merely eliminated all results that do not include 25 heads. So, for instance, this result would be included: TTTTTTTTTTTTTHHHHHHHHHHHHTTTTTTTTTTTTHHHHHHHHHHHHH.
This result would have been eliminated if Tom only reveals the first 25 results. Since the 25 heads can be distributed amongst any of the 50 results, rather than confined to the first 25, we have far more results to consider.
CliveStaples (4c1930) — 8/24/2008 @ 12:16 amOK, so our issue is what happens at 50 coins tossed. Tom turns over 25 coins. If they aren’t all heads, he tries again.
We can event-test this at lower levels (Event testing would have saved a lot of math PhD’s a lot of embarrassment on the Monty Hall problem; there are only six events in that one.)
Let’s look at four coins. Tom will always turn over coins 1 and 2; all of the combos are:
HHHH (Tom keeps)
HHHT (Keep)
HHTH (Keep)
HTHH X
THHH X
HHTT (Keep)
HTHT X
THHT X
HTTH X
THTH X
TTHH X
TTTH
TTHT
THTT
HTTT
TTTT
That’s 16 combinations. Of those, Tom keeps four:
HHHH
HHHT
HHTH
HHTT
Thus, the expected heads in the last two coins are one, with exactly the distribution we expect.
You can do this with larger number of coins, and you will invariably get the same answer. Very large numbers of coins do not change the answer. (If you’ve got the computing firepower, you can event-test the 50-coin problem.)
Xrlq, at comment 26/ 9:17 p.m. is mistaken (which is rare and surprising on probability). He’s right that it’s not bloody likely that the other 25 coins are tails – it’s 1 in 2 to the 25th against that. The expected total heads of the original 50 is 37.5.
Again, you can event-test this at any number of coins, and you will find that the remaining coins are unaffected – so long as Tom is randomly turning over fair coins, and waits until he gets the 25.
Xrlq, if you believe that this analysis is wrong, or you believe something occurs to the odds at very high numbers of coins, my questions are:
1. At what number of heads using this method does the randomness of the remaining coins end?
2. Why?
If you don’t believe my event-testing is a proper method….. well, I think you’re mistaken. (We can do a full event-test by numbering the coins 1-2-3-4, and then run Tom in picking any combination of coins from each, for a total of 96 events with our four-coin problem, but I promise the results will be the same.)
–JRM
JRM (355c21) — 8/24/2008 @ 12:25 amStashiu and CliveStaples: Thanks a lot, and I appreciate your patience with me. This isn’t about Tom or Joe or the rules.
I’ve postulated a scenario where the same toss, with the same heads revealed, seems to be getting two different sets of odds. Given that the coins are all the same for both tosses, and that the same sets of heads are revealed, the only difference is the state of the person uncovering the coins.
This seems incongruous to me. But then again I don’t know statistics at all – which is self-explanatory from my comments.
Apogee (366e8b) — 8/24/2008 @ 12:26 amSince I’m one of the few who got this completely correct, where’s my prize?
I must say its fun watching regular commentators like nk beat their head against their own ignorance.
That being said, this is not a subject matter that can be understood unless you have taken a college-level statistics course. UCLA extension offers some online. I’d recommend taking a statistics course to anyone capable of handling it, i.e. remembers high school calculus.
Cyrus Sanai (4df861) — 8/24/2008 @ 12:26 amThe only results you can consider are the 25 you are wagering on. Either there is no way to determine how many heads there are (best case and still the odds are against you), or Tom has more information than you do about how many tails might be there… if not complete information as to how the 25 you are betting on are distributed. Make the bet and you will lose.
Apple cider is my favorite, how about you? 😉
Yes, I understand what you are saying, but that’s why it’s called the Gambler’s Fallacy. All the numbers seem to come out in your favor, but the house always ends up ahead.
Stashiu3 (460dc1) — 8/24/2008 @ 12:27 amI’ve taken college-level statistics… it’s a required course for nurses (we have to understand what the medical research means and how reliable the studies are, ya know?). I also understand why so many people here have a negative reaction to you.
Whatever. You’re a condescending tool who can’t resist an undeserved cheap shot. Got it. Never had a date (with a real person anyway) until college or later. Understood. Pick up your Bozo Button at the door and don’t let it hit ya on your way out. Buh bye.
Stashiu3 (460dc1) — 8/24/2008 @ 12:40 amAnd just so you know Cyrus… nk is ten times the human being you ever thought you could be. Nearly everyone here would agree. So just for this special occasion, I have three words for you that regular readers here will recognize…
Right.Out.Loud.
Stashiu3 (460dc1) — 8/24/2008 @ 12:46 amCyrus Sanai:
You shouldn’t need to take college classes to understand this. It’s very simple.
JRM:
If he must only reveal 2 “heads”, regardless of whether they were coins 1 and 2, then these are the possible results:
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
THHT
HTTH
THTH
TTHH
6 out of 11 of the results are favorable. Only 5 out of 11 are unfavorable. And the number goes up at the number of coins flipped increases, since there’s more overlap–in this case, there were many results with at least two heads.
If he only flipped 2 coins, and showed you 1 “heads”, then you’d have a 50% chance of winning. With 4 coins, you have a ~54.5% (6/11) chance of winning. You’re more and more likely to win.
CliveStaples (4c1930) — 8/24/2008 @ 12:47 amStashiu3:
It isn’t the Gambler’s Fallacy. It’s simple math.
Yes, you’re wagering on the outcome of 25 tosses. But if some of the “heads” Tom reveals are from tosses other than the first 25, then the question is “which 25 results out of the fifty tosses are still left?”
If he eliminates the first 25 results, then yes, you’re wagering on 25 results. If he eliminates any 25 heads, then you’re wagering on quite a few different possibilities.
Let’s take a look at a smaller example:
Imagine he tossed four coins, rather than fifty, and showed you two heads, rather than twenty-five.
The complete list of possible results from flipping four coins are thus (winning combinations in bold):
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
THHT
HTTH
THTH
TTHH
HTTT
THTT
TTHT
TTTH
TTTT
If he shows you only the first two toss results, and they’re heads, that narrows down the possibility to these:
HHHH
HHHT
HHTH
HHTT
If he shows you two heads from any of the tosses, not just the first two, then the possibilities are these:
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
THHT
HTTH
THTH
TTHH
If he only shows you the results of the first two tosses, then he’s eliminated quite a few permutations–any that begin TH, HT, or TT. In that case, you’re wagering on only the results of the last two tosses.
If he shows you heads from any toss result, then he’s eliminated far fewer permutations–only those that contain fewer than two “heads”. In that case, you’re wagering on the results of all the tosses, since the first two tosses are not guaranteed to be “heads”.
CliveStaples (4c1930) — 8/24/2008 @ 12:56 amI answered Clive, but used a “p o k e r” reference, along with “T e x a s H o l d ‘ e m” and think that’s probably what caught me up in the filter. Too tired right now to rewrite it, so I’ll check back later.
In short, I understand the math. I also understand the wager. You’re focused on the math.
Good night Sir and thanks for the stimulating discussion.
Stashiu3 (460dc1) — 8/24/2008 @ 1:14 amFrom a probability perspective, it doesn’t matter what order Tom displayed the 25 heads or whether he new they were heads in advance. All that matters is the probability that there were at least 14 tails in the orginal 50-coin distribution. That probability works out to 99.95%. Method from http://mathforum.org/library/drmath/view/52217.html. Useful calculator (N choose R) from: http://condor.wesleyan.edu/jfrugale/wescourses/2004f/math132/01/calculator.html.
I actually took the calculator page and modified it to calculate a table of probabilities using the following function:
function calc()
{
var i = 0;
var j;
var numtosses = 50;
var totalCombinations = Math.pow(2,numtosses);
var aRCombinations = new Array();
var aSum = new Array();
var aProb = new Array();
var outdoc = document.getElementById(“out”);
var htmlcode = “”;
//calculate the probability for exactly R tails
for( i=0; i<=numtosses; i++)
{
aRCombinations[i] = Math.round(nCr(numtosses,i)); //possible ways to get i tails
aProb[i] = aRCombinations[i]/totalCombinations;
}
//add up the probabilities for R or more tails
for( i=0; i<=numtosses; i++)
{
aSum[i] = 0;
for( j=i; j <=numtosses; j++)
aSum[i] += aProb[j];
}
htmlcode = “number of coins: ” + numtosses + “” +
mark (3686d4) — 8/24/2008 @ 5:47 am“total possible combinations (2^” + numtosses + “): ” + totalCombinations + “”;
htmlcode += “”;
htmlcode += “R” + numtosses + ” choose R (number of ways to get exactly R tails)probability of exactly R tails(rounded to 6 decimal places)probability of at least R tails(rounded to 6 decimal places)” ;
for( i=0; i<=numtosses; i++)
{
//htmlcode += “” + i + “” + aRCombinations[i] + “” + aProb[i] + “”;
htmlcode += “” + i + “” + aRCombinations[i] + “” + Math.round(aProb[i]*1000000)/1000000 + “” + Math.round(aSum[i]*1000000)/1000000 + “”;
}
htmlcode += “”;
outdoc.innerHTML = htmlcode;
}
From a probability perspective, it doesn’t matter what order Tom displayed the 25 heads or whether he new they were heads in advance. All that matters is the probability that there were at least 14 tails in the orginal 50-coin distribution. That probability works out to 99.95%. Method from http://mathforum.org/library/drmath/view/52217.html. Useful calculator (N choose R) from: http://condor.wesleyan.edu/jfrugale/wescourses/2004f/math132/01/calculator.html.
I actually took the calculator page and modified it to calculate a table of probabilities using the following function:
function calc()
{
var i = 0;
var j;
var numtosses = 50;
var totalCombinations = Math.pow(2,numtosses);
var aRCombinations = new Array();
var aSum = new Array();
var aProb = new Array();
var outdoc = document.getElementById(“out”);
var htmlcode = “”;
//calculate the probability for exactly R tails
for( i=0; i<=numtosses; i++)
{
aRCombinations[i] = Math.round(nCr(numtosses,i)); //possible ways to get i tails
aProb[i] = aRCombinations[i]/totalCombinations;
}
//add up the probabilities for R or more tails
for( i=0; i<=numtosses; i++)
{
aSum[i] = 0;
for( j=i; j <=numtosses; j++)
aSum[i] += aProb[j];
}
htmlcode = “number of coins: ” + numtosses + “<br>” +
“total possible combinations (2^” + numtosses + “): ” + totalCombinations + “<br>”;
htmlcode += “<table border=\”1\”>”;
htmlcode += “<tr><td>R</td><td>” + numtosses + ” choose R <br>(number of ways to get exactly R tails)</td><td>probability of exactly R tails<br>(rounded to 6 decimal places)</td><td>probability of at least R tails<br>(rounded to 6 decimal places)</td></tr>” ;
for( i=0; i<=numtosses; i++)
{
//htmlcode += “<tr><td>” + i + “</td><td>” + aRCombinations[i] + “</td><td>” + aProb[i] + “</td></tr>”;
htmlcode += “<tr><td>” + i + “</td><td>” + aRCombinations[i] + “</td><td>” + Math.round(aProb[i]*1000000)/1000000 + “</td><td>” + Math.round(aSum[i]*1000000)/1000000 + “</td></tr>”;
}
htmlcode += “<table>”;
outdoc.innerHTML = htmlcode;
}
Feel free to correct any math errors (call me names if it makes you feel good).
This does not speak to the issue of whether Joe knows the remaining number of heads in the distribution before he offers the bet, just to the probablility of there being at least 14 tails in the distribution.
mark (3686d4) — 8/24/2008 @ 5:52 amI took a lot of college level statistics courses. I needed them for my senior year research in Psychology.
If I were studying the coin-flipper as to whether he had a tendency to throw heads or tails, I might agree with your approach. But that is not the question as Patterico posed it.
Why don’t you explain to me how each coin flip is not an independent, random event. In what manner any coin flip, anywhere, anytime, influences any other coin flip, anywhere, anytime. Thumb fatigue is all I can think of but, again, that was not in Patterico’s scenario.
nk (3c7a86) — 8/24/2008 @ 6:25 amThis is a sucker-bet.
1) Joe pays $10, you pay $12
2) You cannot prove Tom does not work for/ get a cut of the winnings from Joe. (every con has a straight man.)
Even if it were 50-50 it would not matter, just the last coin decides if you lose $12 or $14.
SteveofTheNorth (8447d8) — 8/24/2008 @ 7:30 am(25/2=12.5 {12-13 or 13-12}) on 12-13 is $130 you $156 for Joe and 13-12 is $130 you and $144 for Joe.
So,can you afford to lose $14 or just $12 (or $300 if Tom had flipped all coins to heads?)
Oops, mixed that…
on 13-12 you $130 Joe $144 you lose $14 and 12-13 you $120 Joe $156 you lose $36
But still a bad game.
SteveofTheNorth (8447d8) — 8/24/2008 @ 7:36 amThe followup question is not a gamblers fallacy, but a Monty Hall selection. Imagine that the 50 coins are always randomly heads-tails. Now imagine you keep uncovering sets of 25, and then re-throwing the coins, until you get all heads. It will take a long time, but it will eventually happen (you may have to hand over the job to your great-great-granchildren).
When you manage to do that, and you offer the rube the choice, you have selected a situation where there is a high probability of there being many tails in the remaining coins. You have performed a Monty Hall selection, albeit in an extremely tedious way!
sherlock (b4bbcc) — 8/24/2008 @ 8:25 amJRM, here’s the problem with your analysis: Tom didn’t select “his” 25 quarters until after “Fred” (Patterico never gave the coin-tosser a name, so I just did) had pitched them. If he had, then two groups of quarters would be truly separate, and we could treat Tom as though he had selected this game at random (since he had chosen non-random criteria that correlate only with the 25 quarters we don’t care about, and not with the 25 we do. But since he chose from all 50, we have to recognize that a thumb was on the scale. If Fred threw heads on all 50 coins, the game will definitely NOT be voided. If he threw heads fewer than 25 times, the game definitely will be. Between those two extremes, it’s enough to say that the more heads Fred threw, the less likely Tom was to have voided the game. That systematically favors heads among all 50 coins, not just the 25 that have been uncovered.
Xrlq (62cad4) — 8/24/2008 @ 9:49 am69/Xrlq/9:49 a.m.:
OK, that logic should work for 10 coins also, then. (Or any other number of coins.)
Here’s my proposal:
I’ll get a computer simulation of 10 coins written to your standards and agreement. The simulation will pick five coins at random. If all are heads, we’ll bet on the remaining coins. We can run a large number of iterations.
After 5000 iterations (or 25000 coins), I’ll pay you $2 for each head over 13000; I’ll designate a charity to take $2 for each head under that. We’ll cap the payout at $2K.
I’ll warn you: You’re going to lose. Of course there were likely more heads than tails in the full group – 37.5 heads total, on average. But the idea that the others are more likely to be heads is mistaken.
If there are other backers of Xrlq’s analysis, I’m prepared to cover substantial wagers (and I’d note further that in my professional legal opinion such a bet is legal in all US jurisdictions I’m aware of; this is not a chance-based bet, even though it’s coin-flipping.)
If you want more coins per flip, I don’t know what numbers will tax the available computing firepower I have, but I’m prepared to work out any deal around these parameters. The logic holds for any number of coins.
Deal?
–JRM, and I really think you want to apply Bayes’ Theorem in redetermining the probability you are right.
JRM (355c21) — 8/24/2008 @ 11:17 amJRM:
Expecting a regression to the mean, eh?
Xrlq:
I had thought this was a statistics question, not a psychology/’prudence in the real world’ dilemma.
If Tom just chooses 25 coins to reveal from among the coins that landed heads, then you should absolutely play the game. The odds are that the division of tails and heads is ~25 apiece; any heads above 25 are unlikely—and unless there were >36 heads from among the fifty coins tossed, you’ll win money.
If Tom chooses 25 coins to reveal from among the coins that landed heads and iterates the tosses until the lay of the remaining 25 coins is to his satisfaction, you should never play the game.
CliveStaples (4c1930) — 8/24/2008 @ 11:59 amCS: if Tom chose randomly, the odds are that the division of heads and tails among the remaining coins will be even.
JRM: Crap, you’re right. I shouldn’t haev been thinking at the game level. I should have been thinking at the level of the individual coin. As long as Tom selects from them in a way that doesn’t make him more likely to pick heads than tails, voiding every game that doesn’t get him 25 heads by coincidence will still leave you with even odds for the remaining 25.
Xrlq (62cad4) — 8/24/2008 @ 1:25 pmvoiding every game that doesn’t get him 25 heads by coincidence will still leave you with even odds for the remaining 25
Unless Patterico means at least 25 heads without limitation as to how many more heads there may be. The p-value for whether a coin toss is fair is by calculating both the heads and tails. 25 heads and 25 tails would be perfectly random. But if Tom does not care how many heads there are as long as there are more than 25, he is creating a preselected sample. A rigged game.
The fair way would be at least X but not more than Y depending on what we agree is the appropriate p-value.
nk (3c7a86) — 8/24/2008 @ 1:54 pmXrlq:
That is simply not true. Look at the four coin example; if he chooses to reveal 2 heads randomly, you’ll win money 6/11 times by playing. If he chooses to reveal the first two coins as heads, you’ll only win money 1/4 times.
CliveStaples (4c1930) — 8/24/2008 @ 1:54 pmThe question is which set of 25 ‘remaining coins’ are we playing with. There are many different ways to get at least 25 heads from 50 tosses; more than half of them have enough tails to make playing the game profitable.
CliveStaples (4c1930) — 8/24/2008 @ 1:55 pmXrlq, in order for your argument to be true, you’d have to have ~12.5 heads (half of the remaining 25 coins) on top of the 25 heads that were revealed. So what you’re really claiming is that the most probably outcome of 50 coin tosses will contain ~37.5 heads. Do you really think there’s a >50% chance of getting “heads” from a coin flip?
CliveStaples (4c1930) — 8/24/2008 @ 2:00 pmnk: Probability theory is an attempt to turn ignorance into a science. Nothing more.
Measuring uncertainty is the chief and foremost goal of all real science.
Daryl Herbert (4ecd4c) — 8/24/2008 @ 2:35 pmCS, what I’m claiming, and what you appear to be missing, is that the 50-50 odds only tell you what to expect from future rolls. You seem to be falling for the gambler’s fallacy, where one expects the original odds to still control even after an intervening event has changed them. Of course the most probable outcome of 50 tosses wasn’t 37.5 heads at the start, but then again, getting heads on the first 25 wasn’t all that probable, either. Now that it happened, the most probable outcome for the remaining 25 is exactly the same as what the most probable outcome was for all 50 before you started: a 50/50 split.
Xrlq (62cad4) — 8/24/2008 @ 2:44 pmNo, the chief and foremost goal of all science is to achieve a verifiable, reproducible result.
nk (3c7a86) — 8/24/2008 @ 2:46 pmXrlq:
I’m not going to insult you by suggesting that you’re lacking in perception or understanding.
The “Gambler’s Fallacy” would be if Tom said, “I’ll flip this coin, and every time it comes up tails I’ll give you $10, and every time it comes up heads you’ll give me $12”, and after a run of heads, I expect a tail.
What we’re talking about here is the lay of 50 coins. The most probably outcome has a 25/25 split between heads and tails; less probably, a 24/26; less probably, a 23/27, etc., etc.
There’s only one combination of tosses that results in a 50/0 heads/tails result; there’s many combinations that result in a 49/1 heads/tails result; even more combinations that result in a 48/2 heads/tails results.
It’s Pascal’s Triangle. While the odds of getting heads or tails is even on each coin toss, we can look at the probabilistic results.
Ever played D&D? If you roll 2d6, you’re more likely to get a result of ~7, since there’s many more combinations of 1-6 that result in 7, even though each particular combination that results in 7 has the same likelihood of occurring.
Look at the results of two coin tosses:
HH
HT
TH
TT
You’ve got a 75% chance of rolling at least one “tails” with two coins…even though the chance of rolling it on a particular toss is only 50%.
CliveStaples (4c1930) — 8/24/2008 @ 3:00 pmHere’s my take.
If Tom selects 25 coins to turn over in a fashion that is independent of the results of the original toss, then the fact that all 25 coins are heads has no effect on the remaining 25 coins. In that case, Tom’s wager is favorable to Tom and unfavorable to Joe.
If Tom knows whether the coins are heads or tails and has selected 25 heads out of the toss, then the bet Patterico is described is a sucker bet — for Tom.
This is because given that there are 25 or more heads out of the 50 tosses, the expected total number of heads is about 28. You can work this out the hard way, as mark does in #65, or you can be lazy like me and use the normal approximation.
A much more interesting version of the bet goes as follows: Joe tosses 25 fair coins. Tom, who gets to see which tosses are heads and which are tails turns over 25 coins. All are heads. Tom then offers to pay Joe $6 for every remaining head if Joe will pay him $1 for each tail. (If there are more tails then heads, turn over 25 tails and make the corresponding bet with heads and tails reversed.)
If Tom selects the coins he turns over at random and repeats the coin-flip-and-reveal until the reveal is all heads, then the reveal mechanism is still independent of the result of the tosses, so the number of heads in the remaining 25 tosses is still independent of the results of the revealed tosses.
W. Krebs (7c80ac) — 8/24/2008 @ 3:05 pmW. Krebs:
It isn’t about whether the uncovered coins have an “effect” on the remaining coins. It’s about what combinations are still possible given 25 heads. If the coins are randomly selected, and all happen to be heads, then most likely there are around 25 tails left; less probably, 24 tails and 1 heads; less probably, 23 tails and 2 heads…etc., etc., until the least probable outcome of 0 tails and 25 heads.
It’s the least probably because all 50 coins would have to be heads for the remaining 25 to be heads; if there were only, say, 26 heads and 24 tails, then 25 of the heads would be revealed, leaving 24 tails and 1 heads.
There are many more combinations of results that include 25 heads with 25 tails than 28 heads with 22 tails.
And if you look at #65, he shows that there’s a 99.95% chance that there would be at least 14 tails, which leaves 11 heads (36 total heads minus the 25 revealed). That’s a 99.95% chance of winning money by taking the bet.
CliveStaples (4c1930) — 8/24/2008 @ 3:13 pmGood, because I don’t. I will, however, gladly insult you by suggesting the same in return, because it’s clear that you do.
Potato, potahto. It doesn’t matter whether the coins have yet to be tossed, or if they have been tossed already but neither you nor Tom know how they landed. The probabilities are exactly the same: all known coin values, plus a 50-50 split among the unknowns. When no coins had been counted, your expectation of 25 heads and 25 tails was reasonable. But after the very first coin was uncovered, it was not; from there you should have expected 24.5 heads and 24.5 tails for the remainder of the game. And after 25 heads in a row, an exceedingly improbable event, you would have to be certifiable to expect an equally improbable string of 25 tails to follow. If ever there was a textbook example of the gambler’s fallacy, this was it.
Note that the opposite is not quite true. After an incredibly improbable run of 25 heads, you would indeed be justified in thinking that the game was somehow rigged, either in favor of heads (all coins have heads only, in which case you should bet on heads for the remainder of the series as well) or in favor of tails (i.e., you had the Monty Hall problem, where Tom saw all the coin values first, and hand-picked 25 he knew to be heads). In that case, you’re betting that Tom is a liar, a bet you would almost certainly win. But those facts are outside the stipulated facts of this game, so you can’t consider them here.
Neato. But again, that 75% probability disappears as soon as you find out how the first coin landed. After that, it’s either 100% (because you just got tails) or 50% (because you just got heads, but the other coin still has a 50-50 chance of getting tails). The only way you can keep that 75% probability is the Monty Hall scenario, where you have reason to believe Tom uncovered the first coin because that coin was heads (or because the other was tails).
Xrlq (62cad4) — 8/24/2008 @ 4:28 pmOK, I think I see where you’re stumbling. You seem to be assuming that only one type of statistical miracle could have occurred here, namely that Fred threw the expected 25-25 split (or close) and Tom just “happened” to land on heads 25 times in a row. But there’s no reason to assume that Tom is any more capable of working miracles than Fred is. Maybe the real miracle was that Fred actually pitched heads 50 times, and Tom’s picking heads 25 times among them was to be expected. Or maybe the real miracle was a combination of two lesser miracles, with Fred pitching less than a perfect game, but still an improbably high percentage of heads, while Tom still had to overcome some odds, but not such astronomical ones, to bat 1.000 among them.
In other words, every coin shown to be heads could either be evidence that you’re one step closer to your original, expected target of 25, or it could be evidence that your original estimate of 25 was off by 1. Until you see all 50, you have no way of knowing which. That is why you must treat an uncovered coin (assuming that you trust Tom hasn’t peeked at it, either) exactly the way you would treat a coin that has yet to be flipped at all.
Xrlq (62cad4) — 8/24/2008 @ 4:35 pmExactly. Individual coins have no memory of what other coins have done. That’s why I said it doesn’t matter what the first 25 coins were… you’re only wagering on the last 25 and they are a 50/50 shot if everyone is legit. A 6/5 bet on 50/50 odds is a sucker bet, and that’s the best case scenario.
Clive, if you use your reasoning to make real wagers, you will quickly go broke in a very logical manner.
Note: I have a comment in the filter using the words “p o k e r” and “T e x a s H o l d ‘ e m” which is a fair illustration.
Stashiu3 (460dc1) — 8/24/2008 @ 4:38 pm[That’s odd. I found a few such comments with your handle in the spam queue, and marked them as not spam. For some reason, they still don’t appear to be displaying. -X]
Xrlq:
Do you understand binomial coefficients?
The chance of getting 14 tails out of 50 tosses–at which point you make money by taking the bet–is 99.95%. It’s 1-(nCr(50,0)+nCr(50,1)+nCr(50,1)…+nCr(50,13)).
nCr(50,0) is the odds of getting zero tails out of 50 tosses. It’s 1/(2^50). nCr(50,1) is the odds of getting 1 tails out of 50 tosses. It’s 50/2^50.
Xrlq:
Right, but we don’t know that the 25 heads shown were the first 25 tosses. It could be any combination of 25 heads, which could be any of hundreds of trillions of combinations.
You’re missing the point. If Tom flips two coins, and reveals one heads, then there’s one of three possibilities:
HH
HT
TH
There’s a 66% chance that the other coin is “tails”, not 50%. Because Tom doesn’t have to show you the first coin; the “heads” he shows you might be the second coin.
CliveStaples (4c1930) — 8/24/2008 @ 4:50 pmstashiu3:
I’d win money 99.95% of the time, assuming that Tom isn’t fixing the game. The odds of getting at least 14 tails out of 50 coin tosses is ~99.95%.
The question is how many variables we’re working with. If you say “The first 25 results are heads”, then of course we’re only playing with the results of the next 25 tosses. The odds of getting 14 tails out of 25 tosses is quite a bit lower–about 34.5%.
Whereas if Tom is simply choosing 25 heads from among the 50 tosses, that means any of the first 25 tosses could possibly be tails.
To use the two-coin example:
If Tom shows you the result of the first coin, and its heads, then these are the possibilities:
HH
HT
Obviously a 50% chance.
If, however, Tom is showing you a “heads” from ANY of the tosses, rather than only the first, these are the possibilities:
HH
HT
TH
That’s a 66% chance of the ‘other’ coin being tails. There’s a HUGE difference between the chance of getting at least 1 “tails” out of two flips, and the chance of the second coin being tails.
CliveStaples (4c1930) — 8/24/2008 @ 5:00 pmClive, you’re wrong. You’re succumbing to the Gambler’s Fallacy for sure. Read that link please. Make sure you look at Bayes’ Theorem at the same time. Each “reveal” is a 50/50 chance no matter what has already happened. So the sum odds of all unrevealed coins is 50/50. You cannot treat it as a 50 coin problem when you’re only wagering on 25 coins. Your wager only makes sense if you say, “ok, out of these 50 coins we’re going to take out 25 heads and then bet on the rest.” But that’s not what you’re doing and treating it the same is what the Gambler’s Fallacy is about.
Stashiu3 (460dc1) — 8/24/2008 @ 5:03 pmYou must love cider. 😉
Don’t go to Vegas, m’kay?
Stashiu3 (460dc1) — 8/24/2008 @ 5:06 pmStashiu3:
…uh, isn’t that exactly what I’ve been talking about? I’ve answered both questions:
1. Tom is revealing the results of the first 25 tosses, which were all heads, in which case you need 14 tails from the last 25 tosses (only a 34.5% chance), which is a gamble you’re most likely to lose.
2. Tom flips 50 coins and chooses to reveal 25 of the coins that landed as heads. In which case you have a 99.95% chance to win.
CliveStaples (4c1930) — 8/24/2008 @ 5:09 pmStashiu3, are you familiar at all with binomial coefficients? Remember Pascal’s Triangle?
CliveStaples (4c1930) — 8/24/2008 @ 5:14 pmSpammy hates me here lately. DRJ copy & pasted the earlier ones when the same thing happened. I’m not getting through to Clive though, so I don’t know than putting it in any other terms is going to make it clearer. He’s looking at the math and not the wager, a very organized way to go broke.
Stashiu3 (460dc1) — 8/24/2008 @ 5:17 pmstashiu3:
The question was posed as a statistics problem, and I answered it as such. It’s like saying that I shouldn’t say “Who’s there” to a knock-knock joke because the knocker could be a terrorist who just wants to know if anyone is home to kill.
And I already accounted for your concerns about five posts ago; if Tom’s fixing the game, you shouldn’t play regardless of the theoretical statistical likelihood of success.
CliveStaples (4c1930) — 8/24/2008 @ 5:22 pmClive, I’m not arguing your math… it’s correct for a 50 coin problem. That’s not what this is however. I’m failing to communicate that somehow, so maybe someone else can give it a shot after this.
Look, if you toss 1,000 coins and turn over 999 of them to find they’re all heads, the odds of the last coin being heads is 50/50. That coin has no memory of what the other coins were revealed to be.
Bottom line is… toss as many coins as you like, reveal (or even discard) as many out of those as you like, and the odds for the remaining unrevealed coins is 50/50. Wagering 6/5 on 50/50 odds is a sucker bet.
Stashiu3 (460dc1) — 8/24/2008 @ 5:24 pmThe 6/5 wager on 50/50 odds is when everything is legit. If Tom’s fixing the game in any way, the odds are even worse. I can’t say it any clearer.
Stashiu3 (460dc1) — 8/24/2008 @ 5:27 pmIn other words, you’re not betting that there will be at least one tails out of the 1,000 coins before they’re tossed, but that’s how you’re treating the problem. You have to disregard anything that has been revealed, or it’s the Gambler’s Fallacy.
Stashiu3 (460dc1) — 8/24/2008 @ 5:31 pmstashiu3:
Except that you aren’t just wagering on how many of the last 25 coins are tails; you’re wagering on how many sets of results from 50 tosses that contain 25 heads also contain at least 14 tails. The answer is 99.95% of them.
The odds of any particular coin coming up tails is always 50/50. The odds of a particular set of tosses containing tails isn’t always 50/50. The odds of flipping “tails” is 1/2 with one coin. With two coins, the odds of getting at least 1 tails is 75%. With three coins, the odds of getting at least 1 tails is 87.5%.
Take a small example. Three coins. You take out 2 heads, and i’ll get $10 for every tail left and you’ll get $12 for every head left.
Here are the possible results:
HHH
HHT
HTH
THH
I’ll win 75% of the time. You can only win if every coin is heads. Similarly, Tom can only win if at least 37 coins are heads.
CliveStaples (4c1930) — 8/24/2008 @ 5:36 pmstashiu3:
No, because you don’t know which set of 25 heads has been revealed; it isn’t necessarily the first 25 tosses. If every even toss was tails, and every odd toss was heads, Tom would still remove 25 heads and there would be 25 tails left (a profit of $250 to me).
CliveStaples (4c1930) — 8/24/2008 @ 5:38 pmYes, it is always 50/50. I thought you answered kind of quick after my Gambler’s Fallacy link, but this shows you didn’t read it. The odds for all unrevealed coins are 50/50.
Stashiu3 (460dc1) — 8/24/2008 @ 5:40 pmFrom the link:
I’m done… take care.
Stashiu3 (460dc1) — 8/24/2008 @ 5:41 pmStashiu3, you’re missing a huge part of the problem: Tom’s removal of “heads”. Each coin has only a 50% chance of being tails, but each time a coin comes up heads, Tom removes it from the pool until he’s removed 25 heads. This EXTREMELY favors “tails” results to be left.
CliveStaples (4c1930) — 8/24/2008 @ 5:44 pmOne last try, lol.
If we toss 1,000 coins and turn over 999 heads, will you wager $2,000 of your money against $20 of mine that the last coin is a tails? What do you figure the odds at?
Stashiu3 (460dc1) — 8/24/2008 @ 5:51 pmstashiu3:
Look at your question. You really think that there’s a 50% chance for 1000 coin flips to come up heads?
You don’t think that there’s a good chance that out of 1000 coin tosses, that 1 of them is tails?
Out of all the possible results with 999 heads, there are 1000 ways for the last unturned coin to be tails. There’s only 1 way for the last unturned coin to be heads.
Look at a smaller example (which I assume you don’t like doing, since it proves you wrong rather easily). Flip two coins, and remove 1 heads. 66% of the time, the remaining coin will be tails. Proof:
HH
HT
TH
Those are the only possibilities with at least one head. If you only turn over the first coin, and it’s heads, then you’ve eliminated “TH”, reducing the chance of the other coin being tails to 50% (obviously). But the “heads” that you turn over might be the second coin from the “TH” result.
CliveStaples (4c1930) — 8/24/2008 @ 6:00 pmI’ll try to explain the difference.
You flip a coin. The result is heads. I hand you $12. I would exhibit the Gambler’s Fallacy if I then thought, “Well, the next flip is more likely to be tails!” That is, if I believed the odds of the second flip being tails was greater than 50%.
Now, imagine that you had flipped two coins. The odds of the first coin being tails is 50%; the odds of the second coin being tails is 50%. But the odds of neither coin being tails is only 25%. So the odds of having ONE of the coins come up tails–not just the second coin, and not just the first coin, but EITHER–is 75%.
CliveStaples (4c1930) — 8/24/2008 @ 6:05 pmNo, it proves you are using faulty logic Clive. Do you make the bet? Using your reasoning, the odds are astronomically in your favor. Yet, the odds of that last coin being heads is 50/50… read the link. That’s such an extreme example I can’t believe you don’t see it yet. Now, I really give up. Maybe someone else will have more luck.
Be well.
Stashiu3 (460dc1) — 8/24/2008 @ 6:05 pmStashiu3:
It’s because your example is unclear. If we’re only betting on whether the 1000th coin is heads or tails, then it’s a 50/50 shot.
If we’re betting on the likelihood that 1000 tosses will yield 999 heads and 1 tails, then it’s practically guaranteed to be tails.
CliveStaples (4c1930) — 8/24/2008 @ 8:27 pmRather, if we’re betting on whether a result set with 999 heads will also contain 1 tails, it’s practically guaranteed to be true. There’s 1000 result sets with 999 heads that contain 1 tail, and only 1 result set with 999 heads that contains 1 head. But that “1 tail” isn’t necessarily the last one; that’s only 1 out of 1000 possibilities that would still result in 1 tail.
CliveStaples (4c1930) — 8/24/2008 @ 8:28 pmFinally!! A breakthrough!
My example was perfectly clear… we’re betting on the last unrevealed coin. All the other coins have been revealed as heads. So now you agree that it’s a 50/50 chance in that case.
Same thing applies to the original problem. You’re only betting on the unrevealed coins. What has been revealed has no bearing on what the unrevealed coins are, or the odds attached to them. The odds for all unrevealed coins are 50/50, individually and as a group. If you take the revealed coins into consideration, whether they were revealed in a completely legitimate manner or not, you’re falling for a sucker bet because even the best case has you betting 6/5 on 50/50 odds.
Taking revealed coins into consideration at any time is the Gambler’s Fallacy because they have no bearing on the odds for the unrevealed coins. Those results are already known and you’re only betting on the unrevealed coins. If you are betting before any of the coins are revealed it would be 50/50. When you bet after coins are revealed, those revealed coins are not included in the bet.
If we were to toss 50 coins and 25 heads were revealed to start, then I were to bet that there were at least 26 heads total, you would say “of course, that’s only one more head.” At what point does it no longer become “of course, that’s only ___ more heads.”? Around 12 or 13, because thats the 50/50 mark where the odds truly lay. And even then it’s a sucker bet unless you’re betting even money, not 6/5. Get it now?
Stashiu3 (460dc1) — 8/24/2008 @ 9:33 pmStashiu3:
Apparently you don’t understand. Let’s use a smaller sample for simplicity’s sake:
5 coins. You reveal 4 heads, and we’ll bet on the remaining coin.
The possibilities that include 4 heads among the 5 coins are:
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
HHHHH
Notice that, regardless of the fact that the 1st, 2nd, 3rd, 4th, and 5th coins each individually have a 50% chance of being tails, the odds that at least one of the five is tails is actually quite high.
If you bet heads, and I bet tails, I’d win 5/6 times.
CliveStaples (4c1930) — 8/24/2008 @ 9:58 pmGambler’s Fallacy defined. I thought you were getting it when you agreed that if we’re only betting on the last coin, it’s 50/50. Now you’re contradicting that. If we flip the same coin four times and it comes up heads each time, the chances on the next flip are still 50/50. Read the link again if you don’t believe me. Just because you’re tossing them all the same time doesn’t change those odds.
Oh well, I tried.
Stashiu3 (460dc1) — 8/24/2008 @ 10:18 pmClive and Stashiu3:
OK, guys, let me see if I can help.
Clive is saying that if Tom takes out heads deliberately, it changes the odds, right? Clive, do you agree that if Tom reveals four out of five coins *at random* and the four revealed are heads that the last one is 50% to be heads?
Stashiu3 is saying that, and he’s right. If Clive is saying (unclearly, I think) that selected heads bias the odds…. well, they do, and Clive’s right, also.
We all agree. Right?
–JRM
JRM (355c21) — 8/24/2008 @ 10:23 pmStashiu3:
You keep using that word; I do not think you know what it means.
I just proved you wrong. You can ignore it all you like, I suppose.
Each coin individually has only a 50% chance of being tails. But if you throw 1000 coins, you’re almost guaranteed to get at least one tails. The 1000th coin isn’t any more likely the be tails, nor is the 999th, nor the 998th, etc. But as a set, the most likely combinations will include at least 1 tails. The only way you won’t get at least 1 tails is if 1000/1000 of the tosses are heads–which is almost impossible.
This isn’t the gambler’s fallacy. I’m not saying that tails becomes more likely after a run of heads; I’m saying throwing more coins means more chances at getting tails. If that’s what you think the Gambler’s Fallacy is, then I guess wikipedia has failed you.
CliveStaples (4c1930) — 8/24/2008 @ 10:31 pmJRM:
I’m saying that the 5th coin that was tossed was 50% likely to be tails. However, so do the 1st, 2nd, 3rd, and 4th. The likelihood that every single toss came up heads is much lower than the likelihood that at least one of the tosses came up tails.
CliveStaples (4c1930) — 8/24/2008 @ 10:35 pmFrom the link:
How is it you proved me wrong again? This is exactly what we just talked about.
Stashiu3 (460dc1) — 8/24/2008 @ 10:49 pmIt makes no difference if you flip one coin five times or flip 5 coins at once.
Stashiu3 (460dc1) — 8/24/2008 @ 10:51 pmThe problem is that Clive and Stashiu3 are talking about two different circumstances. Clive is speaking (conceptually) of the coins thrown *at once*, rather than sequential tosses. Of course, the same result could be obtained by a neutral 3rd party throwing consecutively, recording the result and providing only the pertinent information.
Stashiu3 is speaking of consecutive/sequential coin tosses.
Let me repeat that. Clive’s illustrations are all about when you do not know the sequence, merely the number of heads. Stashiu3’s illustrations are when you do know the sequence, collapsing the probabilities right back down to the 50/50 of the next coin toss.
They are two separate issues, because the amount of information provided is different. With clive, the *only* data you are given are the no of throws, and the no of heads. With Stashiu3, you are given the exact sequence of events.
If you had a look at comment 111 and 112, you will see exactly what I’m talking about. In comment 111, Clive specifically states you only know the number of throws and the number of heads; hence, *all* the *potential* sequences are listed. He’s right; 5 out of 6 possible sequences, he’ll win.
In comment 112, Stashiu3 specifically states the *sequence* of throws *so far* as known (i.e. H-H-H-H-?). If you looked up the list of potential sequences as listed in comment 111, you will notice that there are only *two* sequences starting that way; one ends H, the other ends T. In that instance,
and Clive implicitly agrees.
Gregory (f7735e) — 8/24/2008 @ 11:44 pmNo Gregory, they are not separate issues because no toss of a coin (or any number of coins) can affect the odds of the other coins. It doesn’t matter who throws it, what the sequence is, or how many coins are revealed prior to the wager. The coins have no memory or awareness and the odds are 50/50.
Clive is dead wrong and a victim of the Gambler’s Fallacy. He is being sucked into it by the appearance of previous results which have no effect on future or separate results. (Anyone who understands this can jump in here any time.)
Again, from the link (since some people apparently can’t be bothered to read the thing):
And it doesn’t matter if you’re talking about a sequence or simultaneous toss. Previous results of any type do not affect the odds or the outcome of any other toss.
Stashiu3 (460dc1) — 8/25/2008 @ 12:16 amLet me reemphasize this:
But we’re not. Clive is trying to apply this logic to a subset after the tosses are carried out. That’s Gambler’s Fallacy.
Stashiu3 (460dc1) — 8/25/2008 @ 12:23 amSorry JRM, I missed your comment earlier.
But Clive is saying that they bias the odds in favor of taking the bet. If anything, they do the opposite because the only bias is Tom having more information. If 25 heads are intentionally selected out (assuming there are at least 25 heads), it doesn’t change the odds on the remaining 25. See here for why Clive is wrong.
This can’t be that hard to understand.
Nope. We don’t.
Stashiu3 (460dc1) — 8/25/2008 @ 12:35 amNever mind, Stashiu3. Don’t you know that corrupt judges in the Ninth Circuit are engaged in an ongoing conspiracy to fix the results of all coin tosses?
nk (3c7a86) — 8/25/2008 @ 1:05 amHmmm… after some thought, I see that I’m quite wrong and the 50-50 folk are right regarding the bonus problem.
Another way to look at the followup is to toss the 50 coins, then line them up #1 through 50. Tom then uncovers #1 through 25. This “ordering” is exactly the same as picking 25 at random.
Looked at this way, the fact that Tom (finally) manages to get his 1 in 33 million draw of “25 heads” to come out implies NOTHING about coins #26-50, as their state has had no effect on anything as yet.
For all intents and purposes the remaining 25 coins are still a random toss of 25, and in fact could be thought of as 25 Schroedinger cats, each half heads and half tails.
So, for the followup bonus problem: the odds are indeed 50-50 until someone peeks, and none of the aborted tries to produce a 25-heads initial turn have anything to do with anything.
Kevin Murphy (0b2493) — 8/25/2008 @ 1:25 amDear Stashiu3;
That’s why I said you and clive are talking past each other’s heads. Or at least, you seem to be bloody-mindedly doing so – it seems to be that clive is more willing to agree where agreement is necessary.
Here’s the thing.
Question: What is the probability of the last coin being a tail?
This is clive’s setup.
– You are given ONE piece of information; There were four Hs.
This is your setup.
– You are given ONE piece of information; The previous four coin tosses came out H, i.e. H-H-H-H.
You’re both speaking about TWO different probabilities here, okay? In clive’s setup, you DO NOT KNOW the sequence; hence, the possible setups are as he described, i.e.
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
HHHHH
Therefore, of all the ways the coin tosses could have come up, these are the only 6 ways in which you can get 5 Heads. This is out of the 2^5=32 possible permutations of coin tosses.
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
This is a classic combinatorial setup, for crying out loud. In this instance, 5 out of 6 times, there WILL be a T, because 5/6 permutations will include a Tail.
In your setup, this collapses to the last two permutations, because in your setup, the first four tosses are KNOWN to be H-H-H-H. Hence, the 50% chance of either H or T. By the way, probability theory comes out of combinatorials, so you should have no excuse for not knowing what clive was talking about.
The question is, which mathematical formula do you use? Strangely enough, it depends on how much information you have.
Which, by the way, is what the original question was all about. Your *knowing* whether the coin tosses were revealed *in the sequence* they were thrown or whether they were revealed *randomly* or whether they were revealed according to some other algorithm changes the probabilities *you* work with.
You are speaking of TWO different things here. DIFFERENT. Oi oi, I don’t know how to explain it any clearer than that.
Gregory (f7735e) — 8/25/2008 @ 1:33 amGood gravy… did I become a citizen of Bizzaro World? Seriously, what is so hard to understand about this after I’ve explained it so many different ways? It’s like “Dungeons and Dragons Universe Gone Wild” where magical coins can determine what other coins are going to do. I’ve tried to be patient, but it’s frustrating to be told I’m proven wrong (twice!) when I gave a link proving what I had just said and the link obviously wasn’t even read.Then, two helpful souls jump in trying to say we’re both right… pffft! That just means three out of four of us are wrong because Clive and I are saying the exact opposite thing and can’t both be right. I even linked to where it stated Clive’s exact point was an example of Gambler’s Fallacy (the 5 coin with 4 heads first example) and am told it doesn’t say what I just showed it said.
It does not matter in either case Gregory. What part of:
is so hard to understand? It doesn’t matter how the information came to us… that information cannot change the odds of the remaining coin. I understand Clive’s point… it’s wrong. It’s the Gambler’s Fallacy and apparently you don’t understand it either. The sequence doesn’t matter and a combinatorial setup does not apply.
Stashiu3 (460dc1) — 8/25/2008 @ 1:45 amNo, Gregory. What the *possible* permutations are before the coins are tossed becomes meaningless once the coins are tossed. There is absolutely no reason any coin has to adhere to the permutations. Whether a thousand coins are tossed all at once or whether they are tossed sequentially, and 999 are tails, in either case there is nothing that compels the 1000th coin to be heads. That’s classic Gambler’s Fallacy.
Like I’ve said before, it would be a different story if what you were looking at was whether this was a fair toss. That the coin was not loaded or that the tosser was not manipulating it. In that case, you would set a range of so many heads and so many tails. That’s the p-value. That’s acceptable in statistical theory so I’ll accept it too. But that’s not the scenario here. Here, we assume that the coin tosses are fair.
And to carry it further to Patterico’s bonus question, unless you set a range of at least 25 but no more than 25+X heads, then Tom is rigging the game by finding those fifty tosses of all coin tosses made in the history of the world which favor him. Those tosses are still independent events in and of themselves. The rigging is in Tom’s cherry-picking.
nk (3c7a86) — 8/25/2008 @ 2:03 amKevin Murphy #123:
Glad somebody else mentioned that damn cat so I didn’t have to.
(Although I have to credit the cat with the “A” on the paper that applies to this very problem.)
Clive (and Gregory) are making the mistake of viewing this as a problem of permutations~which it isn’t. The coins really don’t care how they are supposed to behave: in a “fair” toss (which takes a special test rig to do, btw) each coin has a 1 in 2 chance of landing with either face up, until we have looked at it! IOW, it is not the Monty Hall problem, because these are independent probabilities, not dependent as in the Monty Hall problem.
So, you can string “H’s” and “T’s” together in whatever fashion you like, all week long, and we still don’t know what that last coin is, until we look at it.
And that is an even money bet.
Of course, with the one single bit of sleight of hand I know (I’m too much of a klutz to learn any that require any actual skill), if you allow me to flip the coin, it’ll piss in your ear.
EW1(SG) (84e813) — 8/25/2008 @ 3:11 amDear Stashiu3 and nk;
Yes. I agree. When you flip the coin, assuming standard assumptions (fair coin, fair toss, ceteris paribus and all that), it’s ALWAYS a 50-50 chance. Please, don’t get me wrong. I AGREE. And so does Clive.
For longer explanation, look below. But the shorter explanation, here.
What do probabilities measure? Well…
http://www.discover6sigma.org/post/2005/09/basic-probability-theory/
And if that isn’t bloody permutations and combinations, what the hell is it!
Probability theory comes in when (a) You haven’t done the tossing or experiment or whatever and want to know what are the chances of various events happening or (b) You have done the tossing or experiment or whatever and want to estimate the likelihood of a specific outcome based on information you are given from someone who knows the outcome.
Let’s say random reveal of the 5 coin tosses, and four were revealed to be Heads. Well…
The coin tosses could be THHHH, but the random reveal was (2), (3), (4), (5). Or, (3), (2), (4), (5). Or, …
The coin tosses could be HTHHH, but the random reveal was (1), (3), (4), (5). Or, (3), (5), (1), (4). Or, …
And so on and so forth. Can you still say with a straight face that it’s a 50-50 chance? Good heavens.
Not knowing the sequence of reveals (and hence, coin tosses) makes this a combinations/permutations exercise. Even the collapse of the probabilities to a single coin toss is a permutations exercise. *Any* probability theory usage is a permutations/combinations exercise.
—
Longer explanation below…
—
However, in reality, ONCE the coins are tossed, all the probabilities collapse to one single outcome. ONE. Will you agree? Let’s say the last coin came out Heads. The last coin is 100% Heads and 0% Tails – this is not some kind of quirky quantum state, it’s settled. But only the person/people who observed knows this. The rest of us don’t. And it is precisely because we do not know the outcome that we apply probability theory to this. And we use what we know to apply the most appropriate theory.
Again, IF you know that the tosses were H-H-H-H, then obviously the last one can be EITHER H or T, and there’s a 50-50 chance. But what if you DIDN’T know that?
What do the permutations measure? They measure all the possible states of coin toss sequences; i.e. if you toss a coin 5 times, or toss 5 coins at once, there are only 32 possible ways it can turn out.
How many of those ways have at least four Heads in them? Well, 6 out of those 32 ways, as Clive states.
Now, here’s where the information you get changes your estimation of the probabilities. Why estimation? Because once the coins have been thrown, only ONE permutation is correct, and you are estimating the probability of that permutation.
Information known about the throws:
1. zilch.
Then you can guess any of 32 possible ways, and your odds of getting it right are 1/32.
2. There are at least four Heads.
Then you can guess one of 6 ways out of those 32. And further, out of those 6, 5 of them contain a Tail toss. Hence, your chances of getting the ‘final’ Tail is 5/6.
Remember: you are not told which of the tosses gave you the Heads, only that they are Heads.
3. The first four tosses were Heads.
Then you can guess one of two ways, and it goes back to your standard coin toss, because the tosses that went before are immaterial.
—
What Clive is saying, pretty much, is that in the absence of additional information, just knowing that there are four Heads in the 5 tosses narrows down the 32 possible states to 6 possible states (again, only ONE is correct). And in 5 of those 6 potential possible states, a Tail is tossed.
Hence, out of the 6 ways the tosses could have worked out such that you get at least 4 Heads, 5 of them contain a Tail. You are far more likely to have the last coin a Tail than a Head for that one toss sequence.
Go back to the theory behind measuring probabilities. They came about because there was a finite set of permutations to test for, and instead of just counting each one as it happened, work out the mathematics to calculate it.
Now, this is what happens in the first (original) question. You need to ask the questions so that it gives you more information so as to cut down the potential solution space.
The follow-up question is similar, except now, you have a vastly potential smaller solution space, as you cut out all permutations containing less than 25 heads.
And here is where you two commit your fallacy. The probability of a single fair coin toss is always 50-50. But the probabilities involved in a *sequence* of coin tosses differs depending on what you’re looking for. A sequence of all Heads in 5 tosses, for instance, is only 1/32 – which means that the probability that there is at least ONE Tail is 31/32, and you are on fairly good grounds of betting on there being at least one T. If you knew that there were at least four Heads, then the probability of ONE coin toss being Tails reduces to 5/6. But it is still 5/6.
What you’re doing, Stashiu3, is collapsing all of the other probabilities, and narrowing the focus down to the *last* coin toss. Which, obviously, you can do only once you have enough information to be able to collapse the other probabilities. Which, in turn, requires you to settle the order in which the tosses were made. And based on the information given in the question, you cannot do that.
Gregory (f7735e) — 8/25/2008 @ 4:01 amGregory, as Kevin notes above: it doesn’t matter which coin toss you examine; first, last, always.
The math is thus: the probability of the coin landing on one side or the other is 1/2, or 0.5. For the sake of argument, we’ll assume that the probability of the coin landing on a face is 1, and we’ll include this term because it makes the problem easier to see. So far, we have the probability of the coin landing on face A = (.5 x 1), plus the probability of the coin landing on face B = (.5 x 1), which adds up to unity~the probability of the coin landing. So far, so good.
Now, from your link, since these are mutually exclusive events, the rule is P(A or B), or P(A) + P(B). We can also add the rule for independent probabilities…which makes our math look like: (P(A)+ P(B)) x (P(A) + P(B)) …. and so on, simplifying to 1 x 1 x 1 …. where each term describes a single coin.
So far, what we know from examining any individual coin toss, is that the coin will most likely land. And that is all we know!
EW1(SG) (84e813) — 8/25/2008 @ 4:35 amstashiu3:
Really? You don’t think there’s any greater odds of getting 1 tail in 5 tosses than in getting 1 tail in 1 toss? You honestly believe that?
CliveStaples (4c1930) — 8/25/2008 @ 5:41 amIf you flip one coin five times, you’re sequentially eliminating combinations. For instance, if you flip a coin four times and get HHHH, the only possible combinations left for the 5-toss sequence are HHHHT and HHHHH–a 50/50 chance, obviously, since you can only get a “tails” from the final toss.
Whereas if you toss five coins at the same time, any number of combinations are possible with 4 heads. The first coin could be tails; the second coin could be tails; the third coin could be tails; the fourth coin could be tails; or the fifth coin could be tails. Obviously, you’re more likely to get “tails” if you’re considering 5 tosses than if you’re only considering one.
The question is how many variables are at play. When you say that the first four tosses are heads, you’ve eliminated the first four tosses as variables. When you say that the results of the five coins include four heads, there are many more variables at play, since any of the tosses could still possibly have yielded a tail.
I don’t know how to explain it any clearer. The fifth and final toss has only a 50% chance of being tails; but if all you know is that five tosses yielded four heads, then it’s more likely that at least one of the five tosses landed tails than all five tosses landing heads.
We aren’t just talking about whether the last toss was tails, or whether the first toss was tails, or whether the second toss was tails, etc. We’re talking about the likelihood of ANY of five tosses landing tails, not the likelihood of ONE particular toss landing tails.
CliveStaples (4c1930) — 8/25/2008 @ 5:47 amCS:
Irrelevant. Unless Tom peeked, he could have chosen the first 25 coins tossed, the last 25 coin tossed, the odd coins, the even ones, or any other selection criteria he wanted. It makes no difference statistically. For all intents and purposes, the “first” 25 coins are whichever ones Tom selected, and the “last” 25 are the ones he did not.
Ditto for your two coin example. It doesn’t matter if Tom uncovered the first coin or the second one. All that matters is that he (1) unconvered one, (2) didn’t uncover the other and (3) didn’t use the fact that either was heads or tails in deciding which one to uncover (which he couldn’t have done if he didn’t know).
Wrong again. Tom either selected the first coin, or he selected the second, or you don’t know which. If he selected the first coin, he just eliminated the TH and TT options, leaving only HH and HT as viable options: a 50-50 split. If he selected the second coin, he just eliminated the HT and TT options, leaving only TH and HH options on the table, again a 50-50 split. If he didn’t tell you whether he chose the first option (or perhaps he didn’t pay close enough attention to know himself), then he just eliminated the TT option, and reduced the respective probabilities of the HT and TH options by half. More options still on the table, but still 50-50 odds.
If “the second coin” means the second coin that Fred tossed, then your point is irrelevant. If it means the second coin Tom looked at, giving him two potential heads coins to choose from (and he was looking for heads), then it’s a rigged game.
Only if Tom cheated. Otherwise, sequence doesn’t matter. Not knowing whether the remaining coin is Coin #1, Coin #2, etc. up to #5 doesn’t affect the numbers at all. Sure, it gives you five theoretical possibilities rather than one if the uncovered coin were identified. It also makes each of these five possibilities only one-fifth as likely, since each of the five coins now has a 4/5 chance of being among the known heads, and only a 1/5 chance of being the one still in play (and that coin, whichever one it may be, still has only a 1/2 chance of being tails).
In other words, one approach (positively identifying the coin) leaves you with four coins known to be heads, and one identifiable coin with a 50-50 chance of being tails. The other approach (throwing up your hands and saying you don’t know if it was the first, the second, etc.) leaves you with 5 coins, each of which has a 9/10 chance of being heads and a 1/10 chance of being tails. In the end, the probabilities are exactly the same.
Xrlq (b71926) — 8/25/2008 @ 5:50 amThat’s correct. One group toss of five coins is no more or less likely to yield 1 tail among the five coins than are five sequential tosses of the same coin. Either way you have five opportunities to get tails, and either way, each of those five opportunities faces 50-50 odds. And either way, you’re committing the gambler’s fallacy when you take into account the original odds after some of the coins have been revealed.
Xrlq (b71926) — 8/25/2008 @ 6:08 amI keep getting the impression that Clive has the Monty Hall problem down pat, but keeps applying it where Monty is not appearing.
SPQR (26be8b) — 8/25/2008 @ 6:14 am121/Stashiu/12:35 a.m.:
Uh, no.
You’re saying if we flip four coins, look for heads, and remove two heads whenever possible that the odds of the last two unknown coins are 50-50, because the information does not affect the other two coins?
I’m prepared to bet at any stakes you like. I’ll give you 6-5 and bet on tails on the remaining coins. Warning: You’ll lose.
–JRM
JRM (355c21) — 8/25/2008 @ 7:07 amI remember as a young boy experimenting with tossing a penny to see the resulting pattern of heads and tails.
To put a little fly in the ointment one time the penny came to rest on its edge.
I wonder what the odds of that occurrence is?
rab (7a9e13) — 8/25/2008 @ 7:10 amJRM, I don’t think that’s what Stashiu is saying at all. Clive is arguing that if we flip four coins, choose two of them at random and they just happen to be heads, that this somehow does affect the odds for the two remaining coins.
Xrlq (b71926) — 8/25/2008 @ 9:13 amJRM #135,
Stashiu can speak for himself. I’ll just point out that in Patterico’s Tom and Joe scenario, the coins that were eliminated could not occur as tails whereas in yours they can. So if you are telling me that out of four coin flips there will be two where I cannot possibly win but you can, then I want much better odds than 6 to 5. This is how slot machines work to make money for their owners. In scientific research this is how researchers get into trouble by eliminating results they do not like.
nk (3c7a86) — 8/25/2008 @ 10:39 amXrlq/137:
Interpret for me, then:
…. from Stashiu3 at comment 121.
–JRM
JRM (de6363) — 8/25/2008 @ 11:05 amrab #136:
Approximately (Πd²/Πdw); where w is the width of the milled edge of the coin.
EW1(SG) (da07da) — 8/25/2008 @ 11:19 am140 EW1~ Actually, I mistyped above, its:
There are also some terms ignored, which make the number even smaller. (But make the calculation much more complex).
EW1(SG) (da07da) — 8/25/2008 @ 11:29 amWith Xrlq’s indulgence:
Interpret for me, then:
If 25 heads are intentionally selected out (assuming there are at least 25 heads), it doesn’t change the odds on the remaining 25.
Because, unlike your scenario in your comment #135, they do not have a chance to be tails either. They were made “not have happened”.
nk (3c7a86) — 8/25/2008 @ 11:39 amComment by Xrlq — 8/25/2008 @ 6:08 am
either way, you’re committing the gambler’s fallacy when you take into account the original odds after some of the coins have been revealed.
So it seems that Xrlq and Stashiu3 (and some others) are being consistent about the original 25 reveals, regardless of Tom’s action, having nothing to do with the odds on the remaining coins.
My original question in #34 is not directed at them, but at the commenters who say that Tom having seen the results alter the case for taking the bet. In my #34 the same coins produce two different choices. It might be the case that I’m introducing another variable by specifying this, but it seems counterintuitive to get two different results from the same set of data.
Apogee (186a12) — 8/25/2008 @ 1:12 pmJRM/139: If I had come across that quote in a vacuum, I would have interpreted it to mean that Stashiu doesn’t understand the Monty Hall problem. However, I didn’t come across it in a vacuum, nor are there any intervening “Oh crap, I was wrong” posts between the comment you quoted and this earlier one:
Read together, I interpreted Stashiu’s reference to “intentionally selected out” to mean hand-picked without knowledge of which coins were heads vs. tails. But perhaps Stashiu will chime in if my interpretation was wrong.
Xrlq (b71926) — 8/25/2008 @ 1:26 pmApogee, I’m not sure I understand your comment. It sounds to me like the first scenario in #34 had Tom seeing all 50 coins, and deliberately picking out the ones that were heads, while the second scenario had the rules contemplated by this problem, where Tom picked 25 coins at random, and they just turned out to be heads. In that case, the answer is yes, the first scenario is the Monty Hall situation and the second is not.
Xrlq (b71926) — 8/25/2008 @ 1:32 pmXrlq – What my #34 was asking (and probably not too clearly) was:
1) The first scenario has Tom seeing all 50 coins, and deliberately picking out the ones that were heads.
2) The second scenario has Tom picking 25 coins at random, and they not only turned out to be heads, but the same heads.
3) They are not two different throws, but two different actions on the same throw (which is why I stipulated that the single throw was digital and sent to two different screens at two different locations.
What’s confusing to me (and I appreciate your patience) is that the same data results in two different situations – one in which you take the bet and one in which you do not.
As I said, I’m not sure if I’m introducing another variable or not, but it seems confusing to have a positive and a negative action from the same set of data.
Apogee (186a12) — 8/25/2008 @ 1:37 pmApogee, does the same player see both screens? Obviously, if you know that one Tom landed on these 25 coins by chance, and that the other Tom hand-picked them specifically because they were heads, then you’re going to play the game differently. But if one player sees the first screen, then based on what he knows, he should bet on most or all of the other 25 being tails. If a second player sees only the second, he should bet on the remaining 25 coins having a roughly equal distribution. One of these two players will lose the bet, on this round. But both will win over the long haul if they apply the strategy that has the odds in their favor.
Xrlq (b71926) — 8/25/2008 @ 2:02 pmXrlq – 2:02pm
Two different players, who do not know of the other’s existence. The correct answer seems to be related to a longer term plan, as the player taking the 50/50 view would most likely lose, by chance.
This is what I was alluding to regarding the addition of another variable, which, by your answer, would be that restricting the number of plays to this round only skews the results to avoid the incongruity of the same data being affected by the knowledge of the coin selector. This is the rare instance where the player taking the bet despite the random reveals is more likely to win.
I hope I’ve communicated my ideas effectively. As I have read the comments, the one thing that’s jumped out at me in this discussion that has continued to confuse me is the relationship between events that occur once (the 50 coin throw), singular events (each coin’s individual throw), repetetive events (throwing 50 coins many times) and the relative importance and weight given each one.
I appreciate your feedback Xrlq, as well as that of all the commenters, as that interplay is interesting and challenging to think through.
Apogee (186a12) — 8/25/2008 @ 2:30 pmJRM, Clive, and Gregory still don’t understand the Gambler’s Fallacy and apparently never will because it doesn’t “feel right”. At least one of those pesky coins just has to turn up tails. The truth is, they don’t… no matter what the previous results were. Previous results don’t affect the other results.
They just have a good way to go broke in a very organized manner.
Stashiu3 (460dc1) — 8/25/2008 @ 3:04 pmStashiu3:
Your point about the gambler’s fallacy is not well-taken. Your gambler’s fallacy argument applies if the coins tosses are mutually independent; in passing, this requires that the probability of getting a head must be known in advance.
If you toss the coin 50 times and reveal the first 25 tosses or if you toss the coin 50 times and reveal 25 tosses selected at random, then the coins you reveal are independent of the unknown tosses, and the gambler’s fallacy applies.
If, however, you toss the coin 50 times and some third party examines all 50 tosses and selects 25 heads to reveal out of the 50 tosses, then the coins that are revealed are not independent of the coins that are still concealed. After all, you expect to see 25 heads give or take 3 or 4, in the whole set of 50 tosses. If somebody pulls 25 heads out of the full set of 50, then the coins remaining should be overwhelmingly tails.
Per Apogee, #34 et seq, this is precisely the Monty Hall situation.
By the way, here’s a bonus problem. Tom gives you a coin that may or may not be biased. You toss the coin 25 times; all tosses are heads. Joe then offers to pay you $1 for each tail in the next 25 tosses. However, if the next 25 tosses are all heads, then you must pay Joe $1. Should you take this bet? Why or why not?
Clive, I’ll respond to your comment later this evening.
W. Krebs (c7e91b) — 8/25/2008 @ 4:25 pmW.Krebs, in that specific case above I said it’s a sucker bet because they have more information than you do and would not offer the bet if they were likely to lose. That’s the difference. If a third party is the one to remove 25 heads, what you’re still betting on is the original toss of 50 coins and you should go for it.
If the one offering the bet is the one to take out the heads, he knows how many tails he had to return in search of another head until he got to 25. That’s not Gambler’s Fallacy, that’s Monty Hall. If he only saw one tails before getting to 25 heads revealed, the chances of the remaining 24 coins (the 25 revealed heads and the one tail he saw) are 50/50. If he saw 10 tails before he revealed 25 heads, the chances on the remaining 15 coins (take out the 25 heads and 10 tails) are 50/50, but he would know that means at least 10 of the 25 are tails and would need 12 out of the 15 remaining to be heads to win… so why would he offer the bet? He wouldn’t.
Stashiu3 (460dc1) — 8/25/2008 @ 4:43 pmOnly if you like cider. 😉
Stashiu3 (460dc1) — 8/25/2008 @ 4:45 pmThat’s not the difference. The difference is that if someone hand-picked heads from the original sample, he preserves the odds that applied when all 50 coins were covered up.
Xrlq (62cad4) — 8/25/2008 @ 5:14 pmDear Stashiu3;
I understand Gambler’s Fallacy quite well. I would like to ask you; do you understand the question quite well?
Nowhere in here will you find further information as to whether this was revealed randomly, or whether Tom cherry-picked it using some form of algorithm, or even whether Tom knew or not the answer in advance. Now, you are given the option to ask Tom a question, to help clarify the situation.
If we reduced the problem space to 5 coin tosses, then 6 times out of 32, you would expect at least 4 coin tosses to be H, and 5 out of those 6 times, you would expect the last one to be T. Now, if this was a random pick, then yes, you’re right, the calculation would then be *what is the probability that Tom would pick by random exactly the 4 Head tosses when the final toss is a Tail* as you explained above, and the last coin toss would be 50-50. But as it stands, *we cannot make that determination*. Hence, my question would be aimed at giving me more information so that I know which probability calculation to pick.
IOW, I am implicitly assuming that TOM knows the answer when making the 5/6 probability call. Of course, the question does not say one way or the other, true. And making this assumption without checking may lead me to make an ass of myself. But the same applies to you, Stashiu3.
Now, the follow-up is a bit more subtle, as Tom is expressedly stated not to know the H or T from the outset. In that case, I would apply the independent events assumption (and noting that the potential solution space has been cut down quite drastically).
Gregory (f7735e) — 8/25/2008 @ 5:26 pmWhich is what I meant by:
Gregory,
Read the entire thread and you’ll see that we drifted away from the original scenario and my response was to the (then current) comments, not the original problem. But I will gladly withdraw you from the “not understanding” group as you have nailed the Gambler’s Fallacy with your comment. My apologies for misunderstanding what you meant. JRM and Clive continue to apply probabilities to the entire 50 coin set on a wager involving only the last 25 coins (when everything is assumed to be legitimate), which is what the Gambler’s Fallacy is all about. If you assume the first 25 are taken out in a manner that is not legitimate, it becomes a Monty Hall situation and it’s still a sucker bet because they’re the ones who have additional information, not you or a third party.
Stashiu3 (460dc1) — 8/25/2008 @ 5:57 pmBut my dear Stashiu3; I’m a blog commenter. Nowhere does it say that I read all of the comments posted (I did, but MEGO at some point of the process)… 🙂
But setting aside my unforced error for not seeing what happened and just jumping in, again nowhere in the question does it state that Tom and Joe are in cahoots with each other (or not), which is yet another factor to be considered.
Okay, so Patterico is no mathematician, nor probability/game theorist, nor statistician. Can we just say the question could have been worded better. 😉 Cause my head hurts. See you in some other threads, Stashiu3…
😀
Gregory (f7735e) — 8/25/2008 @ 7:40 pmI agree completely, but by Clive’s reasoning, them being in cahoots (the selection being biased) actually increases your chances, when actually it makes them worse than the best-case of a 6/5 wager against 50/50 odds because they have more information than you do. If they know it’s far more likely that a majority of the remaining unrevealed coins are tails, they’re not going to make the bet.
However, if there is no information known about the unrevealed coins, it doesn’t matter what the revealed coins were. The individual and aggregate odds of all unrevealed coins is 50/50. Wagering 6/5 on 50/50 odds is a sucker bet.
Take care… and don’t take any wooden nickels. 😉
Stashiu3 (460dc1) — 8/25/2008 @ 8:14 pm