A Fun Statistics Question
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 Joe’s offer?
2) Is there any question you would like to ask Tom before you decide?
P.S. Xrlq may not answer until I give him permission, as the question came up during an e-mail discussion we were having about statistical matters.
P.P.S. Xrlq is taking a poll about a statistical question. Feel free to weigh in here.
It’s a bum deal. The chances are still 50-50.
Mike K (2cf494) — 8/22/2008 @ 7:58 pmDitto. No wager from me.
Chris (da1e70) — 8/22/2008 @ 8:02 pmI would want to know if the 25 coins Tom uncovered were randomly selected. If so, no deal.
CT LEO (afa957) — 8/22/2008 @ 8:06 pmIt’s atill 50-50 because the coin has no memory. However, if you read The Black Swan by Nassim Taleb you might remember the little parable of Fat Tony and the academic, Dr John. Here’s a synopsis:
http://www.sundaymorningqb.com/2008/6/5/546477/nassim-taleb-got-his-philo
For the non-mathematician, probability is an indecipherably complex field. But Taleb makes it easy by proving all the mathematics wrong. Let me introduce you to Brooklyn-born Fat Tony and academically inclined Dr John, two of Taleb’s creations. You toss a coin 40 times and it comes up heads every time. What is the chance of it coming up heads the 41st time? Dr John gives the answer drummed into the heads of every statistic student: 50/50. Fat Tony shakes his head and says the chances are no more than 1%. “You are either full of crap,” he says, “or a pure sucker to buy that 50% business. The coin gotta be loaded.”Fat Tony vs Dr John
PC14 (ec0516) — 8/22/2008 @ 8:10 pmNo — the odds on every individual coin flip are even money.
Joe’s proposal requires you to lay 6-5 in his favor.
WLS (26b1e5) — 8/22/2008 @ 8:14 pmThis coin game is clearly rigged. I’d tell Joe to get lost.
Official Internet Data Office (066b46) — 8/22/2008 @ 8:19 pmI said:
“Before the toss, each had an equal chance of coming up heads or tails.”
In other words, the game is not rigged in the sense that any one coin had a greater initial chance of coming up heads. The odds were 50/50 for each coin before the toss.
Patterico (1157c3) — 8/22/2008 @ 8:22 pmI have less than a 50% chance of getting any statistical question correct, but I think I’d take the bet. Isn’t this similar to the Monty Hall thingy?
Apogee (366e8b) — 8/22/2008 @ 8:25 pmI bet Joe told you that before the toss, each had an equal chance of coming up heads or tails. And you bought it! If the first 25 coins uncovered are heads, then something is screwy. My advice is, run the other way.
Official Internet Data Office (066b46) — 8/22/2008 @ 8:25 pmIf I know that the coins are honest, then the first 25 coins that were revealed don’t matter in the slightest. I don’t accept the bet; the odds for each coin are 50:50 and I’m paying more when I lose than I gain when I win.
Apogee, you’re falling for the Gambler’s Fallacy. This has nothing whatever to do with the Monty Hall problem.
Steven Den Beste (99cfa1) — 8/22/2008 @ 8:28 pmThe scenarios are almost entirely dissimilar.
In this case, the still-covered coins are just as likely to be heads as tails; the deception is in the notion that after 25 sequential heads, a “tails” result is much more likely. This is false; the question is whether the sequence will be H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-T
or H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H-H.
Both results are equally as likely.
CliveStaples (72cbf9) — 8/22/2008 @ 8:30 pmI would tell joe “Hell No”.
Statisticly, he’s going to lose money…
Scott Jacobs (d3a6ec) — 8/22/2008 @ 8:31 pmSteven – 8:28pm You may very well be right, but they’re not flipping the coins again. They’re already flipped, which to me would signify that, now that I have more information than I did before, my odds are no longer 50/50 as they would be if the coins were to be flipped again
Apogee (366e8b) — 8/22/2008 @ 8:32 pmSteven Den Beste nails it: the coins revealed don’t matter, and it doesn’t even matter that Joe has covered up only coins showing heads. Even if Joe covered up only coins showing tails, unless you knew that a priori, Joe’s offering you a losing proposition.
The one question I would ask Joe is, “How many coins showing heads did you cover up?” From that, you can deduce whether you’d take his bet, but I doubt very much that he’d tell you.
Steverino (1dda08) — 8/22/2008 @ 8:32 pmNobody wants to ask Tom a question?
Patterico (f903b2) — 8/22/2008 @ 8:34 pmNo, Apogee, you don’t have any more information than you did before. No matter what the other 25 coins showed, the chance of each of the 25 hidden coins showing tails is still 50%. Each of the 50 coin tosses was a separate event, wholly independent of the others. So the odds of the 25 covered coins being tails don’t change, no matter what the other 25 coins did.
The fact that Joe rigged the outcome beforehand isn’t relevant.
Steverino (1dda08) — 8/22/2008 @ 8:35 pmIf it’s not rigged, then you’re laying 6 to 5 on an even-money bet, which is dumb, anyway. Forget about it.
Official Internet Data Office (066b46) — 8/22/2008 @ 8:35 pmSure. I’ll ask him one. “You throw away a lot of money. Are you with the Chicago Public School system?”
Apogee (366e8b) — 8/22/2008 @ 8:36 pmI reject Tom’s offer and ask if he has any cold beer.
Ropelight (4a83c9) — 8/22/2008 @ 8:38 pmDid Tom know which were heads or tails before he uncovered them?
How did Tom decide which ones to uncover?
DRJ (a5243f) — 8/22/2008 @ 8:39 pmIn my medical decision making class at Dartmouth, an instructor on decision theory did an experiment. He flipped five coins, lined them up, then covered them. He then asked each member of the class to list the sequence from left to right. He collected the answers then told us this is a standard experiment with undergraduates (we were grad students). 76% of students will list HTHHT. That is the students’ concept of randomness. They start with heads, then tails, then heads, then realize they are following a non-random sequence so they will choose heads twice in a row. That was one of the most enjoyable lectures on decision making.
Mike K (2cf494) — 8/22/2008 @ 8:44 pmafter 25 sequential heads is what has me flummoxed. That would seem to mean that there is an even statistical probability of getting 50 heads as 25/25.
I promise nothing regarding statistics, which is why I don’t gamble.
Apogee (366e8b) — 8/22/2008 @ 8:44 pm…in a single throw of 50 coins.
Apogee (366e8b) — 8/22/2008 @ 8:45 pmPatterico #6:
And they’re still 50/50 for each coin after the toss.
Don’t think I’ll lay in on Joe’s game. But I’m curious how Tom managed to uncover that many heads: did he see them before they were covered?
EW1(SG) (84e813) — 8/22/2008 @ 8:47 pmAssuming that the coin flipping is independent and the results are random and fair (I’m not sure that the initial phrasing did that) I still want to ask Tom about the process of uncovering the coins. Was the number 25 chosen before the flipping, were the 25 uncovered chosen randomly, does he know the Head-Tail status of any of the still covered coins, ….
No bet. I’m not even sure I’d bet if it was even money.
htom (412a17) — 8/22/2008 @ 8:48 pmSo here we are, the Friday night debate club.
DRJ (a5243f) — 8/22/2008 @ 8:52 pmI’d remind “Joe” that the typical lib math is to give $2 in services for every $12 in taxes. Geez! He’s been the presumptive veep for like 5 minutes and he’s already fucking up.
Icy Truth (9e8038) — 8/22/2008 @ 8:55 pmIs there any question you would like to ask Tom before you decide?
Sure. “The FBI annual physical test is grueling. At what age will you retire?”
nk (3c7a86) — 8/22/2008 @ 8:55 pmNot much of a debate on my end. More like stumbling around in search of a flashlight.
It’s the separation of the coins in aggregate vs. the 50% per toss that has me hung up.
Patterico, if you’re waiting for a mathematical formula for this, you won’t get one from me.
Apogee (366e8b) — 8/22/2008 @ 8:57 pmI would ask Tom if he tossed the coins together or individually.
Two things come into play in this scenario:
If tossed together:
1) Independent Events or Dependent Events.
A coin tossed indeed has a probability of 50%, but a group of coins tossed interactively creates a dependent event as the interaction between them introduces variables not included in a single toss. By uncovering and not recovering them, this creates a Dependent Event with different odds than 50%.
If tossed individually:
2) Significance of Expected Value.
The equation is as follows:
E(X)=(10)*1/2+(-12)*1/2=-1
Tom’s answer is very crucial my friends.
Jeff Barea (296843) — 8/22/2008 @ 8:58 pmWithout seeing any of the other comments here, I would ask Tom how many of the remaining coins are heads up? Assuming that Tom does not answer, I would not take the bet.
Ira (28a423) — 8/22/2008 @ 8:58 pmIt depends on how risk-averse you are, really. Is it worth risking $12 to get $10? Even presuming that it’s a 50/50 shot, some people might still take their chances.
CliveStaples (72cbf9) — 8/22/2008 @ 9:01 pmBTW, coin tossing has nothing to do with chance~while I don’t understand the math behind it, a coin tossed so that one face is parallel to the landing surface, and the rotational axis is also parallel to the landing surface will always land with the same face up that was up when it was launched. Knowing this, and having a “mark” that will allow me to catch the coin in the air in the act for the purpose of displaying it allows me to control which face of the coin shows after the toss (provided I paid attention to which face was up when I tossed it).
It takes a little more practice to control which face comes up when allowing them to land on an inelastic surface: if they land perfectly flat, there is a good chance that they will bounce, and the bounce is unpredictable. Not as unpredictable to be random, but less desirable than being able to control each individual outcome.
EW1(SG) (84e813) — 8/22/2008 @ 9:03 pmI would ask Tom: “So, this uncovering of the coins thing – Is it a union gig? How’s the dental?”
Apogee (366e8b) — 8/22/2008 @ 9:05 pmI would ask Tom: “Is your internet name XLRQ, and did you bet Patterico that you both could keep a bunch of people home on Friday night waiting for an answer to a statistical question?”
Apogee (366e8b) — 8/22/2008 @ 9:08 pmThat seems pretty irrelevant. Whether or not the “flipper” can control the specific outcome if he chooses doesn’t matter if the “flipper” just flips the coin without care for which face lands up. Unless the question specified that the “flipper” was acting in such a fashion as to always or usually produce the outcome of his choice, I’d presume that the results can be treated as ‘random’.
I mean, from a physics perspective, the result isn’t “random” at all, whether or not the coin is tossed ‘parallel to the landing surface’. As long as a good physicist knows the original starting position, the final position, and the forces that acted on it, he could determine which face was up.
CliveStaples (72cbf9) — 8/22/2008 @ 9:10 pmAlong with DRJ, my question to Tom is, “Did you selectively uncover only the coins showing heads?”
An affirmative answer reveals a very great deal of new information. The selection of remaining covered coins is no longer random, and assuming the coins and the tosses are honest, as claimed, I’d take the bet.
If, on the other hand, he randomly selected the coins to uncover, and they all came up heads, I’d take that as evidence that something was rotten — either the coins or the tosses or something was dishonest, and I’d expect all or most of the remaining coins to be heads as well, and I wouldn’t take the bet.
DJMoore (3c32ad) — 8/22/2008 @ 9:14 pmCliveStaples #36: Actually, as long as you know which face was up when it was tossed (in a classic spinning toss), then you know what face will be up when it comes to rest.
The point is that coins may make a good (easily understood) model when discussing chance in statistics, but they don’t behave that way in the real world.
EW1(SG) (84e813) — 8/22/2008 @ 9:19 pmFor even 25 out of 50 coins to randomly come up heads and arbitrarily uncovered by Tom to show heads are staggering odds. All the coins are heads up, because they’re heads on both sides of the coins and I’d say see ya later.
Peter (e70d1c) — 8/22/2008 @ 9:29 pmWhy bother asking him anything?
The chance that 25 coins in a row all come up the same is .5^24.
That is to say, about 1 in 16 million.
If he truly did uncover them randomly, and this is the 1-in-16 million event, I stand to lose about $25.
So if I take the bet, the odds of me losing about $25 are 1 in 16 million.
But if he uncovered only ones that he knew to be heads[1][2], most likely the vast majority of remaining coins will be tails. Do a binomial distribution if you don’t believe me. Thus I stand to make a tidy sum, almost certainly at least 15 tails to 10 heads, which is to say 15*10-10*12 = $30 profit for me.
[1] (for example, if he saw them before they were covered up)
[2] (or, say, 20 that he knew to be heads, and 5 by chance)
So I will almost certainly make more than $30, and only in extremely unlikely circumstances (1-in-16 million) I will lose, on average $25 if everything is truly random.
And even if they are cheating, at most I will lose 25 * 12 == $300.
People here are making the mistake of worrying about a very minor problem (the remaining coins being truly random, minor because it will result in, on average, a loss of $25) that is extraordinarily unlikely (1-in-16M)
The only question I would ask is: did you manipulate the coins while you were uncovering them?
Daryl Herbert (4ecd4c) — 8/22/2008 @ 9:34 pmIf you hadn’t explicitly stated that the coins are fair (a 50-50 chance for each coin) at the start of the problem, my question for Tom would be: “Are these coins loaded? Or double-headed?”
But if I know that the coins are fair, my question to Tom becomes: “Did you look at the coins before they were covered up?”
Because if Tom did not look at the coins and was uncovering coins at random, then all I know is that a very improbable event just happened — Tom just did the equivalent of flipping 25 heads in a row. However, it gives me no information about the other coins.
OTOH, if Tom did know ahead of time which coins were heads, then heck yeah, I’ll take that bet with Joe — my expected profit is about $150-200. From the setup, it might look like Joe is offering a sucker bet — but if Tom deliberately selected coins that he knew were heads and uncovered just those coins from a normal distribution of a fair coin flip, then Joe’s the one who’s the sucker.
See, if Tom did have knowledge of which coins were heads, then his uncovering 25 heads is no longer the equivalent of flipping 25 heads in a row. Instead, he’s gone and deliberately pulled 25 heads out of a normal distribution. So that does give me information about the remaining 25 — we’re no longer dealing with a “the next coin has an even chance of being heads or tails” situation here.
Another way to look at it is: in a normal run of 50 flips, you’re going to expect about 20-30 heads. Here, 25 heads have been deliberately selected and pulled out of the bunch. That means there’s probably no more than 5, almost certainly no more than 10, heads left in the group. (If there were more than 10 heads left, that means that perfectly fair coins flipped in a distribution of at least 36-14, or even more skewed. The math for that is pretty easy to work out).
I know this is a pretty unintuitive result for many people. But guess what — I’d be willing to bet real money on this. If you want to offer Joe’s bet to me — flip 50 coins, have someone who knows the result pull 25 heads out and then make Joe’s bet — I’ll gladly accept, and take over $100 dollars from you just about every time.
Robin Munn (2d5754) — 8/22/2008 @ 9:35 pmWhoops – minor mistake in my arithmetic (though not my statistics) in the post above. My expected profit is about $50-100, not $150-200.
Mutter, mutter, stupid “carry the 1” fencepost errors, mutter mutter…
Robin Munn (2d5754) — 8/22/2008 @ 9:39 pmYears ago my brother had this brilliant plan to win the lottery. He tracked numbers that came up for many weeks. Then he bought lottery tickets using the numbers that had not come up. His theory was that if all numbers had an equal chance of being picked at the outset, then the numbers that had not been picked after several weeks were bound to be more likely to come up in the future. He even had some “expert” who verified the statistical logic of the theory in a book.
I explained to my brother that the theory was BS. History could not affect the future randomness of the numbers that were picked. (The little ping pong balls had no idea which of them had been selected previously.) Therefore, regardless of the history of numbers, on each new drawing each number still had an exactly equal chance of being picked. I pointed out that if you flip a coin 49 times and it comes up heads on every toss, on the 50th toss the odds of tossing heads was still 50/50. He swore that the odds would be astronomically high that the next toss would be tails.
My brother was really pissed when I shot down his plan. I gently explained to him that I was a lawyer, so I knew what I was talking about. [snort] That just made things worse.
Cicero (3acd31) — 8/22/2008 @ 9:40 pm1) Sure.
2) The key piece is that there is no mention of Tom uncovering random coins.
It’s a group toss.
Al (b624ac) — 8/22/2008 @ 9:44 pmAnd I never thought I’d say this, but Steven den Beste is wrong in comment #10. If Tom did not know before uncovering the coins which ones were heads, then it is indeed the Gambler’s Fallacy. But if Tom did know, then it’s no longer a case of him picking the first 25 flips in a sequence of 50, but rather cherry-picking the heads out of the sequence, leaving an entirely different set of results after he’s done.
Robin Munn (2d5754) — 8/22/2008 @ 9:44 pmIf Joe looks at all 50 and chooses which 25 to reveal, and they’re all the same, then it’s no game. I won’t bet no matter what the payoffs are.
If Joe reveals 25 at random and they all turn out to be the same, it doesn’t matter. The other 25 are still 50:50 chance of being heads or tails, assuming the coins are honest.
Steven Den Beste (99cfa1) — 8/22/2008 @ 9:44 pmI would ask Tom if he knew anything about the state of a coin before it was revealed.
If he did, then my assumption is that he chose to reveal 25 heads. If he chose to reveal 25 heads, then the remaining population is most likely skewed to tails (using the law of large numbers) and I take the bet. A more rigorous analysis would compare the odds Joe is giving to the population distribution. For example, at 6:5 odds I win if the remaining number of heads is 11 or fewer. Thus, I would look at the following conditions: How many valid combinations of 50 tosses are there? (Note that you have to exclude all combinations that have more than 25 tails.) How many of those combinations have 36 or more heads? If the odds of me losing are less than 5 in 11, I take the bet. My intuition says take the bet.
If the reveals were random, then I do not take the bet under any circumstances. The remaining 25 coin tosses are random and independent from the first twenty five and the odds are really against me.
MartyH (00ac40) — 8/22/2008 @ 9:50 pmIf Tom had had some ability to see which were heads, and uncovers 25 of those, then the likelihood is not even that the rest are heads or tails. Rather it’s pretty likely most are tails.
Once you’ve non randomly taken a bunch of heads out of the total, it’s no longer a gamber’s fallacy to realize that what’s left will be non random as well.
Juan (4cdfb7) — 8/22/2008 @ 9:52 pm#20, 37: Why did that take so long? If the coins Tom showed were chosen by Tom (and Joe has no info on the other 25 before he made his offer), then I’ll crawl over ground glass to take that bet.
The revealed coins indeed convey no information in that case, except that all outcomes involving fewer than 25 heads are eliminated. So the expected number of total heads is somewhat over 25 (due to the lower bound), but not very much, and certainly less than the 12 that would be required for me to lose money.
Kevin Murphy (0b2493) — 8/22/2008 @ 9:53 pmSo the revealed coins indeed convey no information in that case, except that all outcomes involving fewer than 25 heads are eliminated.
It makes it easier for me to state that if Joe didn’t know the outcome, and uncovered another 24 heads, and the bet was on the last coin, it would still be a 50/50 chance. Why that makes it easier, I have no idea.
Apogee (366e8b) — 8/22/2008 @ 10:12 pm#42: The reason you are right is the same explanation I made of independent and dependent events in #30.
When you remove a piece from a group it affects the rest of the group.
So Lotto ball #1 may have X chance, but with Lotto ball #1 removed Lotto ball #2 has a Y chance. Similar with Lotto ball #4 having a Z chance and on and on.
It’s probability theory folks. No mystery to it. Geez.
Someone actually versed in statistical theory should talk now.
Jeff Barea (296843) — 8/22/2008 @ 10:28 pmSeveral people have got part of the answer right: the questions you want to ask are what rule was used to choose (a) the number of coins revealed and (b) the particular coins revealed, and whether this rule was chosen before or after the coins were tossed and the results known.
If the selection was not random, i.e. the coin tosser simply gives you the information, then you have to posit the following question: What percentage of the time do 50 random coin tosses end up with heads 26 to 36 times, and compare it to the odds of of heads coming up 37 to 50 times on a binary distribution table, then calculate the expected value for each of the 25 separate cases. That’s the correct calculation because after a random event the selective revelation of the 25 heads is the same thing as asking the expected value calculations for the portion of the distribution going from 26 heads to 50 heads. This is a classic binomial distribution analysis, and the answer will be non-obvious; my feeling is that the deal is in that case is one that should be taken, but I ain’t running the numbers and could be wrong.
Now let’s assume the revelation is random. Even here there is a question. If it were decided, prior to flipping, that 25 coins were to be taken out, and the selection of those 25 were left to chance via elimination selection (i.e. rolling a 50 sided die, cutting that number out, then a 49 sided die, then a 48 sided die, etc.), then the result of revealing the results and saying you will pay out on the other 25 tosses is no different from simply taking the coins out of play without revealing the results, since every toss is (I am assuming) random. That means the hidden 25 tosses should form their own 25 toss binomial distribution (as randomly selected subsets of a random sample should reflect the same odds, albeit with a different variance). So in that case, the deal should be rejected, since the distribution will be a normal one centered at 0.5, with the expected value of half the results involving you paying more money than if you win on the other half of the curve.
There is another way to represent the random selection, which is tossing another coin, and if it come up heads revealing the result and tails not doing so. In other words, rather than running a 1 in 50 trial to eliminate the first coin, then a 1 in 49 trial to eliminate the second, etc. you run 1 in 2 trials 50 times, and heads comes up 25 times exactly, and every heads coincides with a heads on the first run. This is a statistically different method of elimination which creates its own binomial distribution. I did not take enough statistics to analyze this. I do know that it is a very different question from a pre-set elimination selection. Also, instead of making it a coin flip, you could roll a six sided die and reveal only if it comes up 1; again, a different distribution.
Cyrus Sanai (4df861) — 8/22/2008 @ 10:37 pmI never thought I’d say this, but Steven den Beste is wrong in comment #10.
I concur.
SDB: If Joe looks at all 50 and chooses which 25 to reveal, and they’re all the same, then it’s no game. I won’t bet no matter what the payoffs are.
All due respect Mr. Den Beste, as long as Tom is following the rules of the game, then it is most definitely a game. Anything with rules and payoffs is a game.
Daryl Herbert (4ecd4c) — 8/22/2008 @ 10:41 pmCicero: Years ago my brother had this brilliant plan to win the lottery. He tracked numbers that came up for many weeks. Then he bought lottery tickets using the numbers that had not come up. His theory was that if all numbers had an equal chance of being picked at the outset, then the numbers that had not been picked after several weeks were bound to be more likely to come up in the future. He even had some “expert” who verified the statistical logic of the theory in a book.
That’s not only no better than average (for the reasons you describe), it’s actually worse than average.
There are other statistical illiterates out there following the same technique. So if the lottery does hit those numbers, he would have to share the jackpot with them.
Daryl Herbert (4ecd4c) — 8/22/2008 @ 10:48 pmDaryl Herbert – I have a question. Speaking of astronomical odds – What if you had to play the game, and had to repeat it 10 times, and each time Joe randomly uncovered 25 heads, without prior knowledge. Would you still be playing a 50/50 chance game, or would you have odds of coming away ahead?
Apogee (366e8b) — 8/22/2008 @ 10:51 pmsorry – would you have odds of coming away ahead overall?
Apogee (366e8b) — 8/22/2008 @ 10:53 pmTake the bet. They were all thrown together with a 50/50 chance each. You have a large population of items that could be 50/50 based on that one event. If half of the population is already shown to be in one state, the odds are the other half will be more of the second state.
You could easily replicate this and would seldom find where more than 70% of the coins were all in one state after each mass simultaneous throwing.
Ray (8cfb7a) — 8/22/2008 @ 11:29 pm#51 Seriously, there are statistical equations set up for this scenario. None of which support your position. Go read a basic Statistics or Probability book.
Patterico, did you pose this just to piss me off by the determined attempts to explain statistical questions using non-sequitors?
Jeff Barea (296843) — 8/22/2008 @ 11:37 pmBy the way, this IS the Monty Hall problem.
Kevin Murphy (0b2493) — 8/23/2008 @ 1:18 am(Monty Hall problem: 3 doors, one a prize, two crap. You pick one, Monty shows you one of the other two is a zilch. He asks you if you want to change doors. Assuming he does this regardless of your pick, YOU MUST CHANGE, because your chances go from 1/3rd to 2/3rds).
Kevin Murphy (0b2493) — 8/23/2008 @ 1:24 amThanks Kevin – I’m 37.4% sure that’s what I believed initially.
Apogee (366e8b) — 8/23/2008 @ 1:30 amA simpler to understand version of the coin toss.
Tom flips 50 coins behind a screen. The toss is fair. He then pushes either 25 heads or 25 tails out in front of the screen. His choice. Joe comes into the room, with no knowledge of the actual distribution, and says he’ll pay you $10 for each of the unseen type if you give him $12 for each of the seen type.
How can you not take the bet? The coins you see alter nothing: you are still betting on the distribution of the original toss, whose expectancy is still “about even”.
Kevin Murphy (0b2493) — 8/23/2008 @ 1:37 amIt doesn’t matter whether the coins were tossed in a group or individually. It doesn’t matter whether the flipper chose which coins to reveal.
This is not the Monty Haul problem; there, your initial choice was 66% likely to be a goat. There is no “goat revelation” event in the scenario at hand, and the chance of getting a bad result is 50%, which favors neither outcome.
And I was surprised to see someone using this logic:
BTW, coin tossing has nothing to do with chance~while I don’t understand the math behind it, a coin tossed so that one face is parallel to the landing surface, and the rotational axis is also parallel to the landing surface will always land with the same face up that was up when it was launched. Knowing this, and having a “mark” that will allow me to catch the coin in the air in the act for the purpose of displaying it allows me to control which face of the coin shows after the toss (provided I paid attention to which face was up when I tossed it).
That’s simply not true. Say that he only tossed 4 coins and shows you three heads. The last coin is just as likely to be heads as tails, since H-H-H-T is just as likely as H-H-H-H.
CliveStaples (72cbf9) — 8/23/2008 @ 2:13 amAbout 1/2 of the responses to this fail reading comprehension. The coins are fair; that’s a condition of the post.
I want to know from Tom:
1. What were the criteria for uncovering coins (both number and type)? (DRJ got this first, but this is an elementary question.)
2. How many iterations of the 50-coin flip have been performed by Tom or Joe? (The flipper is unknown here.)
3. Are you buddies with Joe? (Question 2 might be overprecise; by doing a large number of multiple-coin flips, outliers could be selected by grifty logic-problem betters. This is a major problem, of course, somewhere between meth and gangs.)
If Tom tells me the coins were selected at random, Tom is lying. (Bayes’ Theorem and the remoteness of 25 same-side exposures of true coins overwhelms any trustworthiness Tom may have. There is no stipulation to Tom’s veracity in the problem.)
If Tom tells me he selected heads because they are shiny, and I am satisfied that Tom and Joe are not in cahoots (relevent to the multi-iteration problem), the other coins are likely to be tails, and the analysis depends on a couple of Tom’s answers, but under most assumptions, the bet is a big profit-maker for me.
–JRM
JRM (355c21) — 8/23/2008 @ 2:17 amAnyone else here think of the 78-in-a-row-heads coin toss at the beginning of Rosencrantz and Guildenstern are Dead? Funny movie.
Rosencrantz and Guildenstern discuss the law of probability (start at 1:40 to skip all the coin-flipping time)
no one you know (1ebbb1) — 8/23/2008 @ 3:52 am1) Do you accept Joe’s offer?
Probably not statistics say the remaining
25 STILL have a 50/50 chance to be either heads or tails
2) Is there any question you would like to ask Tom before you decide?
Dan Kauffman (b31cae) — 8/23/2008 @ 5:02 amYes do the coins have both a head and a tail?
I am going to answer, then read the comments. Assuming actual randomness, I would take the bet in a second.
Having said that, the likelihood of selecting 25 consecutive heads from the collective group of tossed coins has to be astronomical, which would make me question the randomness.
JD (5f0e11) — 8/23/2008 @ 5:05 amIf someone mentioned this, I apologize. This question is very relevant to human nature, as all casinos are aware. It is my understanding that roulette play increased dramatically once operators posted the electronic history boards. As most everyone here realizes, the game – and the odds – didn’t change a whit. Only the erroneous perception of the odds.
rhodeymark (e96855) — 8/23/2008 @ 6:07 amTest
rhodeymark (e96855) — 8/23/2008 @ 6:08 amApparently my comment was eaten because I referenced a game played in a certain town in Nevada, using words that are spam verboten. This game experienced a large spike in play when the operators posted electronic “history” board at the tables. The odds didn’t change a whit in the player’s favor – only the perception.
rhodeymark (e96855) — 8/23/2008 @ 6:13 amI have to agree that in theory it’s 50/50 for the rest of the coins, assuming the toss was fair, but not knowing how the heads were selected is the problem. If it is fair, and they are “covered-up,” and he truly selected 25 Heads at random (either 3MM:1 or 1.6MM to one, can’t do math in my head), a normal distribution would indeed seem to make more tails likely, and probably make the bet pay off at better than 6:5 for the remaining coins.
However, I have to ask – since this was a group toss, were the coins in a roll together, and did they land vertically? If so, maybe they were packed with ‘heads’ up and they are all in the roll that way, fresh from the mint. Just a thought.
carlitos (39ff54) — 8/23/2008 @ 7:17 amI would ask Tom if he would take the same bet if I offered it to him.
daleyrocks (d9ec17) — 8/23/2008 @ 7:25 amThere are a couple of questions / assumptions that need to be made before I can place a bet:
We’ve been told we know the coins are evenly balanced with standard head/tail markings, so that’s not in dispute.
The problem states the coins were thrown on the ground; it doesn’t state they were thrown in a fair fashion. If the throw was unusual, no bet. We used to play Risk such that we could couch the rolls so they ended up 6’s around 80-90% of the time.
The problem states the coins were covered and Tom has uncovered half of them. We must assume that the act of covering and uncovering them does not change the coin; if we can’t do that, no bet.
We then need to ask Tom about how the coins were uncovered; how many and which ones, specifically. We must assume Tom is trustworthy, or else no bet.
Since this is a probability question and not a psychology question, we must assume he’s telling the truth. If he says he uncovered coins at random, either the remaining 50% are going to be distributed according to probability, meaning our bet is likely a net loser, or there is cheating involved somewhere, which also makes the bet a net loser. If he says he has near photographic memory and saw which ones were heads (or has some other way of knowing which were heads) and revealed only heads but not necessarily all heads, then the remainder are tails or mostly tails and our bet is probably good.
Trying probability puzzles on a gamer with “rules lawyer” tendencies leads to overly specific and complex answers like the one above. When taking probability in college, I was compelled to specify that my answers on dice problems were only valid for six-sided dice.
Civilis (344d5e) — 8/23/2008 @ 7:53 amLet me put the question this way:
A bit generator, alleged to be random, is tested 25 times; each time it issues a one.
What are the chances that the bits are truly random? That each bit is truly independent of the previous bits?
It’s possible that you’ve just watched a 1:2**25 “miracle” — but the way to bet is that the bits are not random, not independent, and that the 25th bit will also be a one.
In the current case, the coin flip is said to be honest — but Tom has the chance to filter the output, and no allegation has been made concerning Tom’s honesty. (Indeed, we are explicitly invited to question his honesty.) With 25 heads in a row, you deserve whatever penalty Joe imposes on you if you assume that Tom has not imposed some bias on the (originally) random bit stream.
===
The whole point of statistical testing is to check for systemic biases. Suppose you are doing quality control at a widget factory. Out of each lot of a thousand widgets, you select 10 samples (I’m making these numbers up.) Normally, you might find one or two bad widgets in a given sample. Today, however, all ten widgets in the sample fail the instant power is applied.
Do you:
A) Assume you’ve just witnessed the one time in the entire history of the plant where the sample consists of all the bad widgets in the lot, and initial the QC report “OK”? (Raise your hand, everybody who said “Don’t take Joe’s bet, and don’t bother to ask Tom any questions.”)
Or
B) Fail the entire lot, and notify the shift supervisor that something has gone awry on the assembly line?
DJMoore (3c32ad) — 8/23/2008 @ 9:13 amJust to be even more explicit:
The bit stream generator in question is not just the coin-flip; Tom is essentially acting as the display for the generator. We are testing the entire system, generator and display, not just the generator.
DJMoore (3c32ad) — 8/23/2008 @ 9:19 amDarn it, “The way to bet is that…the twentySIXTH bit will also be a one.”
I specifically checked for that before posting, and still got it wrong.
DJMoore (3c32ad) — 8/23/2008 @ 9:22 amI echo comment #3.
in all, no deal. He has better information than I do, and it is always a bad decision to bet against someone with better info than you have.
headhunt23 (05932f) — 8/23/2008 @ 10:32 amWhat if you had to play the game, and had to repeat it 10 times, and each time Tom randomly uncovered 25 heads, without prior knowledge. Would you still be playing a 50/50 chance game, or would you have odds of coming away ahead?
There is only a 1-in-32M chance that it would come up all heads, one time, if Tom was uncovering at random 25 coins that were flipped at random (with 50/50 chance of heads vs tails).
There is only a 1-in-32M ^ 10 chance that it would come up all heads, ten times in a row, if Joe was uncovering at random 25 coins that were flipped at random (with 50/50 chance of heads vs tails).
That is so unlikely that even if you had a billion coin flippers on a billion worlds within a billion parallel universes flipping coins for a billion years, assuming they could each flip 500 billion coins every year, the odds of that ever happening are still 1000,000,000,000,000,000,000,000,000,000 to 1.
If that happened, it would be safe to assume that the Tom is not picking coins based on pure chance. In fact, that would be the only sane assumption, despite the fact that what you describe is theoretically possible to happen by pure chance.
* * *
Here is the key assumption: that when the coins were tossed, it was 50/50 for each coin whether it would be heads or tails.
Knowing that 50 coins have been tossed under those conditions, and that 25 heads were removed, the vast majority of the remaining coins are overwhelmingly likely to be tails.
That is true whether the 25 initial coins were chosen accidentally, or if they were taken away on purpose. The motive/knowledge of Tom is irrelevant.
Daryl Herbert (4ecd4c) — 8/23/2008 @ 10:47 am76/Daryl Herbert 10:47 a.m.:
You’re mistaken, if I understand you correctly.
Let’s make it simpler: We have four fair coins, fairly tossed. I uncover two at random that are heads. Are you saying that tails is more likely than heads for the other two? If so, you’re mistaken – and I’m absolutely prepared to back my analysis with a wager. If not, why is that different from what you are asserting?
–JRM
JRM (355c21) — 8/23/2008 @ 11:07 amI haven’t read any of the comments yet, but I don’t take that offer at all. These are independent events, so the fact that the previous 25 came up one way has no bearing on the remaining 25.
Andy (8f1d5e) — 8/23/2008 @ 11:59 amI have not yet read the comments: it’d spoil the fun. But your q, Patterico, sounds like a rephrasing of the great “Let Us Make A Deal” puzzle: do you switch doors after Monty shows you a loser, or stick to your orig pick?
You switch, of course, cuz Monty did not show you any old door at random, now did he? He specifically showed you a loser. Likewise, in your q here, I would ask Tom if the 25 coins he showed us were chosen at random or not, and base my decision on whether or not to accept Joe’s wager on that.
I also recall an excellent discussion on all this over at Dean’s World some time back. Ah yes, here it is.
OK, now I can go back and read what other commenters have said before me.
ras (fc54bb) — 8/23/2008 @ 12:16 pmWhy are my comments being eaten?
ras (fc54bb) — 8/23/2008 @ 12:16 pmEveryone who said “No bet, it’s even odds” is invited to poker at my place.
Kevin Murphy (0b2493) — 8/23/2008 @ 12:26 pmWhat Andy #78 said. It is no different than if the coins had been tossed in fifty different countries, fifty days apart each, and Tom had spent ten years getting back with twenty-five which were heads.
nk (3c7a86) — 8/23/2008 @ 12:27 pmnk, andy, etc. The 25 heads are selected, not chance events. I Tom claimed chance, I’d have nothing to do with it, as he’s either lying or God is helping him.
Given that they are 25 heads purposely selected AFTER THE TOSS, they simply deplete the “heads” distribution of the original 50-coin toss. So, to lose, I have to weigh the odds of a 37 heads/13 tails (or worse) toss against a 6-to-5 bet. That takes about 30 seconds, and I’m in.
Kevin Murphy (0b2493) — 8/23/2008 @ 12:37 pmOK, I dunno why my comments were being eaten, but I’ll try rephrasing what I was saying:
I have not yet read the comments; it’d spoil the fun. But your q, Patterico, sounds like a rephrasing of the Monty Hall puzzle: do you switch doors after Monty shows you a loser, or stick to your orig pick?
You switch, of course, cuz Monty did not show you any old door at random, now did he? He specifically showed you a loser. Likewise, in your q here, I would ask Tom if the 25 coins he showed us were chosen at random or not, and base my decision on whether or not to accept Joe’s offer on that.
I also recall an excellent discussion on all this over at Dean’s World some time back. Ah yes, here it is.
OK, now I can go back and read what other commenters have said before me.
ras (fc54bb) — 8/23/2008 @ 12:42 pmOK, I dunno why my comments were being eaten, but I’ll try rephrasing what I was saying:
I have not yet read the comments; it’d spoil the fun. But your q, Patterico, sounds like a rephrasing of the Monty Hall puzzle: do you switch doors after Monty shows you a loser, or stick to your orig pick?
You switch, of course, cuz Monty did not show you any old door at random, now did he? He specifically showed you a loser. Likewise, in your q here, I would ask Tom if the 25 coins he showed us were chosen at random or not, and base my decision on whether or not to accept Joe’s offer on that.
I also recall an excellent discussion on all this over at Dean’s World some time back.
OK, now I can go back and read what other commenters have said before me.
ras (fc54bb) — 8/23/2008 @ 12:43 pmBTW, the Dean’s World link is here.
ras (fc54bb) — 8/23/2008 @ 12:44 pmOK, I figured it out: Apparently, my comments were being eaten because they contained a link to the Dean’s World discussion that I mentioned, cuz when I removed that link everything worked again. Is DW verboten on this site now?
ras (fc54bb) — 8/23/2008 @ 12:45 pmI’m still waiting to discover the “fun” aspect of this question.
Seems to me that if you go by what is probable you take the bet. You need 14 of the remaining 25 coins to come up tails in order to come out ahead ($140 to $132). It’s reasonable to expect a random throw of 50 coins to come out with a 36-14 ratio of heads-to-tails.
But if you go by statistics and your only data is that each coin has a 50-50 chance of landing on one side or the other, then the best you can reasonably hope for is a 13-12 split one way or the other — and if that’s what happens then either way you lose ($144 to $130, or $156 to $120).
Maybe that is the “fun” of it; using probability, it is more likely but less certain that you will come out ahead.
Icy Truth (784175) — 8/23/2008 @ 12:48 pm“Is DW verboten on this site now?”
Please don’t assume that I have that level of control over what my spam filter does.
Patterico (c0bb5b) — 8/23/2008 @ 1:08 pmUnless you believe that a coin gives orders that are obeyed, no coin toss affects any other coin toss. Whether right next to it, a second apart, or across the world a thousand years apart. Every coin toss is an independent event. If Tom is scamming you it is ONLY because he has seen which of the remaining twenty-five coins are heads and which are tails and you have not.
nk (3c7a86) — 8/23/2008 @ 1:10 pmAnd despite my comment #28, I’m not sure that this is fair way to challenge the product rule. I’m slow, so I need to think about it some.
nk (3c7a86) — 8/23/2008 @ 1:24 pmnk #86:
Essentially, some of us are arguing (Kevin Murphy as articulately as any of us) that a string of 25 heads is very strong evidence that the tosses, as revealed by Tom, are not random. Tom’s filtering of the results means we are not looking at simple statistics.
Patterico explicitly invites us to question Tom, to not blindly accept that he is revealing coins in a neutral, random fashion.
We are presented with two sets of coins, 25 uncovered and 25 still covered. All of the uncovered coins are heads. The fact that all the uncovered coins are heads very strongly implies that either the tosses or the uncovering process was not random. Either most of the remaining coins are tails (if the tosses were truly random, or most of the remaining coins are heads, because the toss was not random, and strongly favored heads.
Not random, folks. Manipulated. Selected. Chosen. Forced. Whatever, but not random.
The chances that you have seen a natural string of 25 heads, and that the next 25 tosses yielded a normal distribution of heads and tails, much less a continuing string of heads, is so close to zero as makes no nevermind.
Either Tom picked his reveals, or the tosses were not random.
How many different ways do we have to say this?
Not! Random!
DJMoore (3c32ad) — 8/23/2008 @ 1:43 pm50 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.
Question #1 to Joe:
So… how do you know Tom?
Seriously though I think to make this more meaningful in a real world way, you make each coin worth a million, or make the loser face death.
Then it gets it out of just the head and into the gut too.
Then questions get asked.
Lets assume the toss has been proven fair and the standard coins tested and fit within statistical expectations.
Question #2
Tom, can you describe to me how the coins were covered or hidden?
(Was the cover something like a curtain that hid all until someone was allowed behind it, or was each coin individually covered from visual ID.
Question #3
Tom, was there any way you could tell which ones were heads when you went to uncover them?
Question #4
Tom, if you answer yes to #3, how many tails would you estimate you had to sort through to get out the 25 heads?
I’m sure there are more… but statistically speaking, 50 isn’t a real big sample for a coin toss experiment where everything is supposed to average out.
If Tom could see all the coins and was able to select out 25 heads then he should have a good idea how many are tails.
If Tom had to uncover them one by one, he should know almost exactly how many tails he turned over to get to 25 heads because he should have needed to uncover most of the coins, so you could ask him how many coins are still covered back there.
I gotta go, but if millions of dollars were riding on it, I’d want to ask as many questions as possible before using my equations
What did the cover look like?
SteveG (71dc6f) — 8/23/2008 @ 1:48 pmDJMoore #88,
My advice is don’t take the bet. But only because Tom might know something you don’t. Not for anything having to do with statistics. For what the actual thing, not the statistical thing, is.
nk (3c7a86) — 8/23/2008 @ 1:59 pmCan’t take it any more!. Yes, the probability of any given coin being heads is 0.5 (50%). However, the basis of the bet is the number of total heads expected out of 50 tosses. As others have said, for Tom to “win” there would have to be 37 heads in the original toss of 50 coins. The probability of that is far less than 0.5. For us to win only requires 14 or more tails out of 50.
Here’s an example of how the math goes: http://mathforum.org/library/drmath/view/52217.html
The example computes the probability of getting at least 4 heads out of 10 flips. This is roughly opposite to the conditions required for “us” to beat Tom (we want 14 tails out of 50 or about 3 of 10). The probability of 4 heads(or tails) out of 10 flips is 0.82 (see link). So it does look like a sucker-bet after all, its just that Tom’s the sucker, not us. And it doesn’t matter whether Tom uncovered only heads intentionaly or by random (extraordinary) chance.
I’ll try to work the math for the 50 toss case but I don’t know if my head can take it.
mark (3686d4) — 8/23/2008 @ 2:05 pmnk #90:
If I’ve convinced myself that the tosses are honest, then I know, as much as I can know anything by statistical inference, that Tom has selected out the heads, and it is safe to bet that almost all the remaining coins are tails.
It’s not a question of the coins “giving orders”, of each coin toss affecting the next sequentially, which is the assertion I was responding to.
It’s a question of Tom “ordering” the coins, of him picking coins from the entire field.
The 25 heads is very, very strong evidence that he does know something I don’t, and if I can trust him not to have flipped the remaining coins when he inspected them, and if I can trust Patterico that the flips were honest, I’ll take the bet, darn betcha.
DJMoore (3c32ad) — 8/23/2008 @ 2:09 pmmark #91:
We’re not betting against Tom; we’re betting against Joe.
Tom is simply part of the betting mechanism. The string of 25 heads is strong evidence that the overall mechanism, end to end, from flip to reveal, is not truly remotely random.
The 25 heads, in concert with other information we may possess about the total system, allow us to make very strong predictions about the states of the remaining coins.
DJMoore (3c32ad) — 8/23/2008 @ 2:16 pm“Uncovers 25 of them”
Uncovers how? Randomly? A half-area approach? Picki and choose?
It matters.
mojo (2303c8) — 8/23/2008 @ 3:15 pmThis has certainly been an interesting test of people’s understanding of randomness and non-randomness in the context of binomial distributions.
A couple points.
Monty Hall involves selective revelation of information about one initially random event, and how that information changes the expected value as to the remaining possible outcomes of the one event, once certain outcomes are intentionally and selectively excluded.
The issue here involves 50 separate random events. The question is whether the information revealed as to half of the 25 event trials should be taken into account. The answer depends on whether the revelation of the information is intentional (i.e. selected), in which case it can, or itself an utterly random elimination of 25 events, that simply happened to have a certain outcome, in which case it makes no difference what the 25 eliminated results were.
If the elimination is predetermined and random (via elimination selection), then it make NO difference what their outcomes were, and the deal sucks. But if the flipper is selectively telling you that 25 were heads, then he is asking a question about the binomial distribution of the 50/50 trial, and that question can be bet upon; and while I have not done the math, my guess is its a good wager.
Again, all this assumes that the flips are random.
Cyrus Sanai (4df861) — 8/23/2008 @ 3:25 pmI’m sorry.
When I wrote “how many tails he turned over”, I should have said uncovered.
If it is life or death, or riches vs ruin, I want to know Tom’s memories of how many tails he uncovered in the course of identifying, seperating and sorting the coins.
What if Tom answered … “I left less than ten coins covered”?
Or: “I had to uncover almost all of the coins to get to 25 heads”
In which case screw the math formulas, put down the calculators.
But if Tom answers that he uncovered 25 heads (each of which was covered individually) right away…. I wouldn’t touch the bet because I wouldn’t trust any of it regardless of what my calcualtor says.
I also wouldn’t take the bet based on my calculator because the sample is so small that the averages mean nothing. I’d expect to see things trend towards equal, but it’d likely shake down as around 10-15 of one or the other in the short term.
“Tom” and what he may know is more important than any formula.
SteveG (71dc6f) — 8/23/2008 @ 3:43 pmWould Tom take the bet?
The important question is, “How did you determine which coins to uncover?”
If the answer to that question is, “I uncovered 25 at random” or, then we’re dealing with a high probability that the remaining 25 regress to the 50/50 mean (~25 being tails). If the answer to the question is, “I uncovered them in the order they were flipped,” the probability of 25 heads in a row is very low—2.98023E-08—which would cause us to question the method of flipping, since the coin itself is heads on one side and tails on the other.
This sounds like a variation of the Monty Hall problem, where a third party introduces an element into the scenario which leads people to incorrectly play the probabilities. In that scenario, Monty reveals what’s behind one of three doors and asks the contestant if they want to switch the door they’ve selected. Since Monty isn’t going to show the grand prize when he opens the door, there is a 2/3 chance that the grand prize is in fact behind the door the contestant hasn’t selected. Switching is the right strategy in that case.
Teflon93 (7e68d0) — 8/23/2008 @ 4:02 pmPatterico – is this ‘fun statistics question’ possibly not about the coins whatsoever, and is actually intended to correspond to XRLQ’s post concerning the degree of speculation allowed regarding statistical information by a jury?
In other words: Jury speculation about the origins, collection, and sequestration of evidence with a statistical basis of relevance can alter the ability to accurately evaluate that evidence?
Apogee (366e8b) — 8/23/2008 @ 4:03 pmIf it’s is possible that Tom uncovered the first 25 heads he found, that means there are nearly none left covered. Half the time Tom wouldn’t even be able to find 25 heads in there. Nearly all of the remaining coins are tails. Accept Joe’s offer.
The question for Tom is did he uncover coins only after seeing they were heads. If not, these two clowns are obviously using two-headed coins.
Wesson (f6c982) — 8/23/2008 @ 4:41 pmWesson, it says “standard coins”.
Icy Truth (784175) — 8/23/2008 @ 4:56 pmIt does not say “Tom uncovers 25 random coins.”
It does say “Tom goes and uncovers 25 of them. All are heads.”
There is no statement about whether Tom got to see the coins before they were covered. The statement is: “Before you see….”
You ≠ Tom.
You don’t need to ask Tom to know it is flipping unlikely to get 25H-in-a-row-randomly. Unlikely enough to persuade me that Tom was, indeed, doing something that is not forbidden in the statement of the problem. So… why bother asking?
The information revealed has eliminated the chances of all distributions involving more than 25T. But… exactly 25T is still the most likely outcome of the remaining possibilities. And it is even more likely in this scenario than in a blind 50-coin toss to end up with exactly 25T. (It’s also more likely to get 24T here than in a straight toss. And more likely to get 23T here….) Because the chance -here- of getting 50T is precisely… zero. As is 49T, 48T, 47T … 26T.
The key bit in crazy statistics problems is correctly parsing exactly which steps are random.
Monty Hall doesn’t open a random door either. Otherwise we’d get “Well, looks like that was the big prize! So sorry!”
Al (b624ac) — 8/23/2008 @ 5:26 pmI don’t know if this has already be discussed(going to finish reading all the comments later).
I would ask Tom if Joe saw all the coins uncovered.
jmel44 (0a4bcc) — 8/23/2008 @ 6:12 pmOK, some of you are failing to read the question and its conditions.
All 50 coins have been tossed. 25 shown are all heads, which is on the bean for expected outcomes. For Tom to make money on the bet, 12 coins of the remaining would need be heads as well. This would be outside of expected outcomes, and just inside of two standard deviations off. (I think. Going by memory on the deviations)
Yes, each coin has 50/50 chance, but they have been thrown already. By showing that 50% was reached, each coin also coming up heads starts to strain credulity. The remaining coins do NOT have a 50% chance of being heads because they are part of a larger set which has been partially exposed. If fewer than 12 coins come up heads, you make money.
Take the bet if Tom can convince the coins are honest.
MunDane (d3328f) — 8/23/2008 @ 6:43 pmI do not take the bet. The only question I would ask would be if he knows the distribution of the remaining 25 coins. If he does, I assume he’s smart enough to know he would come out ahead. If he doesn’t know the distribution, I offer Tom the same bet on a re-do. We drop 50 coins and I get to pull out 25 heads (if there are that many). Then with the remaining 25 coins he will pay me $10 for every tail if I pay him $12 for every head. He just might be stupid enough to take the bet.
Are you going to update this with an analysis/answer?
Stashiu3 (460dc1) — 8/23/2008 @ 6:51 pmOk, you just did. 😉
Stashiu3 (460dc1) — 8/23/2008 @ 6:58 pmDo I have permission to answer yet?
Xrlq (62cad4) — 8/23/2008 @ 7:55 pmI guess not.
Xrlq (62cad4) — 8/25/2008 @ 6:38 pm