A purely abstract decision problem

[jason1990]

There is a bent coin. (Big surprise.) Suppose you really know absolutely nothing about it. You do not even have any information about how coins are generally bent by random people. You really have absolutely no information at all. (This is the abstract part.)

You now get one piece of information. You learn that the coin was just flipped 3 times. The results, in order, were tails, tails, heads.

Here is your decision problem. Your friend offers to lay you 7 to 4 odds if you will bet on heads for the fourth toss. Your friend has the same information (and lack of information) that you do. Do you accept? Why or why not? Is there an objective answer to this question?

[TomCowley]

I take the bet and expect a small +2/55 unit EV because I think P(heads) is greater than 4/11. The objectivity of the answer concerns the objectivity of choosing a prior distribution with no information. I used the uniform prior.

[TomCowley]

I don't claim to be qualified to judge the relative merits of different priors- obviously one for this case must be symmetric around 0.5, but that's the only hard constraint I'm willing to put on. It looks like the uniform prior estimates P=2/5=0.4, the Jaynes prior P=1/3=0.333, and the Jeffreys prior 3/8=0.375 (I think). Given the offered odds of .3636, you'll come up with a different answer depending on the different prior, which was probably Jason's point.

[BluffTHIS!]

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Is there an objective answer to this question?

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It is usually assumed that a coin is fair or not, in stating probability problems. But obviously that is just a convention. The question is that with a random coin, which you don't know is fair or not, how large of a sample size of flips would you need in order to assume it was fair or not? 3 isn't close and I personally wouldn't bet either way at those odds.

[PairTheBoard]

Well, I figure there are conservatively at least a googolplex of different scenarios by which biases might be produced by way of People Bending Coins. ie. There are at least a googolplex of functions
f1(p), f2(p),...,f_googolplex(p) describing prior distributions for the Bent Coin's true long run P(Heads).

Since I know nothing about the Bending System I must conclude all googolplex of the f_i's are equally likely. Since they virtually cover all possible distributions I will just track down each and every one of them, use it to compute the P(Heads | tails, tails, heads, f_i), average those over the googolplex of them and will have my answer.

I'll have to get back to you.

PairTheBoard

[jason1990]

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It is usually assumed that a coin is fair or not, in stating probability problems. But obviously that is just a convention. The question is that with a random coin, which you don't know is fair or not, how large of a sample size of flips would you need in order to assume it was fair or not? 3 isn't close and I personally wouldn't bet either way at those odds.

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Thank you for the reply. Also thank you for bringing the bold part to my attention. It has reminded me that someone might want to randomly choose between 7 to 4 on heads and 4 to 7 on tails.

Let me emphasize that this is not an option in this puzzle. You just need to decide whether or not to accept the bet as given. You can assume your friend has no inside information at all. He just sees the two tails, wants to bet on tails, and is laying you odds. The question is, do you like those odds? Also, do you think this is just a matter of personal taste, or do you think the information in the problem is sufficient to come to a logical decision?

[luckyme]

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do you think the information in the problem is sufficient to come to a logical decision.


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if it isn't then a million flips won't allow us to come to a logical conclusion either.

luckyme

[BluffTHIS!]

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Also, do you think this is just a matter of personal taste, or do you think the information in the problem is sufficient to come to a logical decision?

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This problem would seem to me the equivalent of you turning on the TV to some random game in a sport which I didn't follow and had virtually zero knowledge, excpet that you give me their win/loss stats for their past 3 games only. Which just like your problem I would judge a no bet situation *at those odds*.

The real question here is how good of odds do you need to make a wager in these kinds of situations where you have some very minimal information that either side has won/come up, at least once (making you think, perhaps incorrectly, that the whole game isn't rigged), and for how much of your roll or net worth.

[jason1990]

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do you think the information in the problem is sufficient to come to a logical decision.


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if it isn't then a million flips won't allow us to come to a logical conclusion either.

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Abstractly, you may be right. So does this mean you think it is or it isn't?

[jason1990]

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This problem would seem to me the equivalent of you turning on the TV to some random game in a sport which I didn't follow and had virtually zero knowledge, excpet that you give me their win/loss stats for their past 3 games only. Which just like your problem I would judge a no bet situation *at those odds*.

The real question here is how good of odds do you need to make a wager in these kinds of situations where you have some very minimal information that either side has won/come up, at least once (making you think, perhaps incorrectly, that the whole game isn't rigged), and for how much of your roll or net worth.

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You can assume that the amount your friend wants to wager is small compared to your bankroll -- small enough so that bankroll and utility considerations are not an issue.

[KipBond]

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do you think the information in the problem is sufficient to come to a logical decision.


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if it isn't then a million flips won't allow us to come to a logical conclusion either.

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Abstractly, you may be right. So does this mean you think it is or it isn't?

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??? How can a million flips not give us enough information to determine whether a 7:4 bet on heads is good? I mean, if the next 999,997 flips all end up being tails, I sure do think the 7:4 bet on heads is a bad wager.

[Phil153]

The other point is that if we were offered 100 to 1 in this scenario after tails,tails,heads, we would be insane not to take it.

So clearly there is some objective bound to our confidence, even if we can't pinpoint the exact number given the condition of zero information about the underlying distribution.

[jason1990]

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do you think the information in the problem is sufficient to come to a logical decision.


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if it isn't then a million flips won't allow us to come to a logical conclusion either.

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Abstractly, you may be right. So does this mean you think it is or it isn't?

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??? How can a million flips not give us enough information to determine whether a 7:4 bet on heads is good? I mean, if the next 999,997 flips all end up being tails, I sure do think the 7:4 bet on heads is a bad wager.

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Of course. Any practical person agrees with you. That is why I said he may be right abstractly. But he may be wrong practically. It is at least conceivable that 3 flips is not enough information, but a million flips is -- on a practical level.

Still, though, what decision would you make and why? Do you think the information in the problem can be used to determine an optimal decision?

[luckyme]

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do you think the information in the problem is sufficient to come to a logical decision.


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if it isn't then a million flips won't allow us to come to a logical conclusion either.

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Abstractly, you may be right. So does this mean you think it is or it isn't?

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Well, if we flipped 1M and 666,666 were tails, we'd turn down the 7 to 4 using the same reasoning as we'd use at 3 flips, 33 flips. With such a small sample we'd not bet much at only 3 flips because the variance will be high but I'd take the bet at 9 to 4.

The number of flips changes our confidence level and has no bearing on whether our decision is logical or not.

luckyme

[PairTheBoard]

Isn't the percentage of heads in N flips the maximum liklihood estimator - or something like that - for the true parameter p=P(heads)? Why aren't people just grabbing that, using the "best" estimate for p=1/3, say we should be getting 2-1 odds and decline the bet? Sure we have a wide confidence interval around p=1/3 with a low confidence, but how bad could it be? 7-5 is not even close.

Aren't people hesitating to answer because they can't really believe the condition of "no information" about how coins are bent? They have a hunch that despite this Defined Abstraction, maybe coins don't really get bent that much most of the time so their hunch is outweighing the evidence of the 3 flips? If bending coins generally only makes them slightly bias, then in 3 flips you can't get any closer to 50-50 than 1/3-2/3 anyway. So go on your hunch and take the bet.

What does the posterior distribution after these 3 flips look like assuming a Uniform prior distribution? That math has been done somewhere on the Probability Forum. Does it not shift away from the center as much as we might think from the 33% heads percentage?

Edit: oh. It's 7-4 odds not 7-5. Well that's a little closer but it's still not 2-1.

PairTheBoard

[JayTee]

I have no idea how to think about this probabilistically. I think that it would be OK to wager though. Someone please point out the flaw in my thought process.

I think that before the flip we can assume that for the purpose of determining whether or not to wager that there is an equal chance of all possibilities that favor either heads or tails. This means that for the purpose of wagering we can assume a fair coin. I don't think that three flips is enough information to shift the probability so that the 7:4 wager is bad. My main reason for thinking this is that if a fair coin is flipped 3 times there is no way for it to land heads and tails an equal amount.

Again, I can't back this up. Just my thoughts.

[KipBond]

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The number of flips changes our confidence level and has no bearing on whether our decision is logical or not.

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After those 1M flips, I'd be confident that I would be taking the worst of it in the 7:4 bet on heads (EV = -1/3). After 3 flips, I wouldn't take the bet, even though it might be +EV.

[jason1990]

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Well, if we flipped 1M and 666,666 were tails, we'd turn down the 7 to 4 using the same reasoning as we'd use at 3 flips, 33 flips. With such a small sample we'd not bet much at only 3 flips because the variance will be high but I'd take the bet at 9 to 4.

The number of flips changes our confidence level and has no bearing on whether our decision is logical or not.

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Cool, thanks. So just to clarify, you would take the bet as long as the odds were better than 2 to 1. This is because you saw 2 tails and 1 head. You are also saying that this answer is a logical consequence of the problem statement, not a matter of personal preference. Similarly, if you had seen 1 tail and 2 heads, you would take the bet as long as the odds were better than 1 to 2, and this is also an optimal answer. Have I understood you correctly?

[PLOlover]

if you're a loose player, gamble.

if you think losing will put your friend on tilt, gamble.

if you can hedge successfully, gamble.

if you likie to gamble , gamble.

[luckyme]

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Well, if we flipped 1M and 666,666 were tails, we'd turn down the 7 to 4 using the same reasoning as we'd use at 3 flips, 33 flips. With such a small sample we'd not bet much at only 3 flips because the variance will be high but I'd take the bet at 9 to 4.

The number of flips changes our confidence level and has no bearing on whether our decision is logical or not.

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Cool, thanks. So just to clarify, you would take the bet as long as the odds were better than 2 to 1. This is because you saw 2 tails and 1 head. You are also saying that this answer is a logical consequence of the problem statement, not a matter of personal preference. Similarly, if you had seen 1 tail and 2 heads, you would take the bet as long as the odds were better than 1 to 2, and this is also an optimal answer. Have I understood you correctly?

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no.
I'd have said those things if that's what I meant:-).

luckyme

[jason1990]

Well, I was hoping to wait longer before posting any solutions, but that has turned out to not be possible. I hope to post three different solutions. They all involve calculus. This is a bit of a rehash of something we discussed recently in the probability forum. Here is the first solution.

We use Bayes theorem. Let A be the event that the first three flips are T,T,H. Let B be the event that the fourth flip is H. We want to compute the probability of B given A, which is

P(B | A) = P(B and A)/P(A) = P(T,T,H,H)/P(T,T,H).

Before computing these two probabilities, we need to think a little about this bent coin. Imagine we could flip this particular bent coin over and over again. In the long run, the fraction of heads would converge to some number P. This number depends on the bend in the coin and is completely unknown to us. Since we have no information about P, we might want to assume that P is equally likely to be any number between 0 and 1. In other words, P is uniformly distributed on [0,1]. In other words, P has density f(p) = 1 for all p.

The formula for computing P(T,T,H) in the presence of this unknown, random parameter is given by the following integral:

P(T,T,H) = \int_0^1 P(T,T,H | P = p)*f(p) dp
= \int_0^1 (1 - p)*(1 - p)*p*1 dp.

If you do this integral, you get 1/12. Similarly,

P(T,T,H,H) = \int_0^1 (1 - p)^2 p^2 dp = 1/30.

So P(B | A) = (1/30)/(1/12) = 12/30 = 2/5. In other words, given our observations of T,T,H, our estimate for the probability of H on the fourth flip is 2/5. We should accept the wager at any odds greater than 3 to 2. Since we are offered 7 to 4, we should accept.

This same calculation (in greater abstraction) is carried out in some lecture notes on Persi Diaconis's website at http://www-stat.stanford.edu/~cgates/PERSI/courses/stat_121/lectures/exch/ . Search the page for the text "bent coin". There is also a nice discussion about Adam (of Adam and Eve) witnessing the sun rising n days in a row and estimating the probability that it will rise the next day.

[jason1990]

Okay, here is the second solution. The first solution was based on a single assumption. We assumed that "no information" about the bend in the coin translated into the density f(p) = 1 for all p. f97tosc has pointed out that Edwin Jaynes, in his 1968 article "Prior probabilities," argues for a different density. He suggests that "no information" should be modeled by the density f(p) = 1/(p(1 - p)). If we follow the first solution, but use this new density instead of the uniform density, we have the following:

P(B | A) = P(T,T,H,H)/P(T,T,H)
= (\int_0^1 (1 - p)^2 p^2 f(p) dp)/(\int_0^1 (1 - p)^2 p f(p) dp)
= (\int_0^1 (1 - p)p dp)/(\int_0^1 (1 - p) dp)
= (1/6)/(1/2) = 1/3.

So Jaynes' estimate for the probability of heads on the fourth flip is 1/3. We should accept the wager only if the odds are at least 2 to 1. Since we are only getting 7 to 4, Jaynes recommends we reject the wager.

Jaynes' justification for using this particular density is based on sound, logical reasoning. His result also agrees with the maximum likelihood estimator, which has an excellent track record in producing accurate statistical results. It would certainly be foolish to summarily dismiss Jaynes' recommendation in this case.

[jason1990]

Finally, the third solution. In the first solution, we said we had no information about the numeric value of P. So we assumed that all numeric values were equally likely. But it would have been just as true to say that we had no information about the order of magnitude of P. So we could have assumed that all orders of magnitude were equally likely. This would have given us a different result. Harold Jeffreys proposed a resolution to this conflict by deriving a density which is meant to express "no information" in a more generic sense. The Jeffreys prior is proportional to the square root of the so-called Fisher information. It is f(p) = 1/sqrt{p(1 - p)}. This time, we get

P(B | A) = P(T,T,H,H)/P(T,T,H)
= (\int_0^1 (1 - p)^2 p^2 f(p) dp)/(\int_0^1 (1 - p)^2 p f(p) dp)
= (\int_0^1 (p(1 - p))^(3/2) dp)/(\int_0^1 p^(1/2)(1 - p)^(3/2) dp)
= (3pi/128)/(pi/16) = 3*16/128 = 3/8.

So the Jeffreys density tells us that the probability of heads on the fourth flip is 3/8. We should therefore accept the wager at any odds greater than 5 to 3. The offered odds of 7 to 4 are slightly better than this, so we should accept.

The derivation of the Jeffreys density is based on sound, logical reasoning, and it also has a strong track record in statistical estimation.

In conclusion, we have seen three different methods for translating the concept of "no information" into actual mathematics. None of them are illogical, and all of them are known to have produced good experimental results in applications. Yet each one gives us a different probability in this particular problem, and there is no consensus between them regarding what decision we ought to make. I throw this out there as food for thought. Hopefully someone will understand it enough to get something out of it.

[KipBond]

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Hopefully someone will understand it enough to get something out of it.

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Thanks, Jason. What I got out of this is that in this case, your subjective assumptions about the prior distribution determines what your resulting decision is. The math/probability calculations are a good tool to double check various assumptions to see if your decision is at least rational based off of your assumptions.

[luckyme]

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In conclusion, we have seen three different methods for translating the concept of "no information" into actual mathematics. None of them are illogical, and all of them are known to have produced good experimental results in applications. Yet each one gives us a different probability in this particular problem, and there is no consensus between them regarding what decision we ought to make. I throw this out there as food for thought. Hopefully someone will understand it enough to get something out of it.

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whew. I was starting to worry that the TTH run would be turned into a favorite for heads by expert mathguessians.
It's a bit surprising that in a purely abstract test one of the tactics hasn't been a clear winner and the others just coming in decent range just by having reasonable starts.
I'm happy as long as they come out around 3/8 for heads on the 4th flip since that was my 'gun to head' position( no claim to accuracy as I was using an averaging method).

What do we know about the chances of this coin being biased in a specific direction at this stage? 75% of the time it will be biased in favor of tails ( amount of bias unspecified) is the range my averaging scratching come up with. What does any of Jasons rigorous methods produce?

thanks jason,
you make me wish I wasn't a sklamoron at times like this,

luckyme
( unfortunately it usually passes just after lunch)

[jason1990]

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What do we know about the chances of this coin being biased in a specific direction at this stage? 75% of the time it will be biased in favor of tails ( amount of bias unspecified) is the range my averaging scratching come up with.

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Sorry, but I do not have time to go into details right now, and I will be gone for a couple of weeks after today. There is an old thread in the probability forum (maybe someone can link you to it) in which we discussed computing the distribution of P, given our observations. This is the kind of thing you want to look at if you want to answer this question. The answer, of course, will depend on the prior density you decide to use.

[luckyme]

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What do we know about the chances of this coin being biased in a specific direction at this stage? 75% of the time it will be biased in favor of tails ( amount of bias unspecified) is the range my averaging scratching come up with.

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Sorry, but I do not have time to go into details right now, and I will be gone for a couple of weeks after today. There is an old thread in the probability forum (maybe someone can link you to it) in which we discussed computing the distribution of P, given our observations. This is the kind of thing you want to look at if you want to answer this question. The answer, of course, will depend on the prior density you decide to use.

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thanks jason, I'll poke around in the prob forum, should spend time there anyway,
looking forward to when you're back,

luckyme