A purely abstract decision problem

[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...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]

[ QUOTE ]
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]

[ QUOTE ]
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.

[/ QUOTE ]

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]

[ QUOTE ]
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]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]
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