Quote:
Originally Posted by gumpzilla
I'm curious what your take on this problem is. Somebody posted about it a while ago in probability and I never figured out a good resolution to it.
I assume you are referring to
this thread. There, the problem is described somewhat differently. Mathematically, it is the same, but instead of prisoners living or dying, there are teams of players scoring points. I prefer this formulation, because I think it makes the counterintuitive elements of the problem more transparent.
I see there being two distinct counterintuitive elements here. One has to do with decision theory, the other with probability theory.
Regarding decision theory, the problem presents an example of a well-known situation where, if each player optimizes his or her individual behavior, then the team's performance will nonetheless be suboptimal. While this is interesting, it is not particularly novel. After all, the original prisoner's paradox makes exactly this point, but presents it in
much simpler packaging.
Regarding probability theory, the problem presents a situation which seems to defy some basic intuition we have about the relation between probabilities of individual events and long run frequencies. Intuitively, the question is, how can each player on the team of probabilists have a 50% chance of earning a point, and yet the team, as a whole, only earns 48% (or 47.9%) of the possible points in the long run?
Let us begin by identifying the probabilistic principle that this situation appears to violate. Suppose we have a sequence of events A
1, A
2, A
3, ..., and suppose these events are
exchangeable. (Exchangeability can be viewed as a generalization of independence.) Let S
n be the proportion of A
1, ..., A
n that occur. It can be proven that:
Quote:
If P(An) = p for all n, then
and
(This theorem can be viewed as a generalization of the law of large numbers.)
And yet, in this problem, we seem to have a situation that violates this. Let A
n the event that the n-th player selects the white stone. It seems that P(A
n) = 0.5 for all n. And yet, P(S
n → 0.48) = 0.5 and P(S
n → 0.479) = 0.5, so that E[lim S
n] = 0.4795. How can this be?
As with so many problems like this, I think the confusion arises because of our failure to explicitly account for the information on which the probabilities are based. Since all probabilities are conditional, the above theorem should technically read:
Quote:
If P(An | Ω) = p for all n, then
and
Now let Ω be the information I have after drawing 479 white stones and 520 black stones from the urn. If I am the first player, then it is true that P(A
1 | Ω) = 0.5. However, for every n > 1, we have P(A
n | Ω) = 0.4795.
All the
other players are exchangeable with respect to my information. But there is an informational asymmetry between myself and them. I know that I first drew 479 white stones and 520 black stones. I do not know what the first 999 draws are for any other player.
If we look
only at players who first drew 479 white stones and 520 black stones, then, based on my information, each one individually has a 50% chance of drawing the white stone. Moreover, for this subpopulation, P(S
n → 1 | Ω) = 0.5 and P(S
n → 0 | Ω) = 0.5, so that E[lim S
n | Ω] = 0.5, just as it ought to be.
I think what throws people off is that instead of thinking about P(A
n | Ω), they (implicitly) think about P(A
n | Ω
n), where Ω
n is the information the n-th player has after drawing 999 stones. In that case, it is certainly true that by symmetry, P(A
n | Ω
n) = 0.5 for all n. But this (as the puzzle demonstrates) does not tell us anything about the team's long term point distribution. If we neglect the role of information in probability, and think only in terms of P(A
n), then we leave the door open to this kind of ambiguity and confusion.