Quote:
Originally Posted by PairTheBoard
I think this is quite a bit different from something like the prisoner's paradox. ... Suppose, going into the experiment, they get together and agree to operate in such a way as to maximize thier results as a group.
What do you mean by "their results"? Your post suggests that you would have the probabilists maximize their expected team score. Why? I would prefer to have them maximize the probability that they beat the slots players (who are all picking from Urn 2). I would interpret the final outcome of the game as binary: either the probabilists win or they lose. (Suppose draws are decided by a coin toss, if necessary.) They should try to maximize the probability that they win.
As an individual probabilist, I might reason that for each fixed set of choices and outcomes for my teammates and the opposing team, I can only affect whether we win or lose in the cases where the scores differ by at most 1. In those cases, I maximize the probability (given what I know) that we will win by maximizing the probability that I choose a white stone, and thereby score a point. I do that by choosing from Urn 1.
Following this strategy (which amounts to every probabilist selecting from Urn 1), the probability that the probabilists will win, given the information they have prior to the game, is a little less than 64.3%. (Calculation sketch is below.)
My claim is that the probabilists can instead choose, prior to the game, to act in such a way so that the probability they will win, given the information they have prior to the game, is a little more than 67.3%. In fact, we may have them act according to your suggestion here:
Quote:
Originally Posted by PairTheBoard
The natural criteria described above would have a probabilist switch if he sees 488 white balls or less.
Let us sketch the calculations that lead to the numbers I claimed above. I will use normal approximations. To the skeptics who think the qualitative result will be reversed if the calculations are done exactly, I will leave it to them to do the exact calculations on their own machines.
Let us first consider Strategy A, which is to always draw from Urn 1. All probabilities, unless otherwise noted, will be conditioned on what is known prior to anyone drawing from Urn 1.
Let k be the number of white stones in Urn 1. Let p = k/1000. The probabilists' score will be approximately k + sU, where s
2 = 1000p(1 - p) and U ~ N(0,1). The slots players' score will be approximately 490 + tV, where t
2 = 1000(0.49)(0.51) and V ~ N(0,1). Note that U, V, and k are all independent.
Their score difference is therefore D = k - 490 + (s
2 + t
2)
1/2Z, where Z ~ N(0,1). The probability the probabilists win, given k, is thus
where Φ is the cumulative distribution function of a standard normal random variable. We therefore have
According to Mathematica, this evaluates to 0.642506.
Now let us consider Strategy B, which is to draw from Urn 1 if and only if we see 489 or more white stones after drawing 999 stones from the urn. In this case, if k < 489, then all probabilists will draw from Urn 2. Under the normal approximation, the probability of a draw is zero, so the probabilists win with probability 1/2. (Or, if draws are decided by a coin toss, then the probability the probabilists win is 1/2 without any normal approximation.)
If k > 489, then all probabilists will draw from Urn 1, and P(D > 0 | k) will be the same as above.
Therefore,
According to Mathematica, this evaluates to 0.673515.