Is this some kind of inverse Monty Hall problem? Counterintuitively, the second solution is incorrect.
If everyone pledges to answer “yea” in case they end up as deciders, you get 0.5*1000 + 0.5*100 = 550 expected donation.
This is correct. There are 10 cases in which we have a single decider and win 100 and there are 10 cases in which we have a single non-decider and win 1000, and these 20 cases are all equally likely.
you should do a Bayesian update: the coin is 90% likely to have come up tails.
By calculation (or by drawing a decision tree), this is indeed correct: prob(T|D1)=0.9
Let us assume that every decider follows your reasoning and says “yea” (which is, I suppose, what you intended). Let me first give a non-explanation for why the reasoning above cannot be correct: Notice that every decider will always say “yea” and this is no different from the deterministic strategy of always saying “yea”. It’s irrelevant what non-deciders say or how the deciders came to their decision. Thus the overall expected donation is again 0.5*1000 + 0.5*100 = 550.
Above you correctly calculated the conditional expectation E( donation | D1) = 910. Conditioned on D1, we expect to win 910 if everyone says “yea”. This number is not directly comparable with the 700, the payoff we’d get for saying “nay” (regardless of whether D1 holds or not). Assume we would play a variant of the game in which the players were promised that D1 always holds; then it’d indeed be better for them to always say “yea”. The conditional expectation, however, does not tell us how well this strategy performs in the original game. The original game allows, after all, that D1 does not hold. To find out how well a strategy performs we must take into account that possibility.
Consider the expected donation conditioned on the fact that D1 does not hold. It turns out that E( donation | not D1) = 0.9*100 + 0.1*1000 = 190 is much smaller than 700. Thus, if all players follow the “yea” strategy, we’ll be worse off since D1 might happen to be false. This can be made precise: The probability that D1 holds is exactly 0.5. Thus, using “yea” for every player, our overall performance will be E( donation ) = 0.5* E( donation | D1) + 0.5* E( donation | not D1) = 0.5*910 + 0.5*190 = 550 as before.
Are you claiming that it’s rational to precommit to saying “nay”, but upon observing D1 (assuming no precommitment happened) it becomes rational to say “yea”?
Actually I find this problem quite perplexing, just like an optical illusion that makes you see different things.
Yes, what I am claiming is that if they observe D1, they should say “yea”. The point is that only player 1 knows whether D1 holds, and no other player can observe D1. Sure, there will be some player i for which Di holds, but you cannot calculate the conditional expectation as above since i is a random variable. The correct calculation in that case is as follows:
Run the game and let i be a player that was selected as a decider. In that case the expected donation conditioned on the fact that Di holds is equal to the expected donation of 550, since i is a decider by definition and thus Di always holds (so the condition is trivial).
Is this some kind of inverse Monty Hall problem? Counterintuitively, the second solution is incorrect.
This is correct. There are 10 cases in which we have a single decider and win 100 and there are 10 cases in which we have a single non-decider and win 1000, and these 20 cases are all equally likely.
By calculation (or by drawing a decision tree), this is indeed correct: prob(T|D1)=0.9
T coin came up tails
D1 player 1 is a decider
Let us assume that every decider follows your reasoning and says “yea” (which is, I suppose, what you intended). Let me first give a non-explanation for why the reasoning above cannot be correct: Notice that every decider will always say “yea” and this is no different from the deterministic strategy of always saying “yea”. It’s irrelevant what non-deciders say or how the deciders came to their decision. Thus the overall expected donation is again 0.5*1000 + 0.5*100 = 550.
Above you correctly calculated the conditional expectation E( donation | D1) = 910. Conditioned on D1, we expect to win 910 if everyone says “yea”. This number is not directly comparable with the 700, the payoff we’d get for saying “nay” (regardless of whether D1 holds or not). Assume we would play a variant of the game in which the players were promised that D1 always holds; then it’d indeed be better for them to always say “yea”. The conditional expectation, however, does not tell us how well this strategy performs in the original game. The original game allows, after all, that D1 does not hold. To find out how well a strategy performs we must take into account that possibility.
Consider the expected donation conditioned on the fact that D1 does not hold. It turns out that E( donation | not D1) = 0.9*100 + 0.1*1000 = 190 is much smaller than 700. Thus, if all players follow the “yea” strategy, we’ll be worse off since D1 might happen to be false. This can be made precise: The probability that D1 holds is exactly 0.5. Thus, using “yea” for every player, our overall performance will be E( donation ) = 0.5* E( donation | D1) + 0.5* E( donation | not D1) = 0.5*910 + 0.5*190 = 550 as before.
Are you claiming that it’s rational to precommit to saying “nay”, but upon observing D1 (assuming no precommitment happened) it becomes rational to say “yea”?
Actually I find this problem quite perplexing, just like an optical illusion that makes you see different things.
Yes, what I am claiming is that if they observe D1, they should say “yea”. The point is that only player 1 knows whether D1 holds, and no other player can observe D1. Sure, there will be some player i for which Di holds, but you cannot calculate the conditional expectation as above since i is a random variable. The correct calculation in that case is as follows:
Run the game and let i be a player that was selected as a decider. In that case the expected donation conditioned on the fact that Di holds is equal to the expected donation of 550, since i is a decider by definition and thus Di always holds (so the condition is trivial).