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
So saying "yea" gives 0.9*1000 + 0.1*100 = 910 expected donation.
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"?
The source is here. I'll restate the problem in simpler terms:
You are one of a group of 10 people who care about saving African kids. You will all be put in separate rooms, then I will flip a coin. If the coin comes up heads, a random one of you will be designated as the "decider". If it comes up tails, nine of you will be designated as "deciders". Next, I will tell everyone their status, without telling the status of others. Each decider will be asked to say "yea" or "nay". If the coin came up tails and all nine deciders say "yea", I donate $1000 to VillageReach. If the coin came up heads and the sole decider says "yea", I donate only $100. If all deciders say "nay", I donate $700 regardless of the result of the coin toss. If the deciders disagree, I don't donate anything.
First let's work out what joint strategy you should coordinate on beforehand. If everyone pledges to answer "yea" in case they end up as deciders, you get 0.5*1000 + 0.5*100 = 550 expected donation. Pledging to say "nay" gives 700 for sure, so it's the better strategy.
But consider what happens when you're already in your room, and I tell you that you're a decider, and you don't know how many other deciders there are. This gives you new information you didn't know before - no anthropic funny business, just your regular kind of information - so you should do a Bayesian update: the coin is 90% likely to have come up tails. So saying "yea" gives 0.9*1000 + 0.1*100 = 910 expected donation. This looks more attractive than the 700 for "nay", so you decide to go with "yea" after all.
Only one answer can be correct. Which is it and why?
(No points for saying that UDT or reflective consistency forces the first solution. If that's your answer, you must also find the error in the second one.)