Ordinary probability theory and expected utility are sufficient to handle this puzzle. You just have to calculate the expected utility of each strategy before choosing a strategy. In this puzzle a strategy is more complicated than simply putting some number of coins in the machine: it requires deciding what to do after each coin either succeeds or fails to succeed in releasing two coins.
In other words, a strategy is a choice of what you’ll do at each point in the game tree—just like a strategy in chess.
We don’t expect to do well at chess if we decide on a course of action that ignores our opponent’s moves. Similarly, we shouldn’t expect to do well in this probabilistic game if we only consider strategies that ignore what the machine does. If we consider all strategies, compute their expected utility based on the information we have, and choose the one that maximizes this, we’ll do fine.
So, let me try again to explain why I think this is missing the point… I wrote “a single probability value fails to capture everything you know about an uncertain event.” Maybe “simple” would have been better than “single”?
The point is that you can’t solve this problem without somehow reasoning about probabilities of probabilities. You can solve it by reasoning about the expected value of different strategies. (I said so in the OP; I constructed the example to make this the obviously correct approach.) But those strategies contain reasoning about probabilities within them. So the “outer” probabilities (about strategies) are meta-probabilistic.
[Added:] Evidently, my OP was unclear and failed to communicate, since several people missed the same point in the same way. I’ll think about how to revise it to make it clearer.
Ordinary probability theory and expected utility are sufficient to handle this puzzle. You just have to calculate the expected utility of each strategy before choosing a strategy. In this puzzle a strategy is more complicated than simply putting some number of coins in the machine: it requires deciding what to do after each coin either succeeds or fails to succeed in releasing two coins.
In other words, a strategy is a choice of what you’ll do at each point in the game tree—just like a strategy in chess.
We don’t expect to do well at chess if we decide on a course of action that ignores our opponent’s moves. Similarly, we shouldn’t expect to do well in this probabilistic game if we only consider strategies that ignore what the machine does. If we consider all strategies, compute their expected utility based on the information we have, and choose the one that maximizes this, we’ll do fine.
I’m saying essentially the same thing Jeremy Salwen said.
So, let me try again to explain why I think this is missing the point… I wrote “a single probability value fails to capture everything you know about an uncertain event.” Maybe “simple” would have been better than “single”?
The point is that you can’t solve this problem without somehow reasoning about probabilities of probabilities. You can solve it by reasoning about the expected value of different strategies. (I said so in the OP; I constructed the example to make this the obviously correct approach.) But those strategies contain reasoning about probabilities within them. So the “outer” probabilities (about strategies) are meta-probabilistic.
[Added:] Evidently, my OP was unclear and failed to communicate, since several people missed the same point in the same way. I’ll think about how to revise it to make it clearer.