Here’s a simple question about game theory.

Imagine that you think (a) that coherent credences need not satisfy countable additivity, and you think (b) that sometimes mixed strategies are optimal, and so we should consider all mixed strategies in deciding what an agent should do. I’m not particularly fond of either (a) or (b), but I know that I have readers who endorse each, and I suspect that I have readers who endorse the conjunction of the two.

Now imagine we’re trying to analyse a game where a player has a countable infinity of option in front of them. Should we take one of the player’s options to be the mixed strategy where the player has, for each option, probability 0 of taking that option? If so, then (a) how do we evaluate how good a strategy that is, and (b) are there any cases where this is the optimal strategy?

Here’s a more particular instance of this puzzle. Row and Column are playing a game of infinite matching pennies. Each player has to select a positive integer. Row wins if they select the same number, Column wins if they select a different number. I imagine that the opponent of countable additivity will want to say that the solution to the game is that both players play the strategy of selecting a random number, with each number having probability zero of being selected. But I don’t know how the opponent of countable additivity figures out what the expected return of this strategy (or any other infinitary strategy) is for each player, so I don’t know why they would think it is an equilibrium. So what should someone who believes (a) and (b) say about this game?

Posted by Brian Weatherson in *Uncategorized*