$2$ men are playing a game: they are wearing countably infinitely many hats on their heads. The hats are either black or white with probability $\frac 12$. They see the other's man hats but cannot see the hats on their own head. Without communicating with each other, they simultaneously point to a hat on their own head. They win the game if the two hats they've chosen are both white. The players devise a strategy before hats are put on their heads.
I'm looking for a strategy with a success probability greater than $\frac 13$.
If both players choose a random hat, they win the game with probability $\frac 14$.
If both players choose the position of the first white hat that is on the other player's head, they have probability $\frac 1 3$ of success: if you let $X_i$ and $Y_j$ denote the colors of the $i$-th and $j$-th hats for the first and second player respectively, you're looking for $$P\left(\bigcup_{i,j\geq 1}\left( (X_i=w)\cap \bigcap_{k=1}^{i-1}(Y_k=b)\cap (Y_i=w) \cap (Y_j=w)\cap \bigcap_{k=1}^{j-1}(Y_k=b)\cap (X_j=w) \right)\right)$$
which, under obvious independence assumptions, is $\frac 13$.
Any ideas ? It would also be interesting to extend the game to $n$ players. In that case, the strategy I suggested has probability success $\frac{1}{2^n-1}$ which gets asymptotically close to $\frac 1{2^n}$ (the chance of success of the dumbest strategy).
You can get them to win with probability 7/20. As of today it is still an open problem whether we can do better, but there is an upper bound at 0.361607.
See here: https://mathmondays.com/hat-game and https://arxiv.org/abs/1407.4711.
Also see this related question: Coin Flipping Riddle.