I am currently studying the blocker effect in the game Kuhn Poker Kuhn Poker Wikipedia. Below you can see the game tree. At the first node, both players have 1/3 of one of the three possible cards (K, Q, J). These distributions can now change depending on the strategy of player 1 at the first node. Assuming player 1 always bets with the K, he obviously cannot hold this hand at the second node. At the same time, however, the probability distribution of player 2 also changes. For him, the probability of holding a K increases, since player 1 can only hold Q or J.
So much for the setting. I would now like to formulate the whole thing mathematically, see the second picture below. I could calculate the conditional probability for player 1 at the second node using Bayes' rule. For the probability distribution for player two I am stuck. How do I have to transform so that the initial probability p(h2 = K) = 1/3 appears in the formula? What is written below is more or less: p(h2 = K | h1 = K) = 0 (because both cannot hold the K), p(h2 = K | h1 = Q) = 1/2 and p(h2 = K | h1 = J) = 1/2… But somewhere the initial probability p(h2 = K,Q,J) = 1/3 must still occur, or not?
Best Answer
Thanks to @Henry here's the solution to my problem: