Consider a variant of two-finger morra where the winner is still determined by the parity of the sum of the two numbers thrown (if it’s even, Alice wins money and if it’s odd, Bob wins money), but the amount won or lost is the product of the two numbers.
If Alice plays 1 finger with probability p and 2 fingers with probability 1 − p, what’s the expected payoff she gets even if Bob knows p and can chose his strategy based on this knowledge? How should Alice choose p such that this payoff is maximized?
Here's what I have so far, (assuming P(A=1) = p),
I know that the expected payoff for Alice is at least:
min(p - 2(1-p), -2p + 4(1-p))
And the expected payoff for Bob is at least (assuming P(B=1) = p),
min(p - 2(1-p), -2p + 4(1-p))
Which means that the optimal strategy for Alice would be to play 1 finger with probability 2/3, and the same goes for Bob.
I know how to get the expected payoff using these probabilities, but I don't know how to get the expected payoff for Alice if Bob knows p and can choose his strategy based on this knowledge. And how does Alice choose p such that the payoff is maximized?
You should not assume the chance Bob plays $1$ is the same as the chance that Alice plays $1$, so let Bob play $1$ with probability $q$. Now when Alice plays $1$ her payoff is $2(1-q)-1q$ and when she plays $2$ her payoff is $2q-4(1-q)$. Her equilibrium strategy is to play these in a ratio that makes her expected payoff independent of $q$.