Game theory probability I have a variant of two-finger morra game, where the winner is determined by the parity of the sum of the two numbers thrown, but the amount won or lost is the product of the two numbers. There are two player - Alice and Bob
If Alice plays:
1 finger with probability $p$ 
2 fingers with probability $1 − p$
Alice wins when outcome is even, and Bob wins if it's odd.
What’s the expected payoff that Bob can achieve if he knows $p$? 
How should Alice choose $p$ such that Bob’s payoff is minimized?
 A: I suppose that Alice wins if the sum of the numbers is even. If $a \{1,2\}$ and $b \in \{1,2\}$ are the strategies played by Alice and Bob, respectively. The payoff of Bob is:
$$B(a,b) = \begin{cases}
-1 & \text{if}~a = b = 1 \\
2 & \text{if}~a = 1, b = 2 ~\text{or}~ a=2, b=1\\
-4 & \text{if}~a = b = 2 \\
\end{cases}.$$ 
The expected payoff of Bob knowing $p$ is:
$$\mathbb{E}[B(a, b) | b] =  \begin{cases}
B(1,1)p + B(2,1)(1-p) & \text{if}~b = 1 \\
B(1,2)p + B(2,2)(1-p) & \text{if}~b = 2 
\end{cases}\\=  \begin{cases}
-1p + 2(1-p) & \text{if}~b = 1 \\
2p -4(1-p) & \text{if}~b = 2
\end{cases}\\ = \begin{cases}
2 -3p  & \text{if}~b = 1 \\
6p-4 & \text{if}~b = 2
\end{cases}$$ 
Bob will choose strategy $b = 1$ if 
$$2-3p > 6p-4 \Rightarrow p<\frac{2}{3}.$$
Of course, he will choose $b=2$ if $p>\frac{2}{3}$. If $p=\frac{2}{3}$, there is no difference between $a$ and $b$.
Now, if $p < \frac{2}{3}$, then Bob plays $b=1$ and his average payoff is $2-3p \in [0, 2]$. On the other hand, if Bob plays $b=2$, then his average payoff is $6p-4 \in [0, 2]$. The minimum of the Bob's payoff is $0$ and it is attained for $p = \frac{2}{3}$. This means that Alice must choose $p = \frac{2}{3}$ to minimize Bob's payoff.
Notice that $p = \frac{2}{3}$ is the component of Nash equilibrium for Alice in the space of mixed strategies.
