Consider the 2-finger.
Bob pays Alice $\$(a + b)$ if $a + b$ is even
Alice pays Bob $\$(a + b)$ if $a + b$ is odd
Suppose Alice plays $one\; finger$ with probability $\frac 12$ and $two\; fingers$ with probability $\frac12$.
If Alice plays $1$ then Alice has expected return $$2q − 3(1 − q) = 5q − 3$$
If Alice plays $2$ then Alice has expected return $$−3q + 4(1 − q) = 4 − 7q$$
If Alice plays her best option, she expects to get $$max(5q − 3, 4 − 7q)$$
If Bob plays $1$ then Bob expects to get $$−2p + 3(1 − p) = 3 − 5p$$
If Bob plays $2$ then Bob expects to get $$3p − 4(1 − p) = 7p − 4$$
If Bob plays his best option, he expects to get $$max(3 − 5p, 7p − 4)$$
If Bob knows this and plays optimally, what is Alice's expected reward?
I know the answer is -0.5 (given as the answer by my professor). However, how do I get that. I know that once I put in $\frac12$ for Bob, I get the max to be:
$$max(0.5,-0.5)$$
How does that translate into the answer?
Your work so far is right. This is a zero-sum game, so the payoff for Alice is minus the payoff for Bob. As you found, Bob gets $\max(0.5,-0.5)=0.5$, so Alice gets $-0.5$.