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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?

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    $\begingroup$ Sorry, I thought you were looking for an equilibrium -- I overlooked that you're using fixed probabilities for Alice. I've deleted my answer. $\endgroup$
    – joriki
    Apr 25, 2016 at 14:54

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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 $-0.5$.

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    $\begingroup$ Oh I see. I missed that this was a zero sum game. Thx! $\endgroup$
    – user274065
    Apr 25, 2016 at 14:57

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