| Here's how you solve for an equilibrium. Koufax wants
to choose p, the probability of playing L so that Gehrig is indifferent between playing L
and playing R.
Gehrig's expected payoff from playing L equals pQ + (1-p).
Gehrig's expected payoff from playing R equals p + (1-p)Q.
How do we get these? Well, p of the time Koufax plays L and (1-p) of the time he plays R.
Therefore, if Gehrig plays L, he gets Q p of the time (hence pQ) and he gets 1 (1-p) of
the time (hence (1-p)). Koufax wants to choose p so that Gehrig is indifferent between
playing L or R.
Why? Well, If Gehrig strictly preferred one of the two alternatives, say L, then in
response Koufax would choose the opposite, in this case R, and then Gehrig wouldn't want
to choose L. So, it's no equilibrium. Got it?
Setting these two equations equal gives
pQ + (1-p) = p + (1-p)Q
or
(1-p)(1-Q) = p(1-Q)
which implies that p must equal 1/2. Using the same logic, you can show that Gehrig should
also play L half of the time. Therefore, in equilibrium, they each randomize evenly
between L and R. This means that half of the time, they match, and half of the time, they
do not.
So, the expected number of goals equals (1+Q)/2.
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