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Figure: Maximin strategies of player 1
u1 u1 (p, T ) = 1 − 2 p u1 (p, H) = 2 p − 1
p∗
p
m1 (p) = {u1 (p, T ), u1 (p, H)}
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The thick (kinked) line represents m1 (p), i.e. the worst outcome for player 1 for each value of p ∈ [0, 1]. In order to maximize m1 (p), we have to compute the intersection of u1 (p, H) and u1 (p, T ):
u1 (H, q) and u1 (T, q)
u1 (H, q) = q u1 (H, H) + (1 − q) u1 (H, T ) ⇔ u1 (H, q) = 2 q − 1 u1 (T, q) = q u1 (T, H) + (1 − q) u1 (T, T ) ⇔ u1 (T, q) = −2 q + 1
u1 (p, T ) and u1 (p, H)
u1 (p, H) = p u1 (H, H) + (1 − p) u1 (T, H) ⇔ ⇔ u1 (p, T ) = 2 p + (1 − p) (−1) u1 (p, T ) = 3 p − 1 u1 (p, T ) = p u1 (H, T ) + (1 − p) u1 (T, T ) ⇔ ⇔ u1 (p, T ) = p (−1) + (1 − p) u1 (p, T ) = −2 p + 1