Reason 3
In a game without a pure strategy Nash equilibrium, a mixed strategy may result in a Nash equilibrium.
Consider a modified version of Benjamin Bunny and Al Paca in which there are just two fields, A and B. If Al sprays A and Ben chooses A for foraging then Al wins and Ben loses, and so on. Their wins and losses are shown in the payoff table.
Al | |||
A | B | ||
Ben | A | -1, 1 | 1, -1 |
B | 1, -1 | -1, 1 |
The best responses in this game are
b^{Ben}(A^{Al}) = B | b^{Al}(A^{Ben}) = A |
b^{Ben}(B^{Al}) = A | b^{Al}(B^{Ben}) = B |
The game seems to have no Nash equilibrium. Suppose Al plays the mixed strategy (A,p). When Ben chooses to graze A then his expected payoff is p(-1)+(1-p)(1) = 1-2p. When Ben chooses B then his expected payoff is p(1) + (-1)(1-p) = 2p-1.
Al plays mixed strategy (A, p) | |
Ben's payoff from playing A | Ben's payoff from playing B |
1-2p | 2p-1 |
In response to Al's mixed strategy, Ben's expected payoff from grazing A is better than that from choosing B only when
1-2p>2p-1
1-4p>-1
-4p>-2
p<1/2
When p = 1/2 Ben finds that the pure strategy of A and the pure strategy of B have the same expected payoff. By also playing a mixed strategy of (A, 1/2) Ben is playing a best response. In conclusion, while there was no Nash equilibrium when the two players were confined to pure strategies, there is a Nash equilibrium when they play mixed strategies.