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

 bBen(AAl) = B bAl(ABen) = A bBen(BAl) = A bAl(BBen) = 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.