Reason 1
A pure strategy that is undominated by other pure strategies may be dominated by a mixed strategy.
Consider the Nole - Grace game again:
Grace's Laces | |||||
U | M1 | M2 | D | ||
Nole's Soles | U | 1, 0 | 4, 2 | 2, 4 | 3, 1 |
M | 2, 4 | 2, 0 | 2, 2 | 2, 1 | |
D | 4, 2 | 1, 4 | 2, 0 | 3, 1 |
There is no dominant pure strategy for either Nole or Grace.
If Nole plays U with probability p = 1/2 and D with probability 1-p = 1/2 then when Grace plays U Nole gets 2.5 ( = 1x1/2 + 4x1/2). When Grace plays M1 then Nole gets 2.5 with his mixed strategy. When Grace plays M2, then Nole gets 2 with his mixed strategy. When Grace plays D then Nole will get 3 by playing his mixed strategy. Put the result of the mixed strategy in a new row of the table:
Grace's Laces | |||||
U | M1 | M2 | D | ||
Nole's Soles | U | 1, 0 | 4, 2 | 2, 4 | 3, 1 |
M | 2, 4 | 2, 0 | 2, 2 | 2, 1 | |
D | 4, 2 | 1, 4 | 2, 0 | 3, 1 | |
N's mix | 2.5, * | 2.5, * | 2, * | 3, * |
So the mixed strategy dominates M for Nole. In fact, for any Pr(U) = p < 2/3 the mixed strategy (U, D, p) dominates M for Nole.
If Nole plays U with probability p_{1} and D with probability p_{2} and M with probability 1-p_{1}-p_{2} then this mixed strategy will dominate M as long as .You should be able to prove this.
If Grace plays U, M1, and M2 with probabilities 1/3, 1/3, 1/3 then this mixed strategy beats her pure strategy of playing D.
What is the message here?
When we consider mixed strategies for Nole then M^{N} is dominated so it can be stricken from the table. Similarly for Grace, consideration of mixed strategies allows us to strike D^{G} from the table. Making these changes:
Grace's Laces | |||||
U | M1 | M2 | |||
Nole's Soles | U | 1, 0 | 4, 2 | 2, 4 | |
D | 4, 2 | 1, 4 | 2, 0 |
Grace now sees that M1 dominates U for her. After striking U^{G}, Nole sees that U dominates D for him. Finally, Grace sees that M2 dominates M1. The solution to the game is {U^{N}, M2^{G}}, which we had seen earlier was the Nash equilibrium.