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:

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:

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 p₁ and D with probability p₂ and M with probability 1-p₁-p₂ 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 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.