Game Theory Question

We will first identify the best responses for the player ROW and the player COLUMN.

For Player Row:

Against L:

If Column plays L, Row's payoffs are 1 (U), -2 (M), 0 (D).

Best response: U (since 1 > 0 > -2).

Against C:

If Column plays C, Row's payoffs are -2 (U), 1 (M), 0 (D).

Best response: M (since 1 > 0 > -2).

Against R:

If Column plays R, Row's payoffs are 0 (U), 0 (M), 1 (D).

Best response: D (since 1 > 0).


For Player Column:

Against U:

If Row plays U, Column's payoffs are -2 (L), 1 (C), 0 (R).

Best response: C (since 1 > 0 > -2).

Against M:

If Row plays M, Column's payoffs are 1 (L), -2 (C), 0 (R).

Best response: L (since 1 > 0 > -2).

Against D:

If Row plays D, Column's payoffs are 0 (L), 0 (C), 1 (R).

Best response: R (since 1 > 0).

Now, we find the Nash Equilibrium. By definition, a Nash equilibrium occurs when both players are playing their best responses simultaneously.

From the best response analysis above:

If Row plays U , Column plays C .

If Row plays M , Column plays L .

If Row plays D , Column plays R .

Looking at the pay-off matrix:

(U, C) gives (-2,1), but Row prefers M over U .

(M, L) gives (-2,1), but Row prefers U .

(D, R) gives (1,1) , which is the only mutual best response.

Thus, (D, R) is the unique pure strategy Nash equilibrium.

If you want to use the hint and show there is no mixed strategy Nash Equilibrium, do the following:

Suppose Row plays U and M with positive probability. Then, Column's best response would be C or L, but then Row would have an incentive to deviate to D.

Thus, Row cannot mix between U and M since playing D strictly dominates a mixed strategy involving U and M.

Therefore, no mixed strategy equilibrium exists apart from the pure strategy equilibrium (D, R).