[Math] Finding Nash Equilibria for this Bimatrix Game

game theorynash-equilibrium

Consider the following Bimatrix Game

          a        b       c       d
-----|-------------------------------------------

T       (1,4)    (4,3)   (0,2)   (1,0)

B       (2,1)    (2,4)   (3,5)   (0,6)

Find all Nash Equilibria if the game in pure strategy and mixed strategy.

My Doubt is :

1) I know in the given figure of the pure strategy equilibria the boxed values represent Best responses to the game but how to find them?

2) For pure strategy nash equilibrium find the maximum payoff in each row and each column was the rule so why do they not box the strategies of player II in (B,b) = 4 > 3 and (B,c) = 5 > 2 ?

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Best Answer

In each cell, the lower left payoff is for Player I and the upper right payoff is for Player II. The boxes identify each player's best outcome taking the other player's strategy as given.

So: if Player I plays T, then Player II will get a payoff of 4, 3, 2 or 0 depending on whether s/he plays a, b, c or d, respectively. The 4 is boxed because that is Player II's best payoff given that Player I is playing T. Similarly, if Player I plays B, then Player II's best option is to play d and get a payoff of 6. Only two of Player II's payoffs are boxed because Player II only has to consider two possible strategies played by Player I.

On the other side, for Player I there are four "what if" possibilities, namely what if Player II plays a? What if s/he plays b? c? Or d? For each possibility there is a best option for Player I, and it is boxed.

The point about there being no pure Nash equilibrium is that there is no choice of strategies for Player I and Player II where each one will be content to stick with his/her strategy even after knowing what the other one is doing. Every combination is unstable, in the sense that at least one player will think "Oh, if that's what the other player is doing, I want to change my strategy." This is indicated by the fact that no cell has both Player I's and Player II's payoff boxed as "best."

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