Mixed Equilibrium

Question

Consider a simple card game between you and a single opponent using only six cards from a standard 52 card deck {Ac, Ad, Ah, As, 5c, 5d}: 


Each player is provided one red and one black ace. You are given a red five and your opponent is given a black five. Each player picks a card ignorant of the other's choice and have a showdown. You win if colors match; you lose if colors differ. The amount won is the numerical value of the winner (where aces are equal to one). If the two 5s are shown, the result is a draw.


How would you play this game?

Answer

This is a zero-sum game that can be described by the following payout matrix relative to player 1 ("P1"):

      P2

      rA  bA  b5

   rA  1  -1  -5

P1 bA -1   1   1

   r5  5  -1   0

When analyzing this matrix, it is clear that P1's red Ace (rA) is dominated strategy. (With perfect information when P2 plays rA, P1 responds with r5; when P2 plays bA, P1 responds with bA; when P2 plays b5, P1 responds with bA.) Knowing this, P2's black 5 becomes dominated. (With perfect information when P1 plays bA, P2 responds with rA; when P1 plays r5, P2 responds with bA.) Thus we are left with the following:

       P2

      rA bA

P1 bA -1  1 

   r5  5 -1 

Nash Equilibrium exist with mixed strategies. 

$P(P1\ plays\ bA)=x$,
$P(P1\ plays\ r5)=1-x$,
$P(P2\ plays\ rA)=y$,
$P(P2\ plays\ bA)=1-y$

P2 chooses $y$ such that P1 is indifferent between choosing bA and r5:

$1*y-1*(1-y)=-5*y+1*(1-y)$

$y=\frac{1}{4}$


P1 chooses $x$ such that P2 is indifferent between choosing rA and bA:

$-1*x+5*(1-x)=1*x-1*(1-x)$

$x=\frac{3}{4}$

And the payoff to each player can be computed:

$E[P1|x,y]=\frac{1}{4}*(\frac{3}{4}*-1+\frac{1}{4}*5)+\frac{3}{4}*(\frac{3}{4}*1+\frac{1}{4}*-1)$
$E[P1|x,y]=\frac{1}{2}$


$E[P2|x,y]=-E[P1|x,y]=-\frac{1}{2}$