I have found 2 pure Nash equilibria in the following game which are (Y,B) & (Z,C)
| A | B | C | |
|---|---|---|---|
| X | 6 , 0 | 0 , 3 | 1 , 8 |
| Y | 3 , 4 | 1 , 5 | 1 , 3 |
| Z | 7 , 2 | 0 , 1 | 2 , 3 |
I am looking for mixed strategy Nash Equilibria.
Let A probability be $p_1$, B probability be $p_2$, and C probability be $1−p_1−p_2$.
$ EU(X)=6p_1+1-p_1-p_2 = 5p_1-p_2+1 $
$ EU(Y)=3p_1+p_2+1-p_1-p_2 = 2p_1+1 $
$ EU(Z)=7p_1+2-2p_1-2p_2=5p_1-2p_2+2$
From these 3 equations I get:
$p_1=1/3, p_2=1$ which does not make sense as a probability distribution.
If I follow the same approach for the other player:
Let X probability be $q_1$, Y probability be $q_2$, and Z probability be $1−q_1−q_2$.
$EU(A)= 4q_2+2-2q_1-2q_2=-2q_1+2q_2+2$
$EU(B)= 3q_1+5q_2+1-q_1-q_2=2q_1+4q_2+1$
$EU(C)= 8q_1+3q_2+3-3q_1-3q_2 = 5q_1+3$
Form these 3 equations I get:
$q_1=0, q_2=0.5$
I am not sure how to interpret these results. Is there a Nash equilibrium of mixed strategies here?