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I'm trying to find any and all pure or mixed strategy Nash equilibria for the game

$$\begin{array}{|c|c|c|c|}\hline & L & C & R \\ \hline T & (6,2) & (0,6) & (4,4) \\ \hline M & (2,12) & (4,3) & (2,5) \\ \hline B & (0,6) & (10,0) & (2,2) \\ \hline \end{array}$$

By inspection I see no pure strategy Nash equilibrium. Not having a pure Nash equilibrium is supposed to ensure that a mixed strategy Nash equilibrium must exist.

However, when I go to solve for the mixed strategies I get one set of solutions that has a negative probability and in the set of equations for the other player I get an inconsistent system. My work is below.

If player I plays T, M, B with probabilities $p,q,1-p-q$ then for player II

$$E(L) = 2p+12q+6(1-p-q) = $$ $$E(C) = 6p+3q+0 = $$ $$E(R) = 4p+5q+2(1-p-q)$$

providing the system of equations

$$-4p+6q+6 = 6p+3q$$ $$-4p+6q+6 = 2p+3q+2$$

which simplify to

$$10p-3q=6$$ $$6p-3q=4$$

And subtracting equations you get $4p=2$ which implies $p=1/2$ but then plugging this into the first equation you find $10(1/2)-3q=6$ implies $q=-1/3$.

Did I do something wrong?

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  • $\begingroup$ Yes, the condition for nash is not that the $E(L) = E(R) = E(C)$, you get a minus value, because this condition is not possible for the case. $\endgroup$
    – john
    Jul 14, 2015 at 20:39

2 Answers 2

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Your mistake is not considering mixed strategies that involve some, but not all, of the possible moves.

Suppose the row player plays $T$ and $B$ with probabilities $p$ and $1-p$, but never plays $M$ (note that $M$ is dominated by a $1/2-1/2$ mixture of $T$ and $B$, so this is reasonable). The expected payoffs for the column player are then $2p + 6(1-p) = 6 - 4p$ for $L$, $6p$ for $ C$, $2 p + 2$ for $R$. With $p = 6/10$ these are $3.6, 3.6, 3.2$, so the column player will never play $R$. Now determine probabilities for the column player playing $L$ and $C$ that make the row player's expected payoffs for $T$ and $B$ equal.

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  • $\begingroup$ Hm. I can understand at least guessing that, in a sense, $q=0$ as a kind of boundary solution to the constrained problem. However, I'm not quite getting the choice $p=6/10$ which comes from setting the expectations of L and C equal--you would get $p=4/6$ by setting expectations of L and R equal. Does this just indicate that there are multiple Nash equilibria at this boundary where $q=0$? Also, how do I know that there aren't other boundary solutions where $p=0$? $\endgroup$
    – Addem
    Jul 14, 2015 at 20:26
  • $\begingroup$ When Robert notes that it is reasonable to consider a mix of $T$ and $B$ (but not $M$) because $M$ is strictly dominated by a mix of $T$ and $B$, he could in fact go further and say that it is the only reasonable thing to do. It can be proven generically that there are no Nash equilibria (mixed or pure) in which a strictly dominated strategy is played with probability greater than zero. That is, in any NE (mixed or pure), all strategies being played must not be strictly dominated. $\endgroup$
    – Shane
    Jul 14, 2015 at 22:05
  • $\begingroup$ @Shane, Oh, I'm starting to get the idea here--thanks! $\endgroup$
    – Addem
    Jul 15, 2015 at 16:41
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For the row player strategy $M$ is never BR and for the column player strategy $R$ is never BR. So, exclude these strategies and compute the mixed equilibrium which is $(T,M,B)=(3/5, 0, 2/5)$ and $(L,C,R)=(5/8,3/8,0)$.

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  • $\begingroup$ I'm not seeing why the row player would never play $M$. I see that $M$ is dominated by certain mixtures of $T$ and $B$ like, for instance, 1/2 and 1/2. But what if player I uses some other mixture for which $M$ is not always dominated? $\endgroup$
    – Addem
    Jul 14, 2015 at 20:45
  • $\begingroup$ @Addem: Actually, it is always dominated. $P(T)=P(B)=0.5$ always dominates $M$ (for any mix of the column player)... $\endgroup$
    – user140541
    Jul 14, 2015 at 21:43
  • $\begingroup$ @Addem: Also, in a 2-player game the set of dominated strategies coincides with the set of "never BR" strategies. $\endgroup$
    – user140541
    Jul 14, 2015 at 21:46

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