I am trying to understand how to calculate all Nash equilibria in a 2 player game, but I fail when there are 3 possible options . Also I have I checked answers on this sites and still I dont get it.

\begin{array}{c|rrr} & L & C & R \\\hline T & 3,4 & 4,2 & 2,1 \\ M & 1,7 & 3,8 & 0,0 \\ B & 1,0 & 1,1 &4,3 \end{array}

Can somebody explain to me how to calculate a matrix like this? Maybe a hint will help me.

Edit: I think that I need to calculate $p$ and $q$. For $2\times2 $ its easy but what about $3\times3$ ?

Edit 2: I think I have found Nash equilibrium:

\begin{array}{c|rrr} & L & C & R \\\hline T & \underline{3},\underline{4} & 4,2 & 2,1 \\ M & 1,7 & 3,\underline{8} & 0,0 \\ B & 1,0 & 1,1 &\underline{4},\underline{3} \end{array}

Now how can I calculate $p$ and $q$?

Thank you


1 Answer 1


Indeed, finding a Nash Equilibrium for larger games than $2\times 2$ is usually not a fun process, as there are many conditions to verify and (simple) equations to solve. The general process would involve checking all possibilities: only pure equilibria, equilibria where only two actions are mixed, and equilibria where all actions are played (for each player!). Luckily, when these kind of questions are given to be solved by hand, usually there are tricks that make the solution more simple, such as symmetry or dominance.

Indeed, lets start with dominance. Note that all the payoffs for the row player in the T row are higher than in the M, so T dominates M and can be deleted from the matrix. It makes the work simpler.

More tricky but also quite visible is that $C$ is dominated, this time by a mixed action. If column player mixes $L$ and $R$ with equal probabilities, his payoff is $2.5$ against $T$ and $1.5$ against $B$, which are higher than his payoffs when playing $C$.

Now you have only a $2\times 2$ games where it is easy to see that there are two pure equilibria ($(T,L)$ and $(B,R)$) and one mixed (each player plays each action with probability half).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .