Understand mixed strategy N.E

game theorynash-equilibrium

Problem:
I'm calculating the mixed strategy N.E for the game

$$ \vec g= \begin{bmatrix}(3,3)&(0,1)\\(1,0)&(2,2)\end{bmatrix} $$

for two players (player X plays in rows, player Y plays in columns) who doesn't trust one another.

Solution: The way I do it for player 1, is calculating

$$ \begin{bmatrix}3&0\\1&2\end{bmatrix} \cdot \begin{bmatrix}y\\1-y\end{bmatrix} = \begin{bmatrix}3y\\2-y\end{bmatrix} = \begin{bmatrix}v\\v\end{bmatrix} $$

where $v$ stands for the value of the game. Setting $3y=2-y \Longrightarrow \vec y=(\frac{1}{2},\frac{1}{2})$ and $v=\frac{3}{2}$. The same procedure for player 2 would give $\vec x = (\frac{1}{2}, \frac{1}{2})$ and $v=\frac{3}{2}$.

Question: I'm trying to get an intuitive understanding of why this procedure works. In the solution above, I'm using the mixed strategy of player 2, and pure strategies for player 1 (I think?) to calculate the value of the game for player 1. Why is this solution better than using the mixed strategy of player 1, and pure strategies for player 2, to calculate the value of the game for player 1?

I find it strange that when calculating the N.E value for player 1, I calculate/use the mixed strategy of player 2. If I am player 1, shouldn't I be focusing on my own mixed strategy? This is what confuses me.

Best Answer

In order to find the best response by player $X$ to any strategy $y\in[0;1]$ of player $Y$, we need to compute the expected pay-off $\pi$ of player $X$ for their strategies $x\in[0;1]$, where $x$ and $y$ are the probabilities of $X$ and $Y$ choosing the first option, respectively.

Hence, player $X$ has an expected pay-off of: $$ \pi(x,y) = 3xy + 0x(1-y) + 1(1-x)y + 2(1-x)(1-y) = 4xy - 2x - y + 2 , $$ with $x$ being the probability of player $X$ choosing the first option.

So, fixing $y$, the best response for player $X$ is:

  • Choosing the first option ($x=1$) if $y>1/2$;
  • Choosing the second option ($x=0$) if $y<1/2$; and
  • The choice is indifferent for $X$ if $y=1/2$ (being indifferent, it means that the pay-offs are equal for any $x \in [0;1]$).

Applying the same reasoning for player $Y$, the indifferent case is when $x=1/2$.

So, if both $x=1/2$ and $y=1/2$ we have a mixed (unstable) equilibria. It is unstable because any deviation, even if it is small, will lead the players to deviate from the mixed equilibria (and play one of the pure strategies).

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