Nash equilibrium, a particular matrix game

Question

Suppose that we have the following matrix game:

\begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix}

What is the intuition behind the fact, that the unique mixed Nash equilibrium is $(\frac{1}{3},\frac{2}{3})$ for the first player? I do not want any computation (or at least very little of computation), just intuition why on the lower positive element of this matrix $1$ is put more weight $\frac{2}{3}$ then on the greater element, $2$, on which we put weight $\frac{1}{3}$.

Answer 1

Denote the possible actions of player $1$ by $U$ and $D$ (up and down) and denote the actions of player $2$ by $L$ and $R$ (left and right).

In a mixed strategy equilibrium, actions played with positive probability must give the same expected payoff. Thus if player 2 plays $L$ and $R$ with probabilities $p$ and $1-p$ respectively, then for player $1$ to play $U$ and $D$ with positive probability it must be that 1's expected payoffs from the two actions are equal: $$\underbrace{p(2)+(1-p)(0)}_{\text{1's payoff from $U$}}=\underbrace{p(0)+(1-p)(1)}_{\text{1's payoff from $D$}}\quad \iff \quad 2p=1-p.$$

The solution is $p=1/3$, i.e. player 2 has to put twice the weight on $R$ as $L$ in order for player $1$ to be indifferent between playing $U$ and $D$.

The intuition is:

• the payoff for player 1 from $U$ if player 2 plays $L$ is 2 and 0 otherwise
• the payoff for player 1 from $D$ if player 2 plays $R$ is 1 and 0 otherwise
• since $2=2(1)$, player 2 must play $R$ with twice the probability of $L$ in order for player 1 to be indifferent between $U$ and $D$ (i.e. in order for the expected values from $U$ and $D$ to be the same)