As $w$ tends to infinity what value does the game converge to?

Question

I am not sure whether I have attempted this question right

So we have a matrix $$\begin{matrix} 1 & 4 \\ 5 & w\\ \end{matrix} $$

What value does the game converge to as w goes to $-{\infty}$

So for $w < 5$ we need to employ mixed strategy so as w tends to minus infinity we need to used mixed strategy to find the value of the game.

So I did that and got $-4p + 5 = (4-w)p + w$ where $p$ is the probability associated with Row1 (this is for the row players)

Then I got $w = \frac{(5 - 8p)}{1-p}$

Then I took the reciprocal $\frac{1}{w} = \frac{1-p}{5-8p}$ then took the limits as $w$ tends to minus infinity and I found that $p = 1$ and hence value of the game is $1$ is this right?

Answer 1

You need to use the fact that both players mix. So if $q$ is the probability that the column player plays the left strategy, you also have the condition that $q+4(1-q) = 5q + w(1-q)$. Simplifying both conditions, you have a mixed Nash equilibrium of $$p = \frac{5+w}{8+w}, \quad q = \frac{4-w}{8-w}.$$

As $w \to -\infty$, $(p,q) \to (1,1)$, which implies a payoff of $1$ for the row player.