# Finding a nash equilibrium in pure or mixed strategies

The question I'm dealing with:

Let $x=3$, find any Nash equilibrium in pure or mixed strategies.

I have observed that the pure strategy Nash equilibrium is $(u,r)=(3,3)$.

The answer to this question is:

$(\pi^*,\rho^*)=(1,0)$

Am I correct in interpreting the above as:

Given the PSNE of $(u,r)$, the row player will play $u$ with probability $1$ and the column player will play $r$ with probability $1-0=1$?

Thanks

• The other way around. The row player will play $u$, and column will play $r$. Other than this your interpretation is correct, probability of $u$ is $\pi = 1$, probability of $r$ is $1 - \rho = 1 - 0 = 1$. – Giskard Apr 23 '15 at 22:14
• @denesp, the row/column mistake was a typo. I appreciate the help you've given me with regards to the several game theory questions that I've asked. – Five σ Apr 23 '15 at 22:23
• You could also observe that $u$ and $r$ are strictly dominant strategies when $x=3$. Thus $(u,r)$ is the unique NE of this game. – Herr K. Apr 24 '15 at 17:23
• @denesp You should make your comment into an answer as it answers the question. – The Almighty Bob Apr 25 '15 at 11:07
• I felt that Five $\sigma$ did most of the work so he should get credit for the answer. – Giskard Apr 25 '15 at 18:55

Once again, thanks to denesp (and Herr K.) for the help. I thought that I'd try to answer the question just for reference purposes, and I guess that it's also a step toward improving the "answers per question" metric on Economics SE.

As aforementioned, the PSNE is $(u,r)=(3,3)$.

Player 2

• If $P1$ plays $u$, $P2$ can play either $l$ or $r$.
• $l$ would give $P2$ a payoff of $2$ whilst $r$ would give $P2$ a payoff of $3$.
• Therefore, $P2$ will play $r$ with probability $\pi=1$

Player 1

• If $P2$ plays $r$, $P1$ can play either $u$ or $d$.
• $u$ would give $P1$ a payoff of $3$ whilst $d$ would give $P1$ a payoff of $2$
• Therefore, $P1$ will play $u$ with probability $1-\rho=1$, $\rho=0$.

The result

$(\pi^*,\rho^*)=(1,0)$