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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

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5
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    $\begingroup$ 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 $1 - \rho = 1 - 0 = 1$. $\endgroup$
    – Giskard
    Apr 23, 2015 at 22:14
  • $\begingroup$ @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. $\endgroup$
    – Five σ
    Apr 23, 2015 at 22:23
  • $\begingroup$ 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. $\endgroup$
    – Herr K.
    Apr 24, 2015 at 17:23
  • $\begingroup$ @denesp You should make your comment into an answer as it answers the question. $\endgroup$
    – The Almighty Bob
    Apr 25, 2015 at 11:07
  • $\begingroup$ I felt that Five $\sigma$ did most of the work so he should get credit for the answer. $\endgroup$
    – Giskard
    Apr 25, 2015 at 18:55

1 Answer 1

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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.

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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

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

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