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
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

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