Unable to find Nash equilibria in mixed strategies

Question

Here is the strategic form game:

                          Player 2
                     Left  Middle  Right
            Top      2,2   0,0     1,3
Player 1    Middle   1,3   3,0     1,0
            Bottom   3,1   2,3     2,2

First I performed an IDSDS and deleted Player 1's "Top" strategy since it is strictly dominated by the "Bottom" strategy. The resulting game is as follows:

                          Player 2
                     Left  Middle  Right
Player 1    Middle   1,3   3,0     1,0
            Bottom   3,1   2,3     2,2

Let p denote the probability that Player 1 will choose Middle.
Let r and s denote the probability that Player 2 will choose Left and Middle, respectively.

Now here's my first problem. When I try to equate the expected payoffs of Player 2 choosing Left, Middle, and Right, I can't derive any values:

$ 3p + (1-p) = 3(1-p) = 2(1-p)\\ $
There is so such value for p here.

Moving on to expected payoffs of Player 1, I run into another problem:

$ r + 3s + (1-r-s) = 3r + 2s + 2(1-r-s)\\ s = 2r + (1-r-s)\\ r = 2s - 1 $

But now I am unsure of how to find the values of r and s.

Where am I going wrong?

Answer 1

Player 2 cannot lose if he were to choose Middle strategy instead of Right. However, you correctly didn't removed it, because it is not dominated strictly. On the other hand, since the probability $p$ that first player will choose Middle strategy is not $100\%$, then maximizing expected payoff for the second player, you can set $r+s = 1$, that is discard the Right strategy. Please note that this is ok only because $p \neq 1$ (and we know that $p \neq 1$ because this game does not have pure Nash equilibrium).

                      Player 2
                    Left  Middle
Player 1    Middle   1,3   3,0
            Bottom   3,1   2,3

What we are left with is a standard $2\times 2$ game that you can solve easily (if you don't know how, then google "calculate 2x2 mixed equilibrium" or something similar, the available texts are much better than anything I could write here). We have

$$ \begin{aligned} EP_1 &= 1\cdot pr + 3\cdot (1-p)r + 3\cdot p(1-r) + 2\cdot(1-p)(1-r) \\ &=-3pr+p+r+2 \\ &= (-3r+1)p + (r+2) \\ EP_2 &= 3\cdot pr + 1\cdot (1-p)r + 0\cdot p(1-r) + 3\cdot(1-p)(1-r) \\ &=+5pr-3p-2r+3 \\ &= (+5p-2)r +(-3p+3) \end{aligned} $$

so $r = \frac13$ and $p = \frac25$. Once again, this game has no pure Nash equilibria, therefore if $r \neq \frac13$ then $p = 0$ or $p = 1$ and then $r = 0$ or $r = 1$ which leads to contradiction; similarly $p \neq \frac25$ cannot be true.

Hope that helps :-)