# Unable to find Nash equilibria in mixed strategies

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?

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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 :-)

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Could you explain more about the part where you said "...then maximizing expected payoff for the second player, you can set r+s=1, that is discard the Right strategy." By maximizing expected payoff by discarding the Right strategy, does this mean that Right has the lowest expected payoff? If so, then how do we know that Right has a smaller expected payoff than Left? In other words, why can't I discard Left instead of Right? – Kevin Jul 16 '12 at 5:32
@Kevin Let $r$ and $s$ describe optimal strategy for the second player. Then strategy with $r' = r$ and $s' = 1-r$ will be no worse (just write the formula for the expected payoff). However, we know that $p \neq 1$ so the new strategy will be actually strictly better. Does this answers your question? – dtldarek Jul 16 '12 at 5:54
Sorry I still don't get it. I found the expected payoff of Left is 2p+1, while Right is 2-2p. But for example when p=0.1, the expected payoff of Right is greater than Left. – Kevin Jul 16 '12 at 6:12
@Kevin: The point isn't that Right is (weakly) dominated by Left, but by Middle. As long as Player 1 has a non-zero probability of choosing Bottom, Right is strictly worse than Middle. – joriki Jul 16 '12 at 12:31
@joriki I see, so if I understand correctly, my first step to solving these questions should be to do both an IDSDS and a IDWDS (repeatedly delete strictly and weakly dominated strategies). Is this correct? – Kevin Jul 16 '12 at 14:51