I'm working on an exercise in which I have to find all mixed Nash equilibria. The payoff matrix is the following 
Player one chooses rows with probabilities (p,q,r,s) , player chooses row L with probability y and player 3 chooses matrix U with probability z.
So far I've found that player 1 is indifferent when y=z= 2/3 , player 2 is indifferent when r=s and player 3 is indifferent when p=1/2 q.
So to answer the question of "Find all mixed Nash equilibria", do I then just say that every strategy that satisfies these indifference conditions is an equilibrium? or can I work more on this and would I do that?
Also after this, there is an exercise that asks us to show that player 2 has a continuum of equilibrium payoffs, how should I go about this?
Best Answer
Its clear none of Player 1's pure strategies strictly or weakly dominate any other pure strategy.
Because Player 1 has 4 possible actions, we have to check all the mixed Nash equilibria for each subset of their actions. This is because there may exist dominated mixed strategies which Player 1 would never rationally choose, since another subset gives better net payoff.
The other two players have only 2 actions, so they don't need to consider choosing a subset, since their only strategy is mixing based their probability $y$ (Player 2) and $z$ (Player 3).
So lets focus on what values would make Player 1 indifferent for each possible subset of their actions.
Player 1 Indifference
Case 1: They mix between all their options (N, E, W, S). This means that for the other players to make Player 1 indifferent, the following equations must have the same value. $$E_N(P1) = y*z*2 + (1-y)*z*2 $$ $$E_E(P1) = y*z*1 + y*(1-z)*2 + (1-y)*z*1 + (1-y)*(1-z)*2$$ $$E_W(P1) = y*z*2 + y*(1-z)*2$$ $$E_S(P1) = y*z*1 + y*(1-z)*1 + (1-y)*z*2 + (1-y)*(1-z)*2$$ When we solve for $y$ and $z$ we get your answer of $y=z=2/3$. Now we must check the expected outcome for Player 1 with this strategy, and we see $E(P1) = 4/3$.
Case 2: Player 1 only chooses between (N, E, W). This involves the same process, but now we exclude the last equation from Case 1, and they still must all have the same value. This gives the same $y=z=2/3$ and thus the same expected outcome.
Case 3: Player 1 chooses between (N, E, S). Again we get $y=z=2/3$.
Case 4: (N, W, S), $y=z=2/3$.
Case 5: (E, W, S), $y=z=2/3$
Case 6: (N, E), Here we get that $z=2/3$ but that $y$ is a free variable in this equilibrium. However, this still gives $E(P1) = 4/3$.
Case 7: (N, W), Here we get the relation $y=z$ where they are now a single free variable $c$. Now we see that $E(P1) = 2c$, so now we must analyze the strategies of Players 2 and 3 to determine a definite value of $c$ and be able to calculate $E(P1)$.
In this scenario for Player 2 the pure strategy (L) weakly dominates (R), and thus we see that $y=1$ is their optimal strategy, and thus $E(P1) = 2$! (We also see that Player 3's strategy of choosing (U) every time also weakly dominates (R), and thus $z=1$)
This is better than any strategy Player 1 has had yet, and thus the mixed strategy (N, W) dominates all the ones we've previously analyzed and Player 1 would never rationally choose any of the previous strategies.
Case 8: (N, S), This yields the relation $z+y/2=1$, so again we must analyze the other players' strategies to determine $E(P1)$.
Here we see that Player 2's pure (R) strategy weakly dominates (L) and thus we see that $y=0$ is the optimal strategy. Thus $z=1$. Thus $E(P1) = 2$, and thus this is another current optimal.
Case 9: (E, W), This yields relation $z+2y=2$. Player 2's (L) strategy weakly dominates (R), thus $y=1$, thus $z=0$, thus $E(P1) = 2$. This has the same outcome as Case 7 and we will keep this strategy as one of the current optimal.
Case 10: (E, S), This yields again the relation $y=z$. Player 2's (R) weakly dominates (L) in this scenario, thus $y=0$, thus $E(P1) = 2$, and this strategy is kept as an optimal.
Case 11: (W, S), This yields $y=2/3$ with $z$ free, but still gives $E(P1) = 4/3$, thus discarded.
Conclusion
We have seen through our analysis that Player 1 will choose to either mix between (N, W), (N, S), (E, W), or (E, S) since they all give the same expected outcome.
Thus there are mixed Nash equilibria: $(p=c, q=0, r=1-c, s=0, y=1, z=1)$, $(p=c, q=0, r=0, s=1-c, y=0, z=1)$, $(p=0, q=c, r=0, s=1-c, y=0, z=0)$ and $(p=0, q=c, r=1-c, s=0, y=1, z=0)$
Where $c$ is free, since the table reduces to pure strategies for the other players.