# Mixed Strategy Nash Equilibrium in this game?

         L (q)  R (1-q)
l (p)   [(2, 1), (0, 1)]
r (1-p) [(-1, 0), (1,7)]


I'm having a lot of trouble understanding what the mixed strategy nash equilibrium is exactly in this game. What I've done thus far is found that Player 1 will play "l" with probability p=1 :

$$E_1[L] = E_2[R] \\ 1p 0(1-p) = 1p + 7(1-p) \\ 7p = 7 \\ p = 1$$

Then, when solving for player 2, I get that they will mix with strategy q = 1/4:

$$E_1[l] = E_1[r] \\ 2q + 0(1-q) = -1q + 1(1-q) \\ 2q = -2q + 1 \\ 4q = 1 \\ q=1/4$$

However, in the solution I was given when trying this problem, it says that the Mixed NE is in fact:

$$\textrm {\{l, q*L + (1-q)*R\} where } q \in [1/4, 1]$$

I'm not quite understanding why in this situation q isn't simply 1/4 and instead includes all values of q above 1/4. Also, if player 1 has the strategy of always choosing "l", in that case wouldn't it at least make sense that q could just be any number [0,1] since player 2 would always be indifferent between the payoff of 1 they would receive from (2,1) or (0,1)?

Sorry if this is too simple of a problem to post here, I'm sure there is a simple explanation here that I'm just not getting but this problem has been annoying me for a little while and I'd appreciate any help!

• Is this a cooperative or a non cooperative game? – user228113 May 3 '15 at 22:36
• Non cooperative – user72266 May 4 '15 at 3:26

I think, there is a problem with the method that you chose.

Let me first explain why for any $q\in [1/4,1]$, $(1,q)$ is indeed a Nash equilibrium. First, as you say, if $p=1$ then whatever second player is doing, his gain will be $1$. Second, let us prove that if $q\in [1/4,1]$, then indeed the best response for the first player is to chose $p=1$. I.e. we need to prove that

$2\cdot q\ge 2\cdot p\cdot q+ (1-p)\cdot (-q+1-q)=4p\cdot q-2q+1$

this is equivalen to say that $4q(1-p)\ge 1-p$. But this is indeed true since $q\ge 1/4$.

Also, it is easy to see that the pure strategy $(1,0)$ is not a Nash equilibrium. Indeed if second player plays $R$, the first player should play $r$.

In other words, the answer that was given to you is correct. It seems to me that the problem is that you treat this game as if it were a zero sum game... (in which case your method would work correctly)

In general, one should always look for strictly dominant/dominated strategies first as this can simplify the game considerably especially when there are more actions available to each player. Then pure strategy Nash equilibria should be found as these will also give insight into any possible mixed strategies.

There are no strictly dominant or dominated strategies for either player, but $\{R\}$ is a weakly dominant strategy for Player 2 (and is highly relevant if the game is repeated).

If Player 1 mixes strategies $\{l,r\}$ or plays $\{r\}$ exclusively, Player 2's BR (best response) is $\{R\}$, and so there is a pure strategy NE $(r,R)$.

To get anything different P1 plays $\{l\}$ exclusively, in which case P2 is indifferent to L and R, and so can play a mixed strategy.

P1's utilities for his two actions (depending on P2's mix) are:

$u(l) = 2q$ and $u(r) = -q - (1-q) = 1-2q$ and since $u(l) - u(r) = 4q - 1 < 0$ for $q<\frac{1}{4}$, so in this case Player 1 should play $\{r\}$ and Player 2's BR is then $\{R\}$, i.e. the pure strategy NE seen before. For $q > \frac{1}{4}$, P1's BR is $\{l\}$.

So, it is arguable that mixing strategies with $q > \frac{1}{4}$ is not "really" a BR for Player 2 as he can definitely do at least as well (but not definitely better) by playing $\{R\}$. In real-life there would probably be some exchange of utilities (P2 pays P1) so as to get both player's to readily agree to play $(r,R)$ exclusively as this maximises their total utility. Unfortunately, this game is non-cooperative!

If the game is repeated, however, Player 2 should dig his heels in and insist on playing $\{R\}$ exclusively; Player 1 is then not able to create any credible threats and so is "forced" to play $\{r\}$ exclusively.