# Mixed Strategy Game Theory

$$\begin{array}{|c|c|}\hline &H&T&O\\\hline H&2,-2&-2,2&0,1\\\hline T&-2,2&2,-2&0,1\\\hline O&1,0&1,0&-2,-2\\\hline \end{array}$$

I want to find an unique Nash Equilibrium here. I know that there is pure nash equilibrium since $$BR_1(H) = H$$, $$BR_1(T) = T$$, $$BR_1(O) = H$$ or $$T$$. Also, $$BR_2(H) = T$$, $$BR_2(T) = H$$, $$BR_2(O) = H$$ or $$T$$

To find a mixed strategy N.E., I can use the way that ($$p_1, p_2, 1-p_1-p_2$$) and ($$q_1, q_2, 1-q_1-q_2$$) and I will find $$(1/3,1/3,1/3)$$. I would like to ask that is there easy way to find a mixed strategy N.E. here?

• @HerrK. is there any dominated strategy here? because I could not find it Mar 29 at 5:53
• I guess A = H, B = T, O = C? Indifference of player 1 between H and T gives you $p_1=p_2$, then indifference between H and O gives you $p_2=p_3$. Analogously for player 2. Mar 29 at 10:57
• @SteveJosh: I was mistaken. Sorry. Mar 29 at 13:56

If you find the P1's utility of each pure strategy to P2 playing $$(q_1,q_2,1-q_1-q_2)$$, then you have: $$u_1 (H,(q_1,q_2,1-q_1-q_2)) = 2q_1 - 2q_2, \quad u_1(T, (q_1,q_2,1-q_1-q_2)) = -2q_1 + 2q_2, \quad u_1(O, (q_1,q_2,1-q_1-q_2)) = 3q_1 + 3q_2 -2.$$ Then setting all of these to be equal will give you $$q_1 = q_2 = q_3 = \frac{1}{3}$$.
However, this doesn't give you all the NE. Notice that when P2 plays $$O$$, P1 is indifferent between $$H$$ and $$T$$. So you might have a Nash equilibrium where P1 plays $$p_1 H + (1-p_1) T$$ and P2 plays $$O$$. We now to need to make sure that P2 playing $$O$$ is the best response to $$p_1 H + (1-p_1) T$$. $$u_2 ((p_1,1-p_1,0),H) = 2-4p_1, \quad u_1((p_1,1-p_1,0),T) = 4p_1 - 2, \quad u_2((p_1,1-p_1,0),O) = 1.$$ For $$O$$ to be the best response, we need $$1 \ge 2-4p_1$$ and $$1 \ge 4p_1 - 2$$, which together gives you $$p_1 \in [\frac{1}{4},\frac{3}{4}]$$.
So we have another set of equilibrium which look like $$\left\{ (p_1 H + (1-p_1)T,O), \frac{1}{4} \le p_1 \le \frac{3}{4} \right\}.$$
Apply the same logic to the other player to get $$\left\{ (O, p_2 H + (1-p_2)T), \frac{1}{4} \le p_2 \le \frac{3}{4} \right\}.$$