\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?

  • $\begingroup$ @HerrK. is there any dominated strategy here? because I could not find it $\endgroup$
    – Steve Josh
    Mar 29, 2023 at 5:53
  • $\begingroup$ 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. $\endgroup$
    – VARulle
    Mar 29, 2023 at 10:57
  • $\begingroup$ @SteveJosh: I was mistaken. Sorry. $\endgroup$
    – Herr K.
    Mar 29, 2023 at 13:56

1 Answer 1


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 $$\left\{ (O, p_2 H + (1-p_2)T), \frac{1}{4} \le p_2 \le \frac{3}{4} \right\}.$$


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