0
$\begingroup$

\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?

$\endgroup$
3
  • $\begingroup$ @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

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.