Consider a one-shot, simultaneous chicken game, as described here: https://en.wikipedia.org/wiki/Chicken_(game)
Assume that I'm playing this game against a player that I consider to be of similar intelligence as myself. It's not clear to me what Game Theory recommends me to do.
The game has three Nash equilibria: two pure strategy equilibria where one player goes straight and the other swerves, and one symmetric mixed strategy Nash equilibrium, where both players go straight with probability $p_{NE}$. Let $E(p,q)$ denote my expected pay-off if I go straight with probability $p$ and my opponent goes straight with probability $q$, so my opponent's expected pay-off is $E(q,p)$. When I play the mixed strategy Nash equilibrium, my opponent has no incentive to change their value $q$, so $p_{NE}$ can be obtained by solving the quation
$$\frac{\text{d}E(q,p_{NE})}{\text{d}q}=0\,.$$
(Since $E(q,p)$ is linear in $q$, $q$ drops out of this equation.) Does Game Theory recommend me to choose $p=p_{NE}$? If so, why?
Alternatively, I could reason as follows. Since my opponent is of similar intelligence, they will choose the same probability as I, q=p. I should thus choose the value of $p$ such that $E(p,p)$ is maximal. That is, choose the value of $p$ that solves
$$\frac{\text{d}E(p,p)}{\text{d}p}=0\,,$$
or choose $p=0$ if the above leads to a negative solution. One can show that the value for $p$ obtained by solving this equation is always smaller than $p_{NE}$.
Which of these two values is the one I should choose and why?
Best Answer
Game theory does not make recommendations, per se. Rather, it makes predictions based upon assumptions. When we are solving for Nash equilibria, we are trying to predict what players will play given rationality of all players and common knowledge of that rationality.
Your alternate reasoning is subtly incorrect. The key error is that you are implicitly assuming that you control your opponent's behavior when you try to maximize $E(p,p)$. Remember, the equally intelligent opponent will play the same strategy as you in equilibrium, but not generically. If you are choosing $p=0.25$, for example, $q=0.25$ is not a best response to that.
Your primary reasoning (the first) is correct. There's a slightly easier way to think about it though. In a mixed equilibrium, each player must be willing to mix. To be willing to mix, each player must be indifferent over the actions across which he is supposed to mix. We can use that indifference condition to solve for the mixed NE. For example, let $\alpha_2(D)$ be the weight P2 is putting on $D$. Then, for P1 to be willing to mix:
$$ \text{Expected Utility from playing D} = \text{Expected Utility from playing C} $$
$$ \alpha_2(D)u_1(D,D) + \alpha_2(C)u_1(D,C)= \alpha_2(D)u_1(C,D) + \alpha_2(C)u_1(C,C) $$
$$ \alpha_2(D)0 + \alpha_2(C)7= \alpha_2(D)2 + \alpha_2(C)6 $$
Now, note $\alpha_2(C) = 1-\alpha_2(D)$, so:
$$ \alpha_2(D)0 + (1-\alpha_2(D))7= \alpha_2(D)2 + (1-\alpha_2(D))6 $$
Solving this gives $\alpha_2(D)=1/3$. Since the game is symmetric, we know that mixed Nash equilibrium is at:
$$\left(\left(\frac{1}{3}, \frac{2}{3}\right),\left(\frac{1}{3}, \frac{2}{3}\right)\right)$$