We have seen the payoffs from the two pure strategies in the war of attrition. In both cases, one person fights, and the other drops out; no costs are incurred.
| Player 2 Fight | Player 2 Quit | |
| Player 1 Fight | (-c, -c) | (V, 0) |
| Player 1 Quit | (0, V) | (0, 0) |
However, there is also a mixed strategy equilibrium that we’ll explore in this post. We know how to find the mixed strategy equilibrium. It is by assigning a probability to player 2 to quit or fight in such a way as to make the other player (player 1) indifferent between fighting and quitting.
| Player 2 Fight | Player 2 Quit | |
| Player 1 Fight | (-c, -c) | (V, 0) |
| Player 1 Quit | (0, V) | (0, 0) |
| p | (1-p) |
The table means that there is a probability p that player 2 fights and (1-p) that player 2 quits.
That means,
if player 1 fights, her expected payoff is = -c x p + V x (1-p)
if player 1 quits, her expected payoff is = 0 x p + 0 x (1-p)
Since the p is such a way that player 1 is indifferent to each.
-c x p + V x (1-p) = 0
V = p (c + v)
p = V / (V + c)
The results show that if the cost of fighting at each stage is small, then the probability that the player fights is close to 1.
Subgame perfect equilibrium: wars of attrition: YaleCourses
