“War of Attrition”: Interpretations and Applications

Last week I posed the war of attrition game, and earlier this week I analyzed it. Building on that analysis, in this post I provide some interpretations and applications for the mixed strategy Nash equilibrium solution we found. As a reminder, here’s a short summary of the game in more general notation than originally posed:

You and a competitor will battle in rounds for a prize worth V dollars. In each round each of you may choose to either fight or fold. The first one to fold wins $0. If the other player doesn’t also fold he wins the V dollar prize. In the case you both fold in the same round you each win $0. If you both choose to fight you both go on to the next round to face a fight or fold choice again. Moving on to each round after round 1 costs C dollars per round per player. Assume V > C.

Recall that what we found in the analysis was that there was a mixed strategy Nash equilibrium to fight with probability p=V/(V+C). In the case V=$5 and C=$0.75, p=0.87. What does this mean?

There are multiple ways to interpret mixed strategy Nash equilibria. One way is to interpret the probability as a statement about a population. Applied to the game of attrition this interpretation would say that proportion p of the population are fighters and the rest are folders. That’s certainly plausible. I bet that upon reading the statement of this problem last week some folks immediately thought “I will not fight even one round,” while other folks immediately thought, “I would fight forever.” Even if nobody actually thought the latter, experiments show that people will really fight a very long time, even to the point that the cumulative fight fees exceed the prize. There really are “fighter” and “folder” personality types in the population.

A second interpretation is that each individual will play a mixed strategy. That is, you yourself will “roll the dice” in your head and fight with probability p and fold otherwise. Notice that each round is an independent “roll of the dice.” Past fight fees have no bearing on your probability of fighting in the current round. They are sunk costs. With probability p you will fight on, and on, and on…

What is the probability that this fight will go to round 2? It is the probability that both you and your opponent fight in round 1, or p2. What is the probability the fight will enter round 3? It is the probability that you and your opponent both fight in round 1 and both fight in round 2. Those decisions are independent so the probability of entering round 3 is p4. In general, the probability of fighting to round n+1 is p2n. When p is large (i.e., V is large relative to C) some very long fights can occur. With each round there is hope of earning some money (if you win) so it is rational for you to continue precisely when your expected winnings of doing so are equivalent to those if you don’t. That’s exactly what p promises.

In fact, long wars of attrition have occurred in history, in warefare, in competition between firms, and in politics. Wars of attrition also occur in auctions. Each side is rational to continue the war but not because they wish to recoup past fight fees. Those are sunk costs, they cannot be recovered, and therefore they are irrelevant to current play. Each stage is independent of the next so players fight on because the expected benefit is equivalent to not fighting (in mixed strategy Nash equilibrium play). Eventually one side’s resources are exhausted and the war of attrition comes to an end.

My set of things to say about this game has also been exhausted so this series on the war of attrition game also ends here, at least for now.

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