Game Theory: Solving n Players Helping an Injured Man

[mathlete]

This question has me totally stumped... any help would be appreciated.

"n players in a game all want to help an injured man. They each get a payoff of 1 if someone helps him, and a payoff of 0 if no one helps him. The person that helps the injured man also receives a penalty of c (0 < c < 1), making their payoff 1-c.

If n >= 2 and offers of help are made simultaneously, show that in a symmetric Nash Equilibrium, each player will refuse to help with probability c^(1/(n-1))"

Anyone? I really can't seem to get this

 

[lippyka]

This problem can be solved using the concept of backward induction. First, assume that all players are rational and that they have perfect information about the game. Now, consider the last move of the game: the decision of the nth player to help or not. If he helps, then the payoff is 1-c, whereas if he doesn't help then the payoff is 0. Since the nth player wants to maximize his payoff, it must be the case that his optimal strategy is to not help the injured man if c^(1/(n-1)) <= 1/2 and to help him otherwise. Now, consider the second-to-last player's decision. He knows that if he chooses to help, then the nth player will choose not to help if c^(1/(n-1)) <= 1/2 and will choose to help otherwise. Thus, the second-to-last player's optimal strategy is also to not help the injured man if c^(1/(n-1)) <= 1/2, and to help him otherwise. The same logic can be applied to the third-to-last player, the fourth-to-last player, and so on until the first player. Thus, in a symmetric Nash Equilibrium, each player will refuse to help with probability c^(1/(n-1)).