Question

1) In a town with three people (N-1,2,3]), the only water well is broken. Residents will have no water unless it is fixed. Each resident can either Fix or Not Fix. If at least one person chooses to Fix, the well could be used, which is worth 10 utils for each player. Playing Fix is costly: it reduces the players utility by 1. Namely, players' utility is O if no one chooses Fix, 10 if someone else chooses to Fix (and they do not), and 9 if they choose Fix. d. Is there a Nash equilibrium where the well is fixed with probability greater than O and less than 1? If yes, find it. If no, explain. [15 pts]

Answer 1

Following payoff matrix for 3 players deciding whether to fix the well or not.

In the pure strategy, there are three Nash Equilibria : (F,DF,DF),(DF,F,DF),(DF,DF,F), where DF means doesn't fix and F means fix. Thus, nash equilibria are the ones where only one player fixes the well while other two doesn't.

(d). Since the payoffs are symmetrical for each player,therefore,there probabilities are equal.

Let probability of fixing the well for each player (Pi) be q₁ , q₂, q₃ for P1, P2, P3 respectively.

and probability of not fixing the well for each player (Pi) be (1- q)₁ , (1-q)₂, (1-q)₃ for P1, P2, P3 respectively.

q₁ = q₂ = q₃ =1/3

(1- q)₁ = (1-q)₂ = (1-q)₃ = 2/3

Following is the expected payoff matrix:

There are 3 Nash equilibria : (DF,DF,F), (F,DF,DF), (DF,DF,DF)

Thus, there are 2 nash Equilibria where the well is fixed with probability q₁=1/3 0r q₃= 1/3