Question

Each of persons A, B and C has a gun containing a single bullet. Each person, as long as she is alive, may shoot at any surviving person. First A can shoot, then B (if still alive), then C (if still alive). Denote by pi ∈ (0, 1) the probability that player i hits her intended target. Assume that each player wishes to maximize her probability of survival; among outcomes in which her survival probability is the same, she wants the danger posed by any other survivors to be as small as possible. (This last assumption is intended to capture the idea that there is some chance that further rounds of shooting may occur, though the possibility of such rounds is not incorporated explicitly into the game). Model this situation as an extensive game with perfect information and 1 chance moves. (Draw a diagram. Note that the subgames following histories in which A misses her intended target are the same). Find the subgame perfect equilibria of the game. (Consider only cases in which pA, pB and pC are all different.) Explain the logic behind A’s equilibrium action. Show that “weakness is strength” for C: she is better off if pC < pB than if pC > pB. Consider the variant in which each player, on her turn, has the additional option of shooting into the air (in which case she uses a bullet but does not hit anyone). Find the subgame perfect equilibria of this game when pA < pB. Explain the logic behind A’s equilibrium action.

Answer 1

Answer:

to maximize their chances these peraons prefer to be left with weaker opponent.

probability of A with shots,surviving against B:

p = Pr(A hits B) + Pr(A misses B) * Pr (B misses A) * p

p = 1/3 + 2/3 * 1/3 * p
p = 3/7

Probability of A, without shot, surviving against B is given by:

p = Pr(B misses A) * (Pr(A hits B) + Pr (A misses B) * p)
p = 1/3 * (1/3 + 2/3 * p)
p = 1/7

Probability of A, with shot, surviving against C is given by:

p = Pr(A hits C) + Pr(A misses C) * Pr (C misses A) * p
p = 1/3 + 2/3 * 0 * p
p = 1/3

Probability of A, without shot, surviving against C is given by:

p = Pr(C misses A) * (Pr(A hits C) + Pr (A misses C) * p)
p = 0 * (1/3 + 2/3 * p)
p = 0

So, her probability of surviving from each position is:

B, with shot: 3/7
C, with shot: 1/3
B, without shot: 1/7
C, without shot: 0

So A is best off not killing anyone since the advantage she gains by having the first shot exceeds any possible benefit of facing B rather than C. She should shoot into the air.

Given that A is neither going to shoot at them, or be shot at by them until one is dead, B and C are essentially in a two person duel, the winner to face A. They cannot improve their chances by forgoing a shot, so they shoot. B wins that 2/3 of the time, C 1/3.

A wins 2/3 * 3/7 + 1/3 * 1/3 = 25/63.
B wins 2/3 * 4/7 = 24/63.
C wins 1/3 * 2/3 = 14/63.

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