What's the probability that A wins the duel?

Question

Three men — conveniently named A, B, and C — are fighting a duel with pistols. It's A's turn to shoot.

The rules of this duel are rather peculiar: the duelists do not all shoot simultaneously, but instead take turns. A fires at B, B fires at C, and C fires at A; the cycle repeats until there is a single survivor. If you hit your target, you'll fire at the next person on your next turn.

For example, A might shoot and hit B. With B out of the picture, it would be C's turn to shoot — suppose he misses. Now it's A's turn again, and he fires at C; if he hits, the duel is over, with A the sole survivor.

To bring in a little probability, suppose A and C each hit their targets with probability 0.5, but that B is a better shot, and hits with probability 0.75 — all shots are independent.

What's the probability that A wins the duel?

Answer 1

To solve this it is easier to break this problem down into smaller sub-problems. First, we'll designate how a problem will be presented. Let P(A, B;A) designate the probability that A wins a duel with participants A and B with it being A's shot. What we're looking for is P(A, B, C;A), the probability that A wins a duel with A, B and C with it being A's shot. Let's also designate that A, B, C are the probability that the respective player hits his or her target and !A, !B, !C are the probability of a miss. Let's calculate the probability of the applicable sub problems. For A to win the duel A has to be a participant in the sub-problem. If A shoots B, then the subproblem that applies then is P(A, C; C) the probability that A wins a duel with C, given that it is C's shot. If A misses and B hits, then the problem becomes P(A, B; A). Notice that the other two sub-problems P(A, B; B) and P(A, C; A) cannot occur due to the order in which the players have to shoot and the fact that we're analyzing the subproblems here in a second from immediately from the point that either B or C dies to simplify the problem. The probability that A wins a duel with C, with C shooting first is calculated as such:
P(A, C; C) = (!C)(A) + (!C)(!A)(!C)(A) + (!C)(!A)(!C)(!A)(!C)(A) + .... = Sum from n=0 to infinity {(!A)(!C)}^n (!C)(A) = 1/3
P(A, B; A) = A + (!A)(!B)A + (!A)(!B)(!A)(!B)A + ... = Sum from n = 0 to infinity {(!A)(!B)}^n (A) = 4/7
Now we express the probility of all subproblems in terms of the initial problem
P(A, B, C;A) = (!A)(B) P(A, B; A) + (A) P(A, C; C) + (!A)(!B)(!C) P(A, B, C; A)
Notice how we have disregarded all possibilities of the initial problem where C shoots A and wound up with the problem resetting itself when all 3 people miss, giving P(A, B, C;A) on both sides of the equations. Plugging in these values we end up with:
P(A, B, C; A) = 12/56 + 1/6 + 1/16 * P(A, B, C;A)
P(A, B, C; A) = 128/315
Notice that this problem is different from most dueling problems in that it assumes that the players blindly shoot at the other players without trying to employ a strategy to intentionally miss the opponent.