Duel of probabilities

You win if you shoot your brother first (prob. = 1/3), or if you miss, he misses, then you hit (prob. = 2/3 1/2 (1/3) = (1/3)^2), or if you miss, he misses, you miss, he misses, you hit (prob. = 2/3 (1/2) (2/3) (1/2) (1/3) = (1/3)^3, etc. Therefore, the total probability of you winning is the sum as n goes from 1 to infinity of (1/3)^n = (1/3)*(1/(1-(1/3)) = 1/(3-1) = 1/2.

Let W, M1 and M2 be the events "Win the duel", "You do a good shot" and "Your brother do a good shot". We are given P(M1) = $\frac{1}{3}$ and P(M2) = $\frac{1}{2}$ . First note that if you miss your shot and your brother do the same we will be returned at the beginning of the problem. Thus we have :

$P(W) = P(M1) + [1 - P(M1)][1 - P(M2)] \times P(W)$

Let P(W) = x. We have :

x = $\frac{1}{3}$ + (1 - $\frac{1}{3}$ )(1 - $\frac{1}{2}$ )x

Solving for x gives $\boxed{\frac {1}{2}}$