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## Homework Statement

"Mr. A and Mr. B fire at each other until one of them is hit. Each duelist has the same probability of hitting each other each time a shot is fired. Mr. A has probability ##a## of hitting, and Mr. B has probability ##b## of hitting. Given Mr. A shoots first, calculate the probability that he wins.

Also calculate the distribution of ##X##, the number of shots until the duel ends, and ##E(X)##."

## Homework Equations

## The Attempt at a Solution

Denote the event of A winning by ##\alpha## and ##A## denote the event of A shooting first. So ##P(\alpha|A)=a## if he gets it on his first try. If he misses, and B gets him, then ##P(\alpha|A)=0## since he loses. But if B misses and A hits on his second try, then ##P(\alpha|A)=(1-a)(1-b)a##. In general, for the ##n-th## round, ##P(\alpha|A)=(1-a)^n(1-b)^na##. So as I see it, the probability of A winning is the union of all these events, and follows a geometric distribution:

##P(\alpha|A)=a+(1-a)(1-b)a+...+(1-a)^n(1-b)^na\text { as }n\rightarrow \infty##

So ##P(\alpha|A)=\frac{a}{1-(1-a)(1-b)}##

I'm not sure if this is correct, since my instructor has informed me that the probability should depend on the parity of the round. Anyway, for the second part of the problem, I reason that ##X## is geometrically distributed with some parameter. I reason that it could be ##a+b##, but I know that it fails for ##a+b>1##, since the terms in the pmf for a geometric r.v. have to be non-negative and between 0 and 1.