One day your rival invites you out to an old fashion duel. The rules are as follows:
You each take turns shooting each other.
The duel stops when one person hits the other.
Both of you are honourable enough to not take shots out of your turn.
The probability of one person hitting the other is 1/2 and is independent of the probability of the other person hitting you.
Clearly, whoever shoots first has a distinct advantage. So your friends suggests flipping a coin for it. Little did you know that your rival uses a coin rigged in such a way that the probability of getting heads is only 1/3.
If you chose heads, what is the probability of you winning the duel?
First find the probability that the first shooter will win. He has probability 1/2 of hitting on the first shot. If he misses (w/ prob 1/2), there's a 1/2 probability the opponent will miss and then 1/2 that the original shooter will hit the mark. Thus the overall prob of a hit on the second go round is (1/2)(1/2)(1/2) = 1/8. Each subsequent possibility (e.g., a hit on the third try for the first player) is 1/4 the previous, as two additional shots have to be made. Thus the prob. that the first shooter wins is
p = 1/2 + 1/8 + 1/32 + ...
then
4p = 2 + 1/2 + 1/8 + 1/32 + ...
so
3p = 2; and therefore p=2/3
As the probability of shooting first is 1/3 and the prob of shooting second is 2/3, the prob of winning is these multiplied by the conditional prob of winning given shooting 1st and 2nd respectively:
(1/3)(2/3) + (2/3)(1/3) = 4/9, which is the answer.

Posted by Charlie
on 20030307 04:08:18 