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.
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Posted by Charlie
on 2003-03-07 04:08:18 |
solution
