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Duel (Posted on 2003-03-06) Difficulty: 3 of 5

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?

See The Solution Submitted by np_rt    
Rating: 3.3333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
another method | Comment 2 of 4 |
Looking at the first six terms for each of the 2 shooters, you can see the pattern and easily express it as the sum of an infinite series.

For the "heads" side it is (1/3)∑(1/2)^n+(1/3)∑(1/2)^(2n), and for the "tails" side (1/3)∑(1/2)^n+(1/3)∑(1/2)^(2n-1). With a little mathematical manipulation, and the knowledge that the first sum to infinity is 1, this breaks down to "heads" as (1/3)+(1/3)∑(1/2)^(2n) and "tails" as (1/3)+(2/3)∑(1/2)^(2n). Now, knowing that the two probibilities are mutually exclusive and certain (i.e., they sum to 1), you can determine that the two remaining "sum" parts must total to 1/3.

So then, for the "heads" side you end up with (1/3) + (1/3)*(1/3), or 4/9 as staetd before
  Posted by Cory Taylor on 2003-03-07 04:49:25
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