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Three for all (Posted on 2002-08-27) Difficulty: 4 of 5
You are in a free-for-all shootout involving three people, and by some rotten luck, you are the worst shooter of the group. You, shooter A, hit your target 33% of the time. The other two, B and C, have 50% and 100% accuracy respectively.

At least you get to shoot first, with B going after you, and C going last. Everyone takes turns shooting until all but one are dead.

Who will you fire at in the first round to maximize your odds of survival?

See The Solution Submitted by levik    
Rating: 4.0000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: So close! | Comment 6 of 17 |
(In reply to So close! by TomM)

Well, what I said was that the situations would be identical, so therefore they could be taken off the table. It's a constant term that cancels out when deciding on the better target.
  Posted by friedlinguini on 2002-08-28 03:23:08

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