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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)

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re: Question | Comment 2 of 17 |
(In reply to Question by friedlinguini)

I would think (this being a math problem), that we should as always assume that all involved are super-smart, and always do the logical thing to ensure the best outcome for themselves (in this case, it's the higher chance of survival)
  Posted by levik on 2002-08-27 11:33:30

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