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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.)
So close! | Comment 5 of 17 |
(In reply to In that case... by friedlinguini)

You started to consider what if you missed, but didn't follow through. If you miss, B shoots C, If he misses, C shoots B (and doesn't miss). Either way, you are sure to survive round 1 with only one other opponant.

The math gets a little complicated with multiple iterations of 33% chances, but it turns out your chances are slightly better if you are sure that you will miss on first round than if you attempt to kill either opponant. Your best course of action on the first round is to fire into the cieling!
  Posted by TomM on 2002-08-27 15:52:01

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