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?

(In reply to re: So close! by friedlinguini)

The only way that this shootout can go on over two rounds is if C gets killed by either A or B in round one. Otherwise, C will kill one of the other two, leaving at most two players - one if someone has already been killed - to go on to second round.

In the second round, either C will be shot by the other remainig player, or he will shoot that player. Either way the game ends.

If C survives to shoot in round 1, he can choose to kill B or A. Either way, the surviving contestant will get one shot at C. Since B has a higher chance of hitting C with a shot, it stands to reason that C will shoot B first.

Since B must realize this, he will shoot C when his turn comes, and not A.

Now I need to put numbers to all this...

Posted by levik on 2002-08-28 05:22:08