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 Solution by Lawrence)

I hate when you can do "neither" but the "question" comment said, "maximize chances of winning." For your solution there is a 100% chance of A missing, a .5 chance of B missing in which case B is dead and A has a .33 chance at C, so A has a .5 * .33 chance of this or a .165 chance.

If B hits C, A has that 50% chance of winning again, so A has a .5*.5 = .25 chance of that occuring.

.165+.25 = .415, so this is better than the 36% chance of A shooting at C. Darn it all!

Posted by Lawrence on 2003-08-27 18:34:24