Weekly puzzle 17
Gunfight
Andy, Bob and Charlie agree to a gunfight under the following conditions. After drawing lots to determine who fires first, second and third, they stand on an equilateral triangle. It is agreed they will fire single shots in turn and continue in the same order until only one survives. At each turn, the man who is firing may take aim wherever he pleases. All three duelists know that Andy always hits his target, Bob is 80pc, and Charlie is the weakest shooter at 50pc. All three are infinitely clever.
Part A
What should Charlie do when his turn comes around ?
Part B
Who has the best survival chance ?
Part C
What are the exact survival probabilities for each man.
Part D
Is it possible to change Charlie’s accuracy in order to give a different man (from Part B) the best survival chance ? If so, how? If not, why?


This must be by far, the “worst” problem you presented here, since I started to read your blog, Andrew. Not because it is particularly conceptually difficult but because is too easy to do an error while trying to solve it. I was going to wait until Sunday to present my solution, but changed my mind: whatever I present, I cannot vow for its validity, so it’s pointless to brag some obfuscated answer to not spoil anyone… Instead, my hopes is that others had already done the problem and came to offer some, “Yes, it gave the same to me”, in order to stop my misery. Of course, I can always wait for the solutions, but that would amount to expect no other else is trying to solve the problem, and what does that says from us, puzzle solvers? Surely we are better than that. So, here’s me candidate solution and, please, please, say if you got the same or something different, please! Please? I don’t want to be alone facing this puzzle.
Answering in order to the questions, Charlie must shoot always the standing-up opponent with best shoot accuracy; he is the one with the best chance survival before drawing the sorts; the exact survival probabilities of each one before the game starts is 29/120, 14/45 and 161/360 respectively for Andy, Bob and Charlie; and yes, there are Charlie’s accuracies for which the order of these survival probabilities are different. If for any reason, the accuracy of Charlie’s drops below ~35.2%, Andy will became the one with best survival chances (if it drops below ~34.5%, Charlie will became last). In the other direction and for reasons easy to understand, Bob will became the best bet, if Charlie accuracy surpasses 4/5 AND that is known to the others. The reason is the same as why Bob had to try to shoot Andy first, even knowing the one who has more chances to win the game is Charlie… in the case of Charlie, if his ability to shoot improves above 80%, his chances of winning the game will drop instantaneously to less than half than before… giving new meaning to “the last will be the first”.
Here… I left out enough numbers, so that anyone can prove they did the problem, presenting them. And please, if you solve it, say so… I can’t stand to wait to know if I did mistakes or not.
PS.: that “All three are infinitely clever” are looming over me… my shooters don’t need to be particularly clever to behave as they do. If this is another recursive pirate’s scenario, then, I’m out of the picture.