BMAD 62 Report post Posted June 1, 2013 Three men: Fermat, Galois, and Hilbert, decide to fight a pistol duel. They'll stand at the corners of an equilateral triangle, and each man, in order, will aim and shoot wherever he pleases. They choose randomly who will be shooting first, second, and third, and will continue in order until two of them are dead. All three know Fermat always hits his target, Galois is 80% accurate, and Hilbert hits his mark half the time. Assuming that all three adopt the best strategy and that nobody is killed by a wild shot not intended for him, who has the best chance to survive, and why? Find the survival probabilities for each man. Share this post Link to post Share on other sites

0 Xavier 1 Report post Posted June 2, 2013 A,B and C having 100%, 80%, 50% sucess: if A goes 1st, his best strategy is to shot B and hope to survive C if B goes 1st he need to try take down A as taking C out is suicide Now, what's interesting is if C goes first: Killing B is suicide as A will take C down. Taking down A will let B shot as C.. with less than 20% survival chances. So C shot in the air and let AB Sort thing out (they will shot at each other and at least one of them will die before C turn is back). This gives C more than 50% survival (50% if A survive, 55.56% if B does). Working out the numbers i got for winning chances: A=30.00% B=17.78% C=52.22% So the worse shooter has best survival chances. Cute! Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted June 3, 2013 i am getting the same answer but diff. Probabilities. Fermat-30%, Galois - 27.2%, Hilbert - 42.8%. Share this post Link to post Share on other sites

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