aez-notes

\(A\), \(B\) and \(C\) are to fight a three-cornered pistol duel. All know that \(A\)'s chance of hitting his target is 0.3, \(C\)'s is 0.5 and \(B\) never misses. They are to fire at their choice of target in succession in the order \(A\), \(B\), \(C\) cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unhit. What should \(A\)'s strategy be?

If we accept that \(B\) will always go for \(C\) before they go for \(A\) (which is true), then we can work out that if \(A\) starts by hitting \(B\) then their overall chance of winning is the sum of hitting \(B\) and then \(C\) or missing \(B\) on the first hit and then getting them on the second: about 0.279

load(simplify_sum);

probLose2C : simplify_sum(sum(1/2 * (7/20)^i, i, 0, inf));

startOnBWin : 3 / 10 * (1 - probLose2C) +
 7 / 10 * 3 / 10;
startOnBWin, numer;
/* approx 0.279 */

Going for \(C\) first leads to a win only if they miss on the first go and get \(B\) after they have shot \(C\). This leads \(A\) to victory with probability 0.21.

startOnCWin : 7/10 * 3 / 10;
startOnCWin, numer;
/* 0.21 */

On the face of it, it appears that \(A\) should start firing at \(B\), but if they intentially miss both targets, then they have 0.3 probability of being able to hit \(B\) on the next round which is actually their best strategy, even if it is one of those annoying riddle type solutions.