Deadly Triangle

[BMAD]

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.

[Xavier]

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!

[dark_magician_92]

i am getting the same answer but diff. Probabilities. Fermat-30%, Galois - 27.2%, Hilbert - 42.8%.