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Consider first a simpler duel where there are only two shooters M and N. Assume that M hits with probability m and misses with probability N with probability n. Assume that M starts first. Let q = (1-m)(1-n). Then q is the probability that in a cycle they both miss. Then if M starts this two person duel the probability that he wins is: | Consider first a simpler duel where there are only two shooters M and N. Assume that M hits with probability m and misses with probability N with probability n. Assume that M starts first. Let q = (1-m)(1-n). Then q is the probability that in a cycle they both miss. Then if M starts this two person duel the probability that he wins is: | ||
m + qm + q^2m + q^3m + ...= m(1 + q + q^2 + ... = m/(1-q). | <center><math> m + qm + q^2m + q^3m + ...= m(1 + q + q^2 + ...) = m/(1-q).</math></center> | ||
Revision as of 15:05, 4 March 2007
The solution given by the Car Talk boys has the right idea but they struggle a bit with the mathematics. This is a well known problem and occurs in most math puzzles books. It is problem 20 in Mosteller's famous book "Fifty Challenging Probability Problems".
We shall call the participents A, B, and C and the probabilities that they hit when they shoot a,b,c. Then in the Car Talk problem a = 1/3, b = 2/3 and c = 1
Consider first a simpler duel where there are only two shooters M and N. Assume that M hits with probability m and misses with probability N with probability n. Assume that M starts first. Let q = (1-m)(1-n). Then q is the probability that in a cycle they both miss. Then if M starts this two person duel the probability that he wins is: