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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".
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:
m + qm + q^2m + q^3m + ...= m(1 + q + q^2 + ... = m/(1-q).

Revision as of 15:01, 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:

m + qm + q^2m + q^3m + ...= m(1 + q + q^2 + ... = m/(1-q).