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ValeryOchkov24-Ruby IV
24-Ruby IV
June 19, 2012
Question

Three-Cornered Duel

  • June 19, 2012
  • 1 reply
  • 8477 views

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 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?

Can you check my solution of this task (Three-Cornered Duel) in Mathcad Prime 2.0!

See the attach!

    1 reply

    H
    hilbert1-Visitor
    1-Visitor
    June 23, 2012

    Duelist A should shoot into the ground and let B and C duel.

    This gives A the first shot at who ever is left which gives A a better chance than .3 to win the duel.

    V
    ValeryOchkov24-Ruby IVAuthor
    24-Ruby IV
    June 24, 2012

    Patrick Maxfield wrote:

    Duelist A should shoot into the ground and let B and C duel.

    This gives A the first shot at who ever is left which gives A a better chance than .3 to win the duel.

    Thanks, Patrik!

    It is a well-known answer.

    I have asked - is the attached sheet (the Monte Carlo Method) correct.

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