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Math Help - russian roulette

  1. #1
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    russian roulette

    I was just wondering how to calculate the expected number of trigger pulls required to shoot a single bullet out of a six-shot revolver.
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  2. #2
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    Quote Originally Posted by frankdent1 View Post
    I was just wondering how to calculate the expected number of trigger pulls required to shoot a single bullet out of a six-shot revolver.
    Spinning the cylinder between shots:

    Probability that you die on the n-th shot p(n)= (5/6)^{n-1}(1/6), so the expected number of shots is:

    E(n)=\sum_{n=1}^{\infty} n(5/6)^{n-1}(1/6)=6

    If the cylinder is not spun between shots we have:

    prob of dying on first shot p(1)=1/6

    p(2)=(5/6)(1/5)

    p(3)=(5/6)(4/5)(1/4)

    p(4)=(5/6)(4/5)(3/4)(1/3)

    p(5)=(5/6)(4/5)(3/4)(2/3)(1/2)

    p(6)=(5/6)(4/5)(3/4)(2/3)(1/2)(1)

    and the expected number of shots is:

     <br />
E(n)=1 \times p(1)+2 \times p(2) + 3 \times p(3) +4 \times p(4)+5 \times p(5) + 6 \times p(6) = 3.5<br />

    RonL
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  3. #3
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    Another related question.

    In 2 player russian roulette, alternating and not spinning the cylinder between shots, what's the probability the first guy will die?
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  4. #4
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    Nevermind it's 50/50 because it's 3.5, which is between 3rd shot (player 1) and 4th shot (player 2). Although you'd think the guy who goes first would have a disadvantage.
    Last edited by frankdent1; August 9th 2008 at 02:29 PM.
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  5. #5
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    Quote Originally Posted by frankdent1 View Post
    Although you'd think the guy who goes first would have a disadvantage.
    He would if it you spin the wheel each time. Otherwise the loser is predetermined and there would be a 50-50 chance because there is a 50% chance it's in an odd round, and a 50% chance it's in an even round.
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