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Math Help - Russian Roulette Statistics

  1. #1
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    Russian Roulette Statistics

    I have a problem concerning Russian Roulette. Not a personal problem, mind you, but a statistical one. The game is a sole player with 1 gun, 1 bullet and the chamber is spun each time. I believe I have MOST of what I need to solve the problem, but am unsure.

    Starting with this probability eq:
    P(n)=N!/(n!*(N-n)!)*p^n*q^(N-n)

    I think I have the constants set out to be
    N=# of rounds played
    p= probability of getting the bullet=1/6
    q=probability of getting empty round=5/6

    If the question is what is the probability of still being alive after N rounds, I can't figure what n should be. I think with that, the problem should be solved, but I can't wrap my mind around what it should be.

    ps. that's q to the (N-n) power, btw, I don't think it looks quite right
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  2. #2
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    Quote Originally Posted by pentaquark View Post
    I have a problem concerning Russian Roulette. Not a personal problem, mind you, but a statistical one. The game is a sole player with 1 gun, 1 bullet and the chamber is spun each time. I believe I have MOST of what I need to solve the problem, but am unsure.

    Starting with this probability eq:
    P(n)=N!/(n!*(N-n)!)*p^n*q^(N-n)

    I think I have the constants set out to be
    N=# of rounds played
    p= probability of getting the bullet=1/6
    q=probability of getting empty round=5/6

    If the question is what is the probability of still being alive after N rounds, I can't figure what n should be. I think with that, the problem should be solved, but I can't wrap my mind around what it should be.

    ps. that's q to the (N-n) power, btw, I don't think it looks quite right
    The equation should be expressed as follows.

    If you're alive afer N rounds then the bullet hasn't been shot after N trials. The probability of beng alive is therefore p^N.

    By the way, I don't think the binomial distribution appropriate here. Does it make sense to have more than zero failures in N trials ....?
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  3. #3
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    Ok, I see. I was way over thinking the problem.

    Yeah, I see why the BD wouldn't work now. Thanks!
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