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Math Help - Rigged Game of Heads or Tails

  1. #1
    Senior Member ecMathGeek's Avatar
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    Rigged Game of Heads or Tails

    Hi guys. I've been pretty inactive in the past few weeks due to finals and other things coming up. Anyways, here's my question:

    The problem I'm about to describe might seem a little obscure. I'll give an example after so that it's more clear.

    If there are two events that can occur, event A and event B, and the likelyhood of event A happening is n (where n is some rational number such that 1 < n < 0) and the odds of B happening is 1 - n. What are the odds of event B happening "r" times out of "m" trials, where the order of wins and losses does not matter?

    Example: In a rigged game of "Heads or Tails," the odds of "winning" (you choose heads and it comes heads or you choose tails and it comes tails) is 75% or 3/4 (so that 3 out of 4 times that you play you will win no matter which side of the coin you chose). What are the odds of loosing 10 times (total, not necessarily in a row) if you play the game 12 times?

    Hope that made sense, and thanks for the help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    Hi guys. I've been pretty inactive in the past few weeks due to finals and other things coming up. Anyways, here's my question:

    The problem I'm about to describe might seem a little obscure. I'll give an example after so that it's more clear.

    If there are two events that can occur, event A and event B, and the likelyhood of event A happening is n (where n is some rational number such that 1 < n < 0) and the odds of B happening is 1 - n. What are the odds of event B happening "r" times out of "m" trials, where the order of wins and losses does not matter?

    Example: In a rigged game of "Heads or Tails," the odds of "winning" (you choose heads and it comes heads or you choose tails and it comes tails) is 75% or 3/4 (so that 3 out of 4 times that you play you will win no matter which side of the coin you chose). What are the odds of loosing 10 times (total, not necessarily in a row) if you play the game 12 times?

    Hope that made sense, and thanks for the help.
    I will presume you mean the probability of B happening is 1-n.

    Then the number of wins for B in m trials has a binimial distribution B(m,1-n) so:

    <br />
p(r\mbox{ wins for B}) = b(r,m,n) = \frac{m!}{(m-r)! r!} (1-n)^r n^{m-r}<br />

    RonL
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I will presume you mean the probability of B happening is 1-n.
    That is what I meant. Thank you very much, CaptainBlack.
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