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Math Help - Variation of a coin flip problem

  1. #1
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    Variation of a coin flip problem

    Question:
    A, B, C take turns fliping a coin. Calculate the probability that A wins, B wins, C wins; the first to flip a head. Assume A flips, then B, then C.

    I understand it follows a sequeance {1, 01, 001, 0001.....etc)
    so theres can be infinity amount of turns before 1 gets a head.
    So how would you add the probability of all the A's wining {1, 0001, 0000001......}
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  2. #2
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    First, what is the probability that the first head is flipped at time T=n (for n\geq 1) ? This is the same as saying that the n-1 first throws are tails, and the n-th one is a head. So this probability is, using independence: P(T=n)=\left(\frac{1}{2}\right)^{n-1}\times\frac{1}{2}.

    Then, for instance, the probability that A wins is P(T=1)+P(T=4)+\cdots+P(T=3k+1)+\cdots=\sum_{k=0}^\  infty P(T=3k+1) =\sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(3k+1)-1}\times\frac{1}{2}. Remembering that \sum_{k=0}^\infty a^k=\frac{1}{1-a} when |a|<1, the probability finally becomes: \frac{1}{1-\left(\frac{1}{2}\right)^3}\times\frac{1}{2}=\frac  {4}{7}.

    Changing 3k+1 into 3k+2 and 3k+3, you get the B and C cases.
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  3. #3
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    Here is another proof.

    Let p_A,p_B,p_C denote the probabilities that A, B or C wins. We have of course p_A+p_B+p_C=1.

    In addition, for B to win, the first coin must be a tail, and then it is the same condition as for A (for a shifted sequence of throws), so that p_B=\frac{1}{2}p_A. The same way, for C to win, the first two throws must be tails, and then the same condition as that for A to win must be checked for the following throws, so that p_C=\frac{1}{4}p_A.

    Coming back to the first equation, plugging in the last two relations, you get that \frac{8}{7}p_A=1, hence p_A=\frac{7}{8}.
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  4. #4
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    i got confused now, how come the two probabilities aren't the same??
    one is 4/7, the other 7/8?
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  5. #5
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    Quote Originally Posted by weakmath View Post
    i got confused now, how come the two probabilities aren't the same??
    one is 4/7, the other 7/8?
    Oh yes, thank you ! This is a typo in the second solution : 1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}, so we get \frac{7}{4}p_A=1 and p_A=\frac{4}{7}.
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