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Thread: Coin Toss

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    Coin Toss




    Hi,
    There are three people, A, B ,C, tossing a coin. The first one to get heads wins, they toss in succession in the order ABC, show the probability of A winning is 4/7.
    I thought that the probability of A winning would be the sum from n=1 to inf. of (1/2)^4n.. as each time 3 tails have to be thrown and then either another tail or head.

    Is this correct? If so, how can I get a value of 4/7?
    Thanks

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    Re: Coin Toss

    A tosses on 1st, 4th, 7th, ... etc round. So A wins if and only if the first head appears on 1st, or 4th, or 7th, or ... etc round.

    Assuming the coin is fair and each toss is independent.

    Currently your series is the probability of first head appears on 4th, 8th, 12th, ... etc round; so it is not the solution of the problem. Try to correct it; and the remaining part will just be a simple infinite geometric series.

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    plm (11-17-2013)

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    Re: Coin Toss

    Oh yeah, so I would need (1/2)^3n-2 instead?

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    Re: Coin Toss

    Or, as sum from 0 to inf. of (1/2)^(3n+1)?

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    Re: Coin Toss

    Yes both are the same.

    \sum_{n=1}^{+\infty} \left(\frac {1} {2} \right)^{3n-2}
= \frac {4} {7} = \sum_{n=0}^{+\infty} \left(\frac {1} {2} \right)^{3n+1}

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    plm (11-17-2013)

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    Re: Coin Toss


    Thanks for your help!
    Last edited by plm; 11-17-2013 at 12:04 PM.

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