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Thread: Gambling odds probability

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    Gambling odds probability




    2. A gambler offers you to pay even odds if you get 9, 10 or 11 heads in 20 tosses. Should you take the bet?
    Assume it is a fair coin.
    X=number of heads
    Sample space of X = {0,1, 2, 3, 20}
    Sample space has 2^50 or 1048576 elements.
    Px (X=9) = (20 choose 9) / 2^20 =0.1602
    Px (X=10) = (20 choose 10) / 2^20 =0.1762
    Px (X=11) = (20 choose 11) / 2^20 =0.1602

    You should not take the bet, the probability of getting 9 heads is 16%; 10 heads 18%; and 11 heads is 16%. The probability of 9, 10 or 11 heads is 49.6%.

    BUT am I missing something about non-mutually exclusive events? If you get 11 heads you have 10 and 9 heads also and if you have 10 heads you have 9 also. How do I solve for the intersection of the events of 9, 10, 11 heads in order to subtract this?

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    Re: Gambling odds probability


    X = 9 and X = 10 are mutually exclusive here. Those are the outcomes "Observed _exactly_ n heads" for n = 9, 10, 11. So If you get exactly 10 heads then you didn't also get exactly 9 heads.
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