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Math Help - Coin Tossing Problem: Formula for accumlating more heads than tails?

  1. #1
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    Coin Tossing Problem: Formula for accumlating more heads than tails?

    Hello

    I have a 'simple' problem but having trouble working out a formula.

    For example with 5 tosses & 3 Heads & 2 tails occurring, there are 5!/(3!(5-3)!) = 10 combinations.

    But how many of these combinations can exist for only more or equal heads tallying to occur before the tails tally occurs.

    For example

    HTHTH
    HTHHT
    HHTTH
    HHTHT
    HHHTT

    are all ok as the tally for H's were equal or more than the T's but

    THTHH
    THHTH
    THHHT
    TTHHH
    HTTHH

    is not as the tally of T's occured before H's did.

    Basically if H= + 1 and T = -1 and you summed up the sequences it cannot be <0 at ANY point in the summing up.

    So I need to work out the no. of outcomes from C combos of getting more heads accumulated before tails: given N no of tosses and X no of heads.

    Any help in the right direction would be great.
    Thanks
    Max
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  2. #2
    Super Member
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    Please read up on Bertrand's ballot theorem. You need something called reflection principle here - pretty elementary but nevertheless an ineresting application !
    Last edited by aman_cc; July 7th 2010 at 03:38 AM.
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  3. #3
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    Brilliant thanks, thats very kind of you to respond and so quickly!
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