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Head and tails game

  1. May 29, 2007 #1
    I ve got this simple problem and I can can not find the solution.
    Two men play head and tails. Each man flips the coin n times, every time he has head, he should add one point to his score. At the end, they compare the scores, whose score has more points , the man is the winner.
    A is the probability of a man lose or win, B is the probability when the two have the same points. Calculate A and B?
    Thanks
     
  2. jcsd
  3. May 29, 2007 #2
    The first thing you want to do is show us where you're stuck. As a hint, find the probability that person 1 rolls m heads, then find the probability that the other rolls more than m heads.
     
  4. May 29, 2007 #3
    Thanks for your answering me
    My way is to find the B value meaning the two are equal. So i tried to calculate the probabiltiy of 1st man to have m heads, then the 2nd also must have m heads, which is called Pm. Then I will sum up Pm with m running from 0 to n. But the formula is very long and i think there must be mistakes somewhere..
    P0 = 1/2^(2n)
    P1=2.n/2^2n
    P2=..
     
  5. May 29, 2007 #4

    chroot

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    Your sum is correct. Are you sure you have to actually evaluate the sum, or can you just leave your answer in terms of the sum?

    This is not a trivial sum to compute.

    - Warren
     
    Last edited: May 29, 2007
  6. May 30, 2007 #5
    Can anyone find the general formula for Pm?, at least one that can be computed using a program.
     
  7. May 30, 2007 #6

    chroot

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    "Pm" is simple -- it's a Binomial distribution:

    [itex]
    p \left( m \right) = \left( {\begin{array}{*{20}c}
    n \\
    m \\

    \end{array} } \right)p^m \left( {1 - p} \right)^{n - m} = \left( {\begin{array}{*{20}c}
    n \\
    m \\

    \end{array} } \right)2^{ - n}
    [/itex]

    where p is the probability of heads (1/2), n is the number of flips, and m is the number of heads.

    Remember that the probability of both men getting m heads is actually [itex][p(m)]^2[/itex].

    - Warren
     
    Last edited: May 30, 2007
  8. May 30, 2007 #7
    So I can not calculate P0, when the two men both have no heads ?

    Oh I m sorry, I thought (n/m) is a division!.
     
    Last edited: May 30, 2007
  9. May 30, 2007 #8

    AlephZero

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    You can get an approximate answer by replacing the binomial distribution by a normal distribution. The pdf of the "two scores are equal" distribution is the square of the pdf of the normal distribution.

    Since [tex](e^{-x^2})^2 = e^{-2x^2}[/tex] this is similar to a normal distribution so you can integrate it.

    The approximate probability of equal scores after n tosses is [tex]1/\sqrt{n\pi}[/tex].
     
  10. May 7, 2009 #9
    please explain me how do this
     
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