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Math Help - Flipping Coins

  1. #1
    Foxy Brown
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    Flipping Coins

    As I flip the coin more and more times are my chances of having an equal number of heads and tails getting better or worse?

    FB
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  2. #2
    Site Founder MathGuru's Avatar
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    Interesting question . . .

    There was a german prisoner (or so the story goes) who spent years in a prison flipping a coin and figured out just this. I'll try to find more info about it.

    As a start when you flip 1 time your chances are 0

    2 flips your chances are 50%
    3 flips your chances are 0

    in fact every odd number of flips is 0%.
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  3. #3
    Site Founder MathGuru's Avatar
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    I would like to develop an equation which will show the percentages.

    However until I do that:

    N=1, 0%
    N=2, 50%
    N=3, 0%
    N=4, 37.5%
    N=5, 0%
    N=6, 31.25%
    N=7, 0%

    I beleive your chances of haveing an equal number of heads and tails decreases towards zero the more flips you make. After two flips is your best chance. This is a very interesting idea, that the more you flip the less chances of having equal heads and tails. Work is below.


    2 Coins (2*2=4 possibilities)

    HT
    TH

    HH
    TT

    2/4 Chances = 50%

    4 Coins (2*2*2*2=16 possibilities)

    HHHH
    HHHT
    HHTH

    HHTT
    HTHH
    HTHT
    HTTH

    HTTT
    THHH

    THHT
    THTH

    THTT
    TTHH
    TTHT
    TTTH
    TTTT

    6/16 Chances = 37.5%

    6 Coins (2*2*2*2*2*2 = 64 possibilities)

    HHHHHH
    HHHHHT
    HHHHTH
    HHHHTT
    HHHTHH
    HHHTHT
    HHHTTH

    HHHTTT
    HHTHHH
    HHTHHT
    HHTHTH

    HHTHTT
    HHTTHH
    HHTTHT
    HHTTTH

    HHTTTT
    HTHHHH
    HTHHHT
    HTHHTH

    HTHHTT
    HTHTHH
    HTHTHT
    HTHTTH

    HTHTTT
    HTTHHH

    HTTHHT
    HTTHTH

    HTTHTT
    HTTTHH
    HTTTHT
    HTTTTH
    HTTTTT
    THHHHH
    THHHHT
    THHHTH

    THHHTT
    THHTHH
    THHTHT
    THHTTH

    THHTTT
    THTHHH

    THTHHT
    THTHTH

    THTHTT
    THTTHH
    THTTHT
    THTTTH
    THTTTT
    TTHHHH
    TTHHHT
    TTHHTH

    TTHHTT
    TTHTHH
    TTHTHT
    TTHTTH
    TTHTTT

    TTTHHH
    TTTHHT
    TTTHTH
    TTTHTT
    TTTTHH
    TTTTHT
    TTTTTH
    TTTTTT

    20/64 Chances = 31.25%
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  4. #4
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    n Choose r, also known as nCr(n,r).

    n = number of coins
    r = n/2

    P = nCr(n,n/2) / 2^n


    Just my wild guess from your patterns.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Foxy Brown
    As I flip the coin more and more times are my chances of having an equal number of heads and tails getting better or worse?

    FB
    Is late better than never? Probably not in this case.

    But if the number of trials N=2n then probability of exactly
    m heads has a binomial distribution so:

    P(m|2n)={2n \choose m}0.5^m\ 0.5^{2n-m}={2n \choose m}0.5^{2n}

    Now we can rephrase the question to: is P(n|2n) decreasing?

    I will now use Stirlings formula to derive the asymtotic form for P(n|2n).

    Stirlings formula is:

    <br />
n!\ \sim \sqrt{2 \pi n}\ n^n\ e{-n} <br />
,

    so:

    <br />
{2n \choose n} \sim \frac{\sqrt{2\pi (2n)}(2n)^{2n}e^{-2n}}{\sqrt{2\pi n}\sqrt{2\pi n}\ n^{2n}e^{-2n}}=\frac{ 2^{2n}}{\sqrt{\pi\ n}}<br />

    Hence:

    <br />
P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}}<br />

    So as N=2n \to \infty,\ P(n|2n) \to 0

    RonL
    Last edited by CaptainBlack; June 17th 2006 at 02:29 AM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    :
    :

    Hence:

    <br />
P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}}<br />
    So how large does this n have to be to give a good approximation?

    Here is a table comparing this with the exact calculation:

    Code:
           N        n       exact      asymt
    
       2.0000     1.0000    0.5000    0.5642 
       4.0000     2.0000    0.3750    0.3989 
       6.0000     3.0000    0.3125    0.3257 
       8.0000     4.0000    0.2734    0.2821 
      10.0000     5.0000    0.2461    0.2523 
      12.0000     6.0000    0.2256    0.2303 
      14.0000     7.0000    0.2095    0.2132 
      16.0000     8.0000    0.1964    0.1995 
      18.0000     9.0000    0.1855    0.1881 
      20.0000    10.0000    0.1762    0.1784 
      22.0000    11.0000    0.1682    0.1701 
      24.0000    12.0000    0.1612    0.1629 
      26.0000    13.0000    0.1550    0.1565 
      28.0000    14.0000    0.1494    0.1508 
      30.0000    15.0000    0.1445    0.1457 
      32.0000    16.0000    0.1399    0.1410 
      34.0000    17.0000    0.1358    0.1368 
      36.0000    18.0000    0.1321    0.1330 
      38.0000    19.0000    0.1286    0.1294 
      40.0000    20.0000    0.1254    0.1262 
      42.0000    21.0000    0.1224    0.1231 
      44.0000    22.0000    0.1196    0.1203 
      46.0000    23.0000    0.1170    0.1176 
      48.0000    24.0000    0.1146    0.1152 
      50.0000    25.0000    0.1123    0.1128 
      52.0000    26.0000    0.1101    0.1106 
      54.0000    27.0000    0.1081    0.1086 
      56.0000    28.0000    0.1061    0.1066 
      58.0000    29.0000    0.1043    0.1048 
      60.0000    30.0000    0.1026    0.1030 
      62.0000    31.0000    0.1009    0.1013 
      64.0000    32.0000    0.0993    0.0997 
      66.0000    33.0000    0.0978    0.0982 
      68.0000    34.0000    0.0964    0.0968 
      70.0000    35.0000    0.0950    0.0954 
      72.0000    36.0000    0.0937    0.0940 
      74.0000    37.0000    0.0924    0.0928 
      76.0000    38.0000    0.0912    0.0915 
      78.0000    39.0000    0.0901    0.0903 
      80.0000    40.0000    0.0889    0.0892 
      82.0000    41.0000    0.0878    0.0881 
      84.0000    42.0000    0.0868    0.0871 
      86.0000    43.0000    0.0858    0.0860 
      88.0000    44.0000    0.0848    0.0851 
      90.0000    45.0000    0.0839    0.0841 
      92.0000    46.0000    0.0830    0.0832 
      94.0000    47.0000    0.0821    0.0823 
      96.0000    48.0000    0.0812    0.0814 
      98.0000    49.0000    0.0804    0.0806 
     100.0000    50.0000    0.0796    0.0798
    Not bad?

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    Hence:

    <br />
P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}}<br />
    This exact same approximation can also be obtained by considering the
    normal approximation to the binomial distribution, but I won't give
    that derivation here.

    RonL
    Last edited by CaptainBlack; June 17th 2006 at 02:48 PM.
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