Flipping Coins

Foxy Brown · March 31st 2005, 03:58 PM

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

**MathGuru** · April 6th 2005, 11:04 AM

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%.

**MathGuru** · April 9th 2005, 12:12 PM

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%

**paultwang** · April 9th 2005, 01:29 PM

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.

**CaptainBlack** · June 17th 2006, 01:12 AM

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:

$ n!\ \sim \sqrt{2 \pi n}\ n^n\ e{-n} $ ,

so:

$ {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}} $

Hence:

$ P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}} $

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

RonL

**CaptainBlack** · June 17th 2006, 01:28 AM

Originally Posted by CaptainBlack

:
:

Hence:

$ P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}} $

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

**CaptainBlack** · June 17th 2006, 01:37 AM

Originally Posted by CaptainBlack

Hence:

$ P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}} $

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

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