1. ## 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

2. 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%.

3. 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%

4. 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.

5. 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

6. 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

RonL

7. 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