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
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%.
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%
Is late better than never? Probably not in this case.Originally Posted by Foxy Brown
But if the number of trials $\displaystyle N=2n$ then probability of exactly
$\displaystyle m$ heads has a binomial distribution so:
$\displaystyle P(m|2n)={2n \choose m}0.5^m\ 0.5^{2n-m}={2n \choose m}0.5^{2n}$
Now we can rephrase the question to: is $\displaystyle P(n|2n)$ decreasing?
I will now use Stirlings formula to derive the asymtotic form for $\displaystyle P(n|2n)$.
Stirlings formula is:
$\displaystyle
n!\ \sim \sqrt{2 \pi n}\ n^n\ e{-n}
$,
so:
$\displaystyle
{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:
$\displaystyle
P(n|2n) \sim \frac{ 2^{2n}}{\sqrt{\pi\ n}}\times 0.5^{2n}=\frac{1}{\sqrt{\pi\ n}}
$
So as $\displaystyle N=2n \to \infty,\ P(n|2n) \to 0$
RonL
So how large does this $\displaystyle n$ have to be to give a good approximation?Originally Posted by CaptainBlack
Here is a table comparing this with the exact calculation:
Not bad?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
This exact same approximation can also be obtained by considering theOriginally Posted by CaptainBlack
normal approximation to the binomial distribution, but I won't give
that derivation here.
RonL