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I notice that the function $\binom{C}{x}$, where $C$ is some constant, resembles a Gaussian function; for example, here is the plot for $\binom{20}{x}$:

enter image description here

This corresponds to the Gaussian function $a e^{- { \frac{(x-b)^2 }{ 2 c^2} } }$, where $a$ is $\binom{20}{10}$, $b$ is 10, and $c$ (determined through curve fitting) is ~2.2689.

$\binom{20}{x}$ corresponds to $\frac{20!}{x!(20-x)!}$; how is this related to a Gaussian function?

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It's a bit hard to imagine that this question hasn't come up before, but I did not look very hard for it. – MJD May 24 '12 at 4:30
Agree, Mark. It is also covered in all the textbooks. Yet, it is possible to see this on you own without really studying the topic. I recall a certain schoolboy collecting data on the sum of the rolls of 5 dice... – Jyrki Lahtonen May 24 '12 at 4:37
Sure! I just meant that there might already be an answer on this site that is better than my answer, and we should link to it. – MJD May 24 '12 at 4:38
up vote 4 down vote accepted

${20\choose x}*2^{-20}$ is the probability of getting exactly $x$ heads in 20 coin flips. Or put another way, it is the sum of 20 random variables, each of which has probability $\frac12$ of having the value 0 heads and probability $\frac12$ of having the value 1 head.

The central limit theorem says that the mean of a large number of independent, identical random variables is (subject to a few conditions) close to a Gaussian distribution. That is what we have here, the sum of 20 copies of the same distribution.

The central limit theorem is the reason for the ubiquity of the Gaussian distribution in the natural world, in things like people's heights. If you only know three things about probability, the central limit theorem should be one of them.

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Thanks for the explanation; it's very accessible. – Hypercube May 24 '12 at 4:49

Ignoring normalization, you have noticed an instance of the de Moivre-Laplace theorem (a special case of the central limit theorem) which shows how the binomial distribution can be approximated by a normal distribution for $n$ large. You have $n=20$ and $p=q =1/2$. The theorem tells us the mean for the relevant normal distribution is $n p = 10$ and the standard deviation $\sqrt{n p q} = \sqrt{5}$.

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That's interesting. I'll have to look more into it. – Hypercube May 24 '12 at 4:47
@Hypercube: It's worth it! – user26872 May 24 '12 at 4:54

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