# Probability of getting n heads when n is very large [closed]

How to show that the probability of getting n heads when 2n times coin is toss is very small? Moreover, how to show that the probability of more than n heads is close to 0.5?

## closed as off-topic by Em., user1551, user91500, Claude Leibovici, user223391 Jul 4 '16 at 21:17

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• Is the coin fair? Are you familiar with the binomial distribution? – carmichael561 Jul 4 '16 at 2:35
• yes, the coin is fair. I am familiar with binomial distribution as well as stirling approximation. However, I am not getting any concrete answer. – meta_finance Jul 4 '16 at 2:39
• I think Stirling's formula should do the trick. For the second part, if $X$ denotes the number of heads, then $\mathbb{P}(X<n)+\mathbb{P}(X=n)+\mathbb{P}(X>n)=1$, $\mathbb{P}(X<n)=\mathbb{P}(X>n)$, and $\mathbb{P}(X=n)$ is small. – carmichael561 Jul 4 '16 at 2:41
• Oh. Got it. Thanks. I didn't give it a thought this way. – meta_finance Jul 4 '16 at 2:42
• @WillFisher clearly it is the probability of getting exactly, since that is the probability which is small as $n$ grows large. At least $n$ heads approaches $0.5$ which is not "small" in the context of probability. – JMoravitz Jul 4 '16 at 2:46

Let $X$ denote the random variable counting the number of times heads occurs in $2n$ independent flips of a fair coin.

Via the binomial distribution:

$Pr(X=k) = \dfrac{\binom{2n}{k}}{2^{2n}}$

More specifically, $Pr(X=n) = \dfrac{\binom{2n}{n}}{2^{2n}}$

Via stirling's approximation, $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$ as $n$ grows large, so the above simplifies to:

$Pr(X=n)\sim \dfrac{4^n/\sqrt{\pi n}}{4^n}=\frac{1}{\sqrt{\pi n}}\to 0$ as $n$ grows large.

For the second half, as mentioned in comments above, we recognize that $Pr(X<n)=Pr(X>n)$ due to the coin being fair and that $Pr(X<n)+Pr(X=n)+Pr(X>n)=1$ as the three events form a partition of the sample space.

From the previous part, since $Pr(X=n)\to 0$ and the observation, this implies that $2Pr(X<n)\to 1$, and thus $Pr(X<n)\to 0.5$