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If I toss an even number of coins, how can I calculate the probability to obtain head or tail? This question is different from the other because I can fling the coin a different number of times but the number is always even. What is the formula to compute the probability?

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up vote 2 down vote accepted

We through 2n coins, what is the probability we get n heads and n tails?

In how many ways could this happen? $\binom{2n}{n}=\frac{2n!}{n!^2}$

What is the total number of throws possible? $2^{2n}$

Since all ordered scenarios are equally likely then the probability?


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Assume you toss the coin $2n$ times, then the total number of possible results is $2^{2n}$. How many of those have the same number of heads and tails? You need $n$ heads and $n$ tails, so to count this you choose $n$ spots for heads and you put tails in all the other spots. This can be done in $\binom{2n}{n}$ ways.

In this way you get $$P(\textrm{"same number of heads and tails with $2n$ flips"})=\frac{\binom{2n}{n}}{2^{2n}}$$

EDIT: If you use Stirling's Approximation, you can get the asymptotic behavior of the probability, as $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$ you get $$P(\textrm{"same number of heads and tails with $2n$ flips"})\sim \frac{1}{\sqrt{\pi n}}$$

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Probability of getting exactly $n$ heads fron $2n$ throws of fair coin = $\frac{(2n)!}{n!n!} 0.5^{2n}$

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If the number of coins is $2n$, then the probability of flipping exactly half heads is $$2^{-2n}\binom{2n}{n},$$ where $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$.

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For $2n$ coins it is $${2n \choose n}{1\over 4^n}.$$ If you use Stirling's formula, it is asymptotic to $1\over \sqrt{n}$.

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Isn't there an extra factor of $\frac{1}{\sqrt{\pi}}$ in the asymptotic equivalent? – Clement C. Feb 4 '14 at 14:10

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