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Say we have Tom and John, each tosses a fair coin $n$ times. What is the probability that they get same number of heads?

I tried to do it this way: individually, the probability of getting $k$ heads for each is equal to $$\binom{n}{k} \Big(\frac12\Big)^n.$$ So, we can do $$\sum^{n}_{k=0} \left( \binom{n}{k} \Big(\frac12\Big)^n \cdot \binom{n}{k}\Big(\frac12\Big)^n \right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac12\right)^{2n},$ but the equality was verified by WolframAlpha -- it's not obvious when you look at it. So I think there's a much easier way to solve this, can someone point it out? Thanks.

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2 Answers 2

up vote 9 down vote accepted

The probability John gets $k$ heads is the same as the probability John gets $n-k$ heads since the coin is fair.

So the answer to the original question is equal to the probability that the sum of Tom's and John's heads is $n$.

That is the probability of $n$ heads from $2n$ tosses which is indeed $\frac{1}{2^{2n}}{2n \choose n}$.

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As you have noted, the probability is $$ p_n = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{k} = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \frac{1}{4^n} \binom{2n}{n} $$ The middle equality uses symmetry of binomials, and last used Vandermonde's convolution identity.

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