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Question: A fair coin is independently flipped $n$ times, $k$ times by $A$ and $n − k$ times by $B$. Show that the probability that $A$ and $B$ flip the same number of heads is equal to the probability that there are a total of $k$ heads.

I know the probability of getting heads or tails is the same for each because the coin is fair. I also know the probability of an arbitrary number, say, $m$ heads is equal to probability of getting $m$ tails.

So I know $P(A$ gets $x$ tails) = $P(B$ gets $x$ heads)

However, I'm confused as to where to go and how to apply this to the problem. Any help appreciated!

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By symmetry we could assume $k \le n-k$. The probability that $A$ and $B$ flip the same number of heads would be $\sum_{i=0}^{k}{\binom{k}{i}\binom{n-k}{i}(\frac{1}{2})^n} = (\frac{1}{2})^n\sum_{i=0}^{k}{\binom{k}{k-i}\binom{n-k}{i}} = (\frac{1}{2})^n\binom{n}{k}$, which is exactly the probability to get $k$ heads.

The second equation comes from a basic combinatorics formula: Choosing $k$ from $n$ balls could be done by choosing $k-i$ from the first $k$ balls and then choosing $i$ from the rest $n-k$ balls.

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  • $\begingroup$ PS: This is called Vandermonde's identity: $$\sum_{i=0}^k\binom mi\binom n{k-i}=\binom{m+n}k$$ $\endgroup$ – Graham Kemp Oct 31 '18 at 2:31

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