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Alice tosses a fair coin $n$ independent times and Bob tosses a fair coin $m$ independent times. Find an elegant or clever argument to compute the probability that they have equal numbers of Tails. It had better not involve any lengthy sums.

The only way I can think of to approach this problem is to express the probability as a sum: $$\sum_{k = 1}^{\min(m, n)} P(\text{Alice gets $k$ tails})*P(\text{Bob gets $k$ tails}) \\ = \sum_{k = 1}^{\min(m, n)} \binom{n}{k} 0.5^n * \binom{m}{k} 0.5^m \\ = 0.5^{n + m}\sum_{k = 1}^{\min(m, n)} \binom{n}{k}\binom{m}{k} $$ However, I am not supposed to involve lengthy sums, but I cannot think of a more elegant way to solve this.

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Alice tosses her coin $n$ times and records $1$ for a tail & $0$ for a head.

Bob tosses his coin $m$ times and records $0$ for a tail & $1$ for a head.

This will generate a binary string of length $n+m$ with $m$ ones (if there are an equal number of tails) \begin{eqnarray*} \underbrace{a_1 a_2 \cdots a_n}_{k \text{ ones and } n-k \text{ zeros}} \,\, \underbrace{b_1 b_2 \cdots b_m}_{k \text{ zeros and } m-k \text{ ones}} \end{eqnarray*} So the probability is $\color{red}{\binom{n+m}{m} \frac{1}{2^{n+m}}}$.

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