Say we have Tom and John, each tosses a fair coin $n$ times. What is $P$ 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 $$\binom{n}{k}\left(\frac{1}{2}\right)^n.$$ So, we can do $$\sum^{n}_{k=0}\left(\binom{n}{k}\left(\frac{1}{2}\right)^n \times \binom{n}{k}\left(\frac{1}{2}\right)^n\right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac{1}{2}\right)^{2n},$ 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.

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.

Say we have Tom and John, each tosses a fair coin n times. What is P 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 $$\binom{n}{k}(\frac{1}{2})^n.$$ So, we can do $$\sum^{n}_{k=0}(\binom{n}{k}(\frac{1}{2})^n \times \binom{n}{k}(\frac{1}{2})^n)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}(\frac{1}{2})^{2n},$ 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.

Say we have Tom and John, each tosses a fair coin $n$ times. What is $P$ 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 $$\binom{n}{k}\left(\frac{1}{2}\right)^n.$$ So, we can do $$\sum^{n}_{k=0}\left(\binom{n}{k}\left(\frac{1}{2}\right)^n \times \binom{n}{k}\left(\frac{1}{2}\right)^n\right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac{1}{2}\right)^{2n},$ 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.