A fair coin is tossed $n$ times by two people. What is the probability that they get same number of heads? 
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.

The problem with this answer is that its not correct to begin with. Assume the problem pertains to two individuals throwing 2 coins then
the total outcomes is indeed $ \frac{1}{2^m} , m= 2n$ however the amount of possible outcomes is not $\binom{2n}{n}$ so in the case of each having 2 coins the resulting outcomes is:
1 head and two tails, this can be done 2 ways $2*2 = 4$
2 heads , this can be done only 1 way $1*1=1$
This is a total of $5$ and $\binom{2*2}{2} = 6$ this is of course not equal and difference of 1 holds as you continue. 
