Two people toss a fair coin $n$ times 
Two people each toss a fair coin $n$ times. Find the probability that they toss the same number of heads.

My approach: Person $A$ flips a coin $n$ times. Person $B$ flips a coin $n$ times. So we have a total of $2n$ coin flips. Therefore the total number of outcomes of choosing $n$ flips that land heads is $${2n\choose n}$$
Since we flip a coin $n$ times and the coin is fair. The probability person $A$ and person $B$ toss the same number of heads is thus $$\left(\frac{1}{4}\right)^n {2n\choose n}$$
To be more clear: The probability person $A$ lands heads after $n$ flips is $\left(\frac{1}{2}\right)^n$ like wise for person $B$ thus we have $$\left(\frac{1}{2}\right)^n \left(\frac{1}{2}\right)^n = \left(\frac{1}{4}\right)^n $$
Second approach with helo from Joeb: Let $E_k$ be the event where each person flips exactly $k$ heads we want to calculate $$\sum_{k=0}^{n}P(E_k)$$
Individually we have a probability of $${n\choose k}\left(\frac{1}{2}\right)^n$$ Thus together $$\sum_{k=0}^{n}P(E_k) = \sum_{k=0}^{n}{n\choose k}\left(\frac{1}{2}\right)^n{n\choose k}\left(\frac{1}{2}\right)^n = \left(\frac{1}{4}\right)^n\sum_{k=0}^{n}{n\choose k}{n\choose k} = \left(\frac{1}{4}\right)^n{2n\choose n} $$
I am just wondering if my reasoning makes sense, I got the correct answer but I just want to know if I explained it clearly enough. I know this is a question that has been posted on here, but I am asking if my approach is correct. The algebra is a bit tricky I didn't solve it myself just used a calculator.
 A: It seems to me like you are only considering the case where each person flips exactly $n/2$ heads out of $n$ flips.
What if instead you calculated the probability of the event $E_k$ where each person flips exactly $k$ heads ($k=0,1,2,...,n$)? Then you can just take the sum
$\sum_{k=1}^n \mathsf{P}(E_k)$.
A: Your reasoning does not explain the result correctly as you pick $n$ out of $2n$ tosses that produce heads, when actually the number of heads can be $\ne n$ and only the number of heads tossed by $A$ shall be the same as te number of heads tossed by $B$.
Note that of $A$ tosses $k$ heads, this means that $B$ tosses $n-k$ tails. Thus all valid outcome can be produced by picking a subset of $n$ of the $2n$ tosses and assign heads to those ($k$, say) of these picked tosses that are among $A$'s tosses as well as to the $k$ tosses among $B$'s tosses that were not picked (and assign tails to the rest). This gives us $2n\choose n$ favorable outcomes, each of which occurs with probability $2^{-2n}$. Hence the result $\frac1{4^n}{2n\choose n}$.
