A fair coin is tossed $n$ times by Adam and $n$ times by Andrew. What is the probability that they get the same number of heads?
Now since there are a total of $2n$ flips, $n$ from each person, we would need to choose $k$ flips that they both have heads correct? So since this is a binomial$\sim (2n,\frac12)$ I came to find that we have ${2n \choose k }\left(\frac12\right)^{2n}$ However my textbook says it differently. I'm wondering why the book says it is: ${2n \choose n }\left(\frac12\right)^{2n}$ Is it the same thing? I'm just confused as to why it would say choose $n$. Should I be assuming assume that half of the total flips are are going to be the same?