$\sum_{i=n}^{2n-1}\binom{i-1}{n-1}2^{1-i}$
For $i = n,n+1,\ldots, 2n - 1$, the sum above computes $P(E_i)$, the probability that i tosses of a fair coin are required before obtaining $n$ heads or $n$ tails.
I am not asking for a solution. I am confused by how that summation computes the probability described in the hint. Shouldn't the binomial coefficient be $\binom{i}{n}$? That's the only question I have.
The earliest that you can get $n$ heads or $n$ tails is the $n$-th toss, and the latest is the $(2n-1)$-st toss. How many ways are there to get it on the $i$-th toss, where $n\le i\le 2n-1$? Let’s say that you get the $n$-th tail on the $i$-th toss. That means that you must have got $n-1$ tails in the first $i-1$ tosses, and they could any $n-1$ of those tosses. Thus, there are $\binom{i-1}{n-1}$ strings of heads and tails that would allow you to get the $n$-th tail on the $i$-th toss. The probability of getting any one of those strings is $2^{-(i-1)}=2^{1-i}$. The probability of getting a tail on the $i$-th toss is $\frac12$, so the probability of a string of length $i$ that has exactly $n$ tails and ends with a tail is $\frac12\cdot\binom{i-1}{n-1}2^{1-i}$. The probability of getting the $n$-th head on the $i$-th toss is identical, so the probability of getting one or the other of these outcomes is
$$2\cdot\frac12\binom{i-1}{n-1}2^{1-i}=\binom{i-1}{n-1}2^{1-i}\;.$$