What is the expected number of coin tosses it would take to get N many heads OR N many tails? Where we do NOT require that the heads or tails be consecutive (though they may be!)
Obviously, this expectation, $E[T]$, is bound as follows: $N < E[T] < 2N - 1$
And obviously $E[T] = \sum_{i=N}^{n=2N-1}i*P[Game \ Ends \ On \ i^{th} \ Round]$, where $\sum_{i=N}^{n=2N-1}P[Game \ Ends \ On \ i^{th} \ Round] = 1$
But how would one find such a probability for an arbitrary $i \in \{N, N+1, ..., 2N-1\}$?
 A: The probability that the game ends at round $i$ is:$$\binom{i-1}{N-1}2^{1-i}$$If e.g. the game ends with a tail at $i$-th round then among the $i-1$ results in the former rounds must be $N-1$ tails. There are $\binom{i-1}{N-1}$ configurations for that and each of them has probability $2^{1-i}$ to occur. Further we must multiply with the probability that a tail will appear at $i$-th trial which is $\frac12$. Same outcome if the game ends with a head at $i$-th round so a multiplication with $2$ is also needed.
So to be found is:
$$\sum_{i=N}^{2N-1}i\binom{i-1}{N-1}2^{1-i}=2N\sum_{i=N}^{2N-1}\binom{i}{N}2^{-i}$$
A: Preliminary Formula
$$
\begin{align}
a_n
&=\sum_{m=n}^{2n}\frac1{2^m}\binom{m}{n}\\
&=\sum_{m=n}^{2n}\frac1{2^m}\left[\binom{m-1}{n}+\binom{m-1}{n-1}\right]\\
&=\frac12\sum_{m=n-1}^{2n-1}\frac1{2^m}\left[\binom{m}{n}+\binom{m}{n-1}\right]\\
&=\frac12\left[a_n-\frac1{2^{2n}}\binom{2n}{n}+a_{n-1}+\frac1{2^{2n-1}}\binom{2n-1}{n-1}\right]\\[3pt]
&=\frac12(a_n+a_{n-1})\\[9pt]
&=a_{n-1}\tag1
\end{align}
$$
and since $a_0=1$, we have
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{m=n}^{2n}\frac1{2^m}\binom{m}{n}=1}\tag2
$$

The Answers
The number of ways to get to $n$ heads on flip $m$ and not on flip $m-1$ is 
$$
\binom{m}{n}-\binom{m-1}{n}=\binom{m-1}{n-1}\tag3
$$
This is also the number of ways to get to $n$ tails on flip $m$ and not on flip $m-1$. Thus, the probability of getting to $n$ heads or $n$ tails on flip $m$ and not on flip $m-1$ is
$$
\frac2{2^m}\binom{m-1}{n-1}=\bbox[5px,border:2px solid #C0A000]{\frac1{2^{m-1}}\binom{m-1}{n-1}}\tag4
$$
Note that $(2)$ shows that
$$
\sum_{m=n}^{2n-1}\frac1{2^{m-1}}\binom{m-1}{n-1}=1\tag5
$$
That is, the probability of getting $n$ heads or $n$ tails by flip $2n-1$ is $1$. The expected duration is then
$$
\begin{align}
\sum_{m=n}^{2n-1}\frac{m}{2^{m-1}}\binom{m-1}{n-1}
&=2n\sum_{m=n}^{2n-1}\frac1{2^m}\binom{m}{n}\\
&=\bbox[5px,border:2px solid #C0A000]{2n\left(1-\frac1{4^n}\binom{2n}{n}\right)}\tag6
\end{align}
$$
