Why is $\sum_{y=0}^n \binom{n+y}{y}0.5^{n+y} = 1$ in terms of tossing a fair coin? Apparently you can deduce that $\sum_{y=0}^n \binom{n+y}{y}0.5^{n+y} = 1$ from thinking about a  fair coin, where $n+y$ is the number of tosses and $y$ is the number of heads.
I'm having a hard time wrapping my head around this intuition.
I know that
$$
\sum_{i=0}^n \binom{n}{i}0.5^n = 1
$$
can be viewed as the sum of probabilities of getting 0,1,2,3,4,...,n heads from $n$ tosses. Since these are disjoint events that make up the entire sample space, these probabilities therefore must sum to 1.
But in my case, the number of tosses is also varying rather than being fixed like in the preceding example, so I don't know what the analogous intuition is for the case where the number of tosses is also varying.
 A: Suppose that you toss a fair coin until you get either $n$ heads or $n$ tails, where $n\ge 1$. Say that you get $n$ tails first. Then your last toss was a tail, and you got $n-1$ tails and $y$ heads in the first $n-1+y$ tosses; the probability of such a sequence of tosses is
$$\frac12\binom{n-1+y}y\left(\frac12\right)^{n-1+y}=\binom{n-1+y}y\left(\frac12\right)^{n+y}\,.$$
Thus, the probability of getting $n$ tails before you get $n$ heads is
$$\sum_{y=0}^{n-1}\binom{n-1+y}y\left(\frac12\right)^{n+y}\,.$$
Of course you are equally likely to get $n$ heads first, so the probability of each of these events must be $\frac12$, and we conclude that
$$\sum_{y=0}^{n-1}\binom{n-1+y}y\left(\frac12\right)^{n+y}=\frac12\,,$$
which after multiplication by $2$ becomes
$$\sum_{y=0}^{n-1}\binom{n-1+y}y\left(\frac12\right)^{n-1+y}=1\,.\tag{1}$$
And $n$ was any positive integer, so we can replace $n-1$ by $n$ throughout to conclude that
$$\sum_{y=0}^n\binom{n+y}y\left(\frac12\right)^{n+y}=1$$
for all integers $n\ge 0$.
A: Note that
$$\sum_{y=0}^n\binom{n+y}y\left(\frac12\right)^{n+y}= 1 =\sum_{y=0}^n \sum_{k=0}^y \binom{n}k\binom{y}{y-k} \left( \frac12 \right)^n \left( \frac12 \right)^y$$
by Vandermonde's identity.
Also
$$=\sum_{y=0}^n \sum_{k=0}^y \binom{n}k\binom{y}{y-k} \left( \frac12 \right)^n \left( \frac12 \right)^y$$
$$=\sum_{y=0}^n \sum_{k=0}^y \binom{n}k\binom{y}{k} \left( \frac12 \right)^n \left( \frac12 \right)^y$$
$$=\sum_{y=0}^n  \binom{n}k \left(\sum_{k=0}^y\binom{y}{k} \right) \left( \frac12 \right)^n \left( \frac12 \right)^y$$
$$= \sum_{y=0}^n \binom{n}k \left( 2^y  \right)\left( \frac12 \right)^n \left( \frac12 \right)^y$$
$$= \sum_{y=0}^n \binom{n}k \left( \frac12 \right)^n  =1 $$
