Finding expected value of a binomial I'm trying to solve the following question: You and n other people (so n+1 people) each toss a probability-p coin, with $0\le P \le 1$. Then each person who got a head will split some arbitrary amount of prize money, K, equally. If nobody gets a head, then each person receives nothing. Whats the expected prize you receive?
I think I want to first find the expected number of people who will toss heads, and in order to start I came up with the following summation: 
$\sum{\frac{k}{x}(1-p)^{(n+1)-x}p^x}$, with $x$ from $1$ to $n+1$ where $n+1$ is the number of people. 
However, I have no idea how to solve this sum, or even if I'm setting it up correctly (I think I may be missing something?); any tips would be greatly appreciated!
 A: Assume that you receive nothing when nobody got a head. For every $0\leqslant i\leqslant n$, call $A_i$ the event that you and exactly $i$ other people got a head. On $A_i$, you receive $K/(i+1)$ hence the mean amount of money you receive is
$$
M=\sum_{i=0}^n\frac{K}{i+1}\mathbb P(A_i).
$$
Now, for every $i$, $A_i$ is realized if you got a head, which happens with probability $p$, and if exactly $i$ amongst the $n$ other players got a head, which happens with probability $p^i(1-p)^{n-i}$ for each subset of $i$ players, times ${n\choose i}$ the number of such subsets. Finally,
$$
M=\sum_{i=0}^n\frac{K}{i+1}\,p\,{n\choose i}\,p^i(1-p)^{n-i},
$$
that is,
$$
M=\frac{K}{n+1}\sum_{i=0}^n\,{n+1\choose i+1}\,p^{i+1}(1-p)^{(n+1)-(i+1)},
$$
hence
$$
M=\frac{K}{n+1}\,(1-(1-p)^{n+1}).
$$
Sanity checks: If $p=0$, $M=0$. If $p=1$, $M=K/(n+1)$. The function $(p,n)\mapsto M$ is increasing with respect to $p$ and decreasing with respect to $n$.
A: Although did already has provided an answer, there's a simpler way to deduce the same result.
The chance that the prize will be paid is $1-(1-p)^{n+1}$, so the expected total reward being paid is $A=K\cdot\{1-(1-p)^{n+1}\}$.
If a reward is paid, it is divided between those who got a head. Since the problem is symmetric in the $n+1$ players, each of them has the same expected reward, which much therefore be $A/(n+1)$.
Hence, the expected reward for any given player is $\frac{K}{n+1}\cdot\{1-(1-p)^{n+1}\}$.
