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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!

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hint: Your Expected Share = $E[K.Prob(k=numOfheads)/(N+1)]$ – jay-sun Oct 31 '12 at 23:45
Thanks for the hint! Unfortunately, I'm don't quite understand what it means, notably the numerator in the $E$ - assuming K is denoting the prize money, what is (lowercase) k? Is it the probability of getting 1 head, 2 heads, 3 heads etc.? Thank you so much for your help! – JuliannaQ Nov 1 '12 at 2:27
k is just the number of heads. You should use the property of the expectation operator when constants are involved. – jay-sun Nov 1 '12 at 5:11
A missing piece of information is how much money you receive if nobody got a head. – Did Nov 1 '12 at 7:30
Yeah, my apologies! Fixed that - makes a big difference! – JuliannaQ Nov 1 '12 at 17:28
up vote 1 down vote accepted

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 $n$.

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Thank you so much! This makes a great deal of sense - certainly a better way of thinking about it than what i started with! My only question - my algebra is a bit shaky at this point, I'm not sure if I follow what happened when you went from step 1 to step 2 in reducing your summation? (the 2nd to 3rd was just an application of the binomial theorem, right?) Thanks again! – JuliannaQ Nov 1 '12 at 17:14
"Step 1" uses $\frac1{i+1}{n\choose i}=\frac1{n+1}{n+1\choose i+1}$, an identity you can probably find a proof of. – Did Nov 1 '12 at 17:37
Whoops, just realized that. Final question (I swear!). At the final step, where you apply the binomial identity, how does it come out to be $(1-(1-p)^{n+1})$ instead of $(p+(1-p))^{n+1})$? I think it has to do with having i+1 instead of i, but I'm not sure i see why? – JuliannaQ Nov 1 '12 at 17:44
Rewrite the last sum (from $i=0$ to $n$) as a sum from $k=1$ to $n+1$, using the change of variable $k=i+1$. You get the full binomial identity, except that the $k=0$ term is missing. And this $k=0$ term is... – Did Nov 1 '12 at 17:47
Ah, perfect! Thanks again for the help - I really appreciate it. – JuliannaQ Nov 1 '12 at 17:54

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 $\frac{K}{n+1}\cdot\{1-(1-p)^{n+1}\}$.

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This is very clever! Definitely preferable to all that nasty algebra :) Thanks! – JuliannaQ Nov 4 '12 at 3:16

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