Finding expected value of a Binomial

[juliannaq]

Homework Statement

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<= P \ <=1. Then each person who got a head will split some arbitrary amount of prize money, K, equally. If nobody gets a head, then nobody gets the prize. Whats the expected prize you receive?

So variables:
N+1 - number of people
P - probability coin lands on heads
K - Prize money

Homework Equations

Relevant equations may be the binomial distribution formula? E(x) (expected value).

The Attempt at a Solution

I think I want to first find the expected number of people who will toss heads. Dividing K by E(#heads) and multiplying by P(the chance of you getting heads, so the chance of you being among the winners). In order to start I came up with the following summation, by looking at the probability of getting various 'K's (i.e. k/1, k/2):

$\sum{\frac{k}{x}(1-p)^{(n+1)-x}p^x}$, 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!

 

[lurflurf]

You do not need it sum if you realize
k=(n+1)Expectation
since the expectation of each player is equal

If sum you must, it should be

$E=\sum{\frac{k}{x}\frac{x}{n+1}\binom {n+1} {x}(1-p)^{(n+1)-x}p^x}$
Simplify and use binomial theorem

Your probability assumed falsely that you are always a winner if anyone is. In fact if there are x (0<=x<n+1) winners you are only among them with probability x/(n+1)
Also you falsely assume all winners are in a certain order, by including the binomial coefficient one can account for the winners being in any order.

 

[juliannaq]

Thank you! I see the errors I made, and the new sum makes sense to me (especially I can't believe I forgot the $\binom[n+1][x]$ ... jeez). I do not, however, have any desire whatsoever to go about simplifying that sum, but I'm not sure if I understand exactly what expectation means here? The expectation of winning?
It's probably obvious and I'm just being stupid, apologies in advance :x

 

[lurflurf]

Expectation is the average outcome of many trails, how much you expect to win in the average trial.

 

[juliannaq]

Alright, that makes sense, since each person will (on average) make the same amount of money.
In fact, my initial solution to the problem was to notice that E(x) with a binomial distribution is np, giving my E(x) = (n+1)p. so K/((n+1)p) should give how much the average payout is. Multiply by p for the chance of you winning, and you end up with K/(n+1).
From your post, it appears this is the result I would get as well, since k = (n+1)expectation
so (k/(n+1))=expectation.
When I think about this, however, I don't think its correct? I believe the payout needs to depend on p. If two people are playing, and the chance of getting heads is .01, neither of them should expect to make half of the prize money each time they play.
Perhaps my question was unclear at the beginning and I failed to mention this?
I'm not entirely sure how to account for the fact that its possible for nobody to win?

(Essentially, as long as p>=.5, this formula will give the proper answer; however when p<.5, it will give expected earnings higher than they should be, since it does not account for the fact that its possible for nobody to win)

 

[Ray Vickson]

Alright, that makes sense, since each person will (on average) make the same amount of money.
In fact, my initial solution to the problem was to notice that E(x) with a binomial distribution is np, giving my E(x) = (n+1)p. so K/((n+1)p) should give how much the average payout is. Multiply by p for the chance of you winning, and you end up with K/(n+1).
From your post, it appears this is the result I would get as well, since k = (n+1)expectation
so (k/(n+1))=expectation.
When I think about this, however, I don't think its correct? I believe the payout needs to depend on p. If two people are playing, and the chance of getting heads is .01, neither of them should expect to make half of the prize money each time they play.
Perhaps my question was unclear at the beginning and I failed to mention this?
I'm not entirely sure how to account for the fact that its possible for nobody to win?

(Essentially, as long as p>=.5, this formula will give the proper answer; however when p<.5, it will give expected earnings higher than they should be, since it does not account for the fact that its possible for nobody to win)

The formulas above are incorrect for computing YOUR expected winnings. Your expected winnings are W = E(your winning|you get heads)*P{you get heads}. Given that you get heads let X = number of OTHER heads ~ Bin(n,p). If k others get heads your actual winnings are K/(k+1), so W = p*K*sum_{k=0..n} b(k)/(k+1), where
b(k) = C(n,k)*p^k*(1-p)^(n-k) is the binomial probability of k heads in n tosses. Basically, this is W = p*K*E[1/(X+1)], where X~Bin(n,p). [The sum is do-able and the answer is not too bad, but I cannot say more now.]

RGV

 

[juliannaq]

So if I'm not mistaking, E(binom(n,p)) = np, since each trial is independent. So the amount of money I expect should be (K*p)/(np+1), where np+1 is the expected number of people who flip heads given that I flip heads, and K*p is the amount I would expect to win playing by myself?

 

[Ray Vickson]

So if I'm not mistaking, E(binom(n,p)) = np, since each trial is independent. So the amount of money I expect should be (K*p)/(np+1), where np+1 is the expected number of people who flip heads given that I flip heads, and K*p is the amount I would expect to win playing by myself?

No. When you write the expression above that you are falling into the trap of the so-called "fallacy of averages". If you have a nonlinear function f, you do *not* generally have Ef(X) equal to f(EX). So, for f(x) = 1/(x+1) you do not have E[1/(X+1)] = 1/(EX + 1) = 1/(np + 1).

RGV

 

[juliannaq]

Again, sorry for all the questions; not sure why I'm having so much trouble wrapping my head around this!
Am I correct in thinking that E[1/(x+1)] is equal to:

$\sum$(1/x+1)*(n choose x) p^x (1-p)^(n-x)

 

[Ray Vickson]

Yes (if by (1/x + 1) you actually mean (1/(x+1)).

RGV