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)
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?
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...
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...
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...