Hi all,You have n coins. You toss each of them and get $1 if you get head, 0 for tail.First basic question, what's the expected value ?Assume you're allowed to toss a coin again in case you don't get all of them as heads. What's the expected value now ?Thanks

The first one is easy so I'll leave that one out. I'll reword the second one. You toss n coins, with the aim of getting as many heads as possible. After all n coins have been tossed, you are allowed to toss one of the coins again. (Obviously you would only do that with a tail). What is the expected value of the number of heads (translated into dollars)? (However, the original wording is somewhat ambiguous -- it could mean that every coin can be tossed twice.)There are two distinct cases -- either you initially get all heads or you don't. Begin by considering the situation before retossing. The expected value given the all-heads outcome is $ n. The expected value given the not-all-heads outcome is harder to calculate. Call the expected value, given not-all-heads NTH. We have (by the first easy omitted question) 1/2^n * n + (1 - 1/2^n) * NTH = n/2. Hence NTH = (n/2 - n/2^n)/(1 - 1 / 2^n) = n * (2 ^ (n-1) - 1) / (2 ^ n - 1).However, the NTH needs to be increased by 0.5 to allow for the retoss allowance. Hence final answer = $ n / 2^n + (1 - 1/2^n) * ( n * (2 ^ (n-1) - 1) / (2 ^ n - 1) + 1/2). I won't bother to simplify this but I will check for small values of n. For n = 1, the answer should be 3/4. That checks. A much harder question than it looks. If used as an interview-brainteaser, I'd bet the majority would get this wrong. (And I'm not completely confident I haven't made an error). Easy to get right but easy to get wrong.CommodityQuant.

Don't need to calculate NTH, the proba of NTH= 1-1/2^n is sufficient.Final answer is n/2+P(NTH)*0.5

Another one: what the expected value if you 're allowed to toss again p coins of your choice instead of one?

Vanubis1,You are correct that your method is neater and simpler. I hope that I hit the right answer, anyway. As a job-seeker, I wonder if someone could tell me how my approach would come across in a job interview. I was very much doing the question job-interview-style, tackling the problem as soon as I met it. (Of course, in an interview, I wouldn't omit the first part of the question on the grounds of it being too simple.)For the p toss problem, again start out with n/2. Then add 0.5 * ( prob(exactly one tail) + 2 *prob(exactly two tails) + ... p * prob(exactly p tails)) + p/2 * prob(more than p tails). Closed-form formulas for the above probably exist and perhaps you can enlighten us by revealing such.CommodityQuant