A related problem to the last post. You toss a fair coin, and keep going until the number of tails you’ve seen is *n* more than the number of heads. On average, how many times do you toss the coin before you stop?

Unlike the last problem, I don’t know the answer to this one, so I’d be interested in hearing anyone else’s approach. I believe that I have a good method for getting at the answer, but that I’m missing one crucial step that I haven’t yet been able to fill in.

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September 16, 2010 at 7:52 am

A random manOn average, you will toss infinite number of times!

December 10, 2011 at 2:01 am

balafor n=1, expected throws = 1

for n>1, expected throws = inf

possible proof: consider a state where T-H = N-1 (where N is the expected number of steps). The next throw would either be a Head or Tail with equal probability (this assumption is not really necessary). If tail, we have success in N steps. If Head, then we have wasted 1 step, and we have N + 1 more steps to go. So, N = 0.5*(N) + 0.5*(N+1). This is possible only if N is inf.