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Attn dennismv, coins	by DesiredUsername	2012-01-08 01:48:43
I posted a numeric answer yesterday. I think I just worked it out analytically. I get the same answer as you said for the average number of flips starting from zero. Call that x. In the case of two heads, x=6. If you flip the first coin and get heads, then you have two options for the next flip. Either heads (and your number of flips is two) or tails. If you get tails, you go back to the starting point and you have six flips expected. However, you only need to do that half the time. The expected number of flips will be = 0.5(1) + 0.5(x+1), or 4 flips more. This can easily be extended to N heads in a row, starting with one head. I'm still working on the final analytic solution, but I'm getting tired..
[ Reply ]
Ok, here is my guess	by DesiredUsername	2012-01-08 02:15:37
Plugging in a few numbers by hand	by DesiredUsername	2012-01-08 02:38:09

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