The third head on the seventh toss

Question

You flip a fair coin repeatedly. What is the probability that the third head appears on exactly the seventh toss? Follow-up: on average, how many tosses until the third head?

Accepted Answer

This is the same trick as Banach's matchbox: fix the decisive event as the final trial, then count how the earlier trials can be arranged. For the third head to fall on toss 7: - toss 7 must be a head, and - exactly 2 of the first 6 tosses were heads (so the running count is 2 going into toss 7). The number of ways to place those 2 heads among the first 6 tosses is \binom{6}{2} = 15. Every specific sequence of 7 tosses has probability (	frac12)^7 = 	frac{1}{128}, so P(	ext{3rd head on toss } 7) = \binom{6}{2}\left(	frac12ight)^7 = \frac{15}{128} \approx 0.117. This is the negative binom…

The third head on the seventh toss

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