# Fair coin, three consecutive heads, p(TAILS=1)

Stuck on this question.

The experiment of tossing a fair coin until three consecutive heads appear is performed. Let X be the number of tosses, and Y be the number of tails that appear. Find the probability p(Y = 1).

I tried listing out the possibilities where Y=1:

THHH HTHHH HHTHHH

So I thought maybe it's 3/(2^6) but that's not listed as an answer.

If the first toss is a T, then the event $Y=1$ happens if the first $4$ tosses are THHH. This has probability $\frac{1}{16}$.
If the first toss is a H, then $Y=1$ can happen in $2$ ways: (i) the second toss is a T and then we get HHH or (ii) the second toss is a H, the next is a T, and then we get HHH. Note that (i) has probability $\frac{1}{32}$ and (ii) has probability $\frac{1}{64}$.
• The unnecessary definition of the random variable $X$ makes me wonder whether what you have is the full question. But the probability that $Y=1$ is clear. If while solving we forgot about HHTHHH, we would get $3/32$, one of the given answers. But of course that would be wrong. Mar 20, 2015 at 5:06
• @SeanBollin: Yes, the interpretation as a conditional probability problem is fully reasonable. However, the probability that sometime or other there will be $3$ consecutive heads is $1$, so the answer is $7/64$. Mar 21, 2015 at 4:30