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You are given a fair coin and flip it until you get two heads in a row. What is the probability that this takes 5 tosses?

Note: HT, TH, TT are all good for the first and second tosses. All that matters is that the 3rd is a T and the 4th and 5th toss are both heads.

I got 3/32, but I know that's not right as that's the probability that you know there are exactly five tosses, but this question is asking what is the probability that the consecutive heads happens on the 4th and 5th toss on an arbitrarily long sequence of tosses.

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    $\begingroup$ How can your event happen? The fourth and fifth must be H, and the third must be T. Now what about the first two? Any of HT, TH, TT is good. $\endgroup$ – André Nicolas Feb 4 '15 at 4:15
  • $\begingroup$ Yes HT, TH, TT are all good for the first and second tosses. All that matters is that the 3rd is a T and the 4th and 5th toss are both heads. $\endgroup$ – Eric Feb 5 '15 at 4:39
  • $\begingroup$ So now we can compute the probability. It is $3/32$, precisely as you thought. $\endgroup$ – André Nicolas Feb 5 '15 at 4:48
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The chance of not getting two head in the first two tosses is $\frac34$. The probability of getting THH for the next three is then $\frac18$. So, we get that you were correct with a $\frac34\times\frac18=\frac3{32}$ chance of getting the first pair of heads being the fourth and fifth tosses.

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