I have a pretty basic probability question, but I'm just having difficulties remembering what distribution this is.

A coin is tossed until a head appears two times in a row. Given that we are using a fair coin, what is the probability that we toss the coin exactly 4 times such that the two consecutive heads are the 3rd and 4th trials?

I know that this would be fairly easy to solve by brute force and listing out the different probabilities, but I'm a little confused as to how to use a "shortcut" such as identifying this as a binomial distribution. I also don't know how to account for the probability that we are using 4 tosses.

Any help is appreciated. Thanks!


There are only two such scenarios:

  • TTHH, for which the probability is $\dfrac{1}{2^4}$
  • HTHH, for which the probability is $\dfrac{1}{2^4}$

Hence the overall probability is $\dfrac{1}{2^4}+\dfrac{1}{2^4}=\dfrac{1}{2^3}$

  • $\begingroup$ Do we not need to consider the fact that there might be the situation such that the coin is flipped more than 4 times? TTTHH would be a valid way to satisfy the conditions. $\endgroup$ – Taylor Jan 18 '16 at 23:38
  • $\begingroup$ @Taylor: You asked "exactly $4$ times", didn't you??? $\endgroup$ – barak manos Jan 19 '16 at 6:08
  • $\begingroup$ Ohh. Exactly 4 times SUCH THAT blah blah blah. The question is asking what the odds are the we end up only needing to toss it 4 times AND it's TTHH and not HTHH. $\endgroup$ – Taylor Jan 19 '16 at 16:14
  • $\begingroup$ @Taylor: Sorry, I cannot make any sense of what you're saying in this comment. $\endgroup$ – barak manos Jan 20 '16 at 9:27

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