Probability of success on third and fourth trials?

Question

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!

Answer 1

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 $\dfrac{1}{2^4}+\dfrac{1}{2^4}=\dfrac{1}{2^3}$