Question:
A fair coin is tossed repeatedly until the sequence $HH$ of two consecutive heads appears, and the total number of tosses $n$ is recorded. Find $\Pr \left( {\left\{ {n = 2018} \right\}} \right)$? Hint given: Note that $n$ does not have a geometric/negative binomial distribution. Use the Law of Total Probability and condition on the outcome of the first two tosses and use the mutually exclusive and exhaustive events $\{HH\}$, $\{HT\}$, and $\{TH,TT\}$.
I know a similar question was asked here, but this question asks to apply the law of total probability. Hopefully I'm not wrong in thinking this would have a different approach.
If we suppose the first two tosses is $HH$. Then how do we apply the law of total probability to find $\Pr \left( n \right)$? Would it look like: $$\Pr \left( n \right) = \Pr \left( {n \mid HH} \right)\Pr \left( {HH} \right) + \Pr \left( {n \mid HT} \right)\Pr \left( {HT} \right) + \Pr \left( {n \mid TH} \right)\Pr \left( {TH} \right) + \Pr \left( {n \mid TT} \right)\Pr \left( {TT} \right)$$
If so, how does one find $\Pr \left( {n \mid HH} \right)$, $\Pr \left( {n \mid HT} \right)$, etc.?