Assume a fair coin. The expectation of the number of tosses $X$ until getting $k$ consecutive heads is well known.
What is the PMF of $X$? ($k$ might not be $2$) I searched high and low but there is no satisfactory answer.
The **distribution** of number of tosses until getting $k$ consecutive heads
combinatoricsprobabilityprobability theory
Related Solutions
Let $X$ be the random variable that counts the number of tosses until we get two successive tosses of the same type. Although you did not say so explicitly, I assume that you want the probability mass function of $X$.
What is the probability that $X>n-1$? We need to have a sequence of tosses of length $n-1$ and of type HTHTHT and so on (alternating heads and tails, starting with head), or THTHTH and so on. The probability that the first type of sequence is followed up to and including the $(n-1)$-th toss is $\frac{1}{2^{n-1}}$. The second type of sequence has the same probability, for a total of $\frac{2}{2^{n-1}}$.
The number $X$ of tosses is equal to $n$ if $X>n-1$ and the last toss matches the next to last toss. The probability of that is $\frac{1}{2}$, so $$P(X=n) =\frac{2}{2^{n-1}}\cdot\frac{1}{2}.$$ This simplifies to $\dfrac{1}{2^{n-1}}$. Note that $n$ ranges over the integers that are $\ge 2$.
Another way: The first toss doesn't matter. After that, it is just a matter of matching the previous toss. So it is very much like tossing a fair coin and winning if you get (say) heads. The only difference is the "wasted" toss at the beginning. If we are tossing a fair coin, and $Y$ is the number of tosses until the first head, then $P(Y=n)=\frac{1}{2^n}$ ($n=1, 2, \dots$). But $X$ has the same distribution as $Y+1$. So $P(X=n+1)=P(Y=n)=\frac{1}{2^n}$, or equivalently $$P(X=n)=P(Y=n-1)=\frac{1}{2^{n-1}}\quad (n=2,3,\dots).$$
Generalization: Suppose that the probability of a head is $p$, and the probability of a tail is $q=1-p$. Either analysis of the problem can be extended to the more general situation. We win on the $n$-th toss if we did not win on the first $n-1$, and then the $n$-th toss matches the previous one.
It is convenient to split the analysis into two cases, $n$ odd and $n$ even. We do a full analysis of the odd case. So let $n=2m$. As is the first analysis, there are two ways to win on the $(2m+1)$-th toss. Either (i) we went HTHTHT and so on, with a tail on the $(2m)$-th toss, and then a tail again on the $(2m+1)$-th toss, or (ii) we went THTHTH and so on, with a head on the $(2m)$-th toss, and a head again on the $(2m+1)$-th toss.
The probability of (i) is $p^mq^m q$ and the probability of (ii) is $q^mp^mp$. Add them together. We get $p^mq^m(p+q)$, which is $p^mq^m$ ($m=1,2,\dots)$.
Next we deal with the case $n$ is even, say $n=2m$. The calculation is very similar to the previous one, so it will be omitted. The probability is the slightly more complicated-looking $p^mq^{m-1}p+q^mp^{m-1}q=p^{m-1}q^{m-1}(p^2+q^2)$, where $m$ ranges over the positive integers.
Let $e$ be the expected number of tosses. It is clear that $e$ is finite.
Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus $$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$ Solve this linear equation for $e$. We get $e=62$.
Best Answer
Suppose $n>k$. To get $P(X=n)$ recursively we note that your string must begin with $H^iT$ for $i\in \{0, \cdots, k-1\}$. Then, after that initial block of length $i+1$ you are back where you started. Thus $$P(X=n)=\sum_{i=0}^{k-1} \frac 1{2^{i+1}}P(X=n-i-1)$$
This provides a handy way to compute $P(X=n)$, though it falls well short of a convenient analytic expression. I'm not sure there's any reason to expect such an expression to exist.