= Solution
Each waiting time has the <geometric distribution> on $1,2,\ldots$ with success probability $1/2$: a wait of length $n$ consists of $n-1$ failures followed by the prescribed success, so its probability is $2^{-n}$. Hence
$$
\boxed{G_{H_j}(s)=G_{T_j}(s)=\sum_{n\ge1}\left(\frac s2\right)^n=\frac{s}{2-s},\qquad
\mathbb E H_j=\mathbb E T_j=G'_{H_j}(1)=2.}
$$
Disjoint successive blocks of independent tosses give independent waiting times. More explicitly, any specified list of block lengths determines exactly one allowed sequence of heads and tails and has probability $2^{-\sum_j(h_j+t_j)}$, the product of the individual block probabilities.
There are $2r$ waiting times. Multiplication of their <probability generating functions> and addition of their <expectations> give
$$
\boxed{G_{Y_r}(s)=\left(\frac{s}{2-s}\right)^{2r},\qquad \mathbb E Y_r=4r.}
$$
To recover its <probability mass function>, expand by the <negative binomial series>:
$$
G_{Y_r}(s)=\frac{s^{2r}}{2^{2r}}(1-s/2)^{-2r}
=\frac{s^{2r}}{2^{2r}}\sum_{m\ge0}\binom{m+2r-1}{2r-1}(s/2)^m.
$$
The coefficient of $s^n$ is therefore
$$
\boxed{\mathbb P(Y_r=n)=2^{-n}\binom{n-1}{2r-1},\qquad n\ge2r,}
$$
and it is zero below $2r$. Equivalently there are $\binom{n-1}{2r-1}$ compositions of $n$ into $2r$ positive block lengths, each with probability $2^{-n}$. This is the <negative binomial distribution> for a sum of geometric waiting times, even though the desired outcome alternates between blocks.
For $r=1$,
$$
\boxed{\mathbb P(Y_1\ge5)=1-\left(\frac14+\frac28+\frac3{16}\right)=\frac5{16}.}
$$
Since $\mathbb E Y_1=4$, <Markov's inequality> gives $\mathbb P(Y_1\ge5)\le4/5$, and indeed $5/16<4/5$.
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