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A phylogenetic epoch is a time interval between successive branching events during which the number of ancestral branches is constant. Read backward from the four sampled leaves: there are four branches until the first merger, then three until the next, then two until their common root. Write the random durations as $T_4,T_3,T_2$, with expected values $E_k=\mathbb{E}[T_k]$.

The original solution diagram below places time horizontally and labels each branch by its number of descendant leaves. Type A is the pectinate topology; type B is the balanced topology. The coloured time strips identify the three phylogenetic epochs; their drawn widths are schematic, not estimated times.

Count the number of descendant leaves on each branch during each phylogenetic epoch. From the tips backward, both topologies have branch-size lists $(1,1,1,1)$ during $T_4$, then $(2,1,1)$ during $T_3$. During $T_2$, type A has $(3,1)$ and type B has $(2,2)$.

Let $L_j$ be the total length of branches carrying $j$ descendant leaves. Adding the relevant phylogenetic epoch durations givesFor example, type A has one external branch and one three-descendant branch throughout $T_2$, while type B has two two-descendant branches. In either case the total expected branch length is $4E_4+3E_3+2E_2$, a useful accounting check. The stem above the common root is excluded because mutations there would be shared by all four samples.

For a haploid population with $N$ possible parents, each pair of ancestral lineages merges backward at rate $1/N$ in the Kingman's coalescent approximation. With $k$ lineages there are $\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ pairs. Hence the mean phylogenetic epoch duration isThus $E_2=N$, $E_3=N/3$ and $E_4=N/6$. For an autosomal sample from $N$ diploid individuals, replace the number of gene copies by $2N$, giving $4N/[k(k-1)]$. A mitochondrial sample naturally uses a haploid maternal effective population size; the common factor cancels in the ratios below.

In the Yule process, each of $k$ branches gives birth independently at rate $\lambda$ per generation. The next birth therefore has total rate $k\lambda$. Its waiting time has an exponential distribution, so the mean phylogenetic epoch duration isFor the ensuing calculation use $E_2=1/(2\lambda )$, $E_3=1/(3\lambda )$ and $E_4=1/(4\lambda )$. This is the inter-event waiting-time convention requested in the question, including an observation interval with four branches. Stopping the forward process immediately when the fourth branch is born would instead leave a zero terminal interval and describe a different sampling convention.

A phylogenetic epoch is an interval between successive branching or merging events during which the number of extant ancestral branches is constant. Its expected duration with $k$ branches is denoted $E_k$.