The answers already provided to this and to the equivalent post about cars in queue
deal with the expected number of groups, and that is what the post asks.
Since the problem is quite interesting, I got curious about the underlying PDF but I did not succeed to find a satisfactory
hint about (admittedly, I might have overlooked something) .
However I tried and develop a different approach
which shows which is the probability distribution behind it.
Consider to "quantize" the speed into $n$ classes.
Then we can represent the $q$ sheeps in queue at time $0$ onto a diagram speed vs. position as in the sketch.
It is clear that the first ship will block all the following ones
with higher or equal speed.
That means all those before the first (n. $4$ in the sketch)
that has a speed lower than n. $1$.
That in her turn will block the following ones with same or higher speed, etc.
Note that we are individuating the resulting groups as the blocks that along the time get
separated by an even larger distance. So in the sketch n. $1$ and n. $3$ are in the same group.
Now the total possible ways of arranging the diagram is $T(n,q)=n^q$.
The number of ways to arrange a group, with leader speed $v$ and $m$ members, is:
$$
G(v,m) = \left( {n - v + 1} \right)^{\,m-1}
$$
Therefore the number of ways $N_{1}(n,q)$ to arrange the sheeps
in such a way that they finally make up only one group will be:
$$
\begin{gathered}
N_{\,1} (n,q) = \sum\limits_{1\, \leqslant \,\,v_{\,1} \, \leqslant \,n} {\left( {n - v_{\,1} + 1} \right)^{\,q - 1} } = \sum\limits_{1\, \leqslant \,\,k\, \leqslant \,n} {k^{\,q - 1} } = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,q - 1} \right)} {\left\langle \begin{gathered}
q - 1 \\
j \\
\end{gathered} \right\rangle \left( \begin{gathered}
n + 1 + j \\
q \\
\end{gathered} \right)} = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,q - 1} \right)} {\;j!\;\left\{ \begin{gathered}
q - 1 \\
j \\
\end{gathered} \right\}\left( \begin{gathered}
n + 1 \\
j + 1 \\
\end{gathered} \right)} = \hfill \\
= \frac{1}
{q}\sum\limits_{0\, \leqslant \,j\, \leqslant \,q - 1} {\left( \begin{gathered}
q \\
j \\
\end{gathered} \right)\;B_j \;\left( {n + 1} \right)^{\,q - j} } \quad \left| {\;1 \leqslant \text{integer }q,n} \right. \hfill \\
\end{gathered}
$$
where $ \left\langle {} \right\rangle $ indicate the Eulerian N., $\left\{ {} \right\}$ the Stirling N. 2nd kind, and $B_j$ the Bernoulli N.
Then the number of ways to arrange them as to finally have two groups is
$$
\begin{gathered}
N_{\,2} (n,q) = \sum\limits_{\left\{ \begin{subarray}{l}
1\, \leqslant \,v_{\,2} \, < \,v_{\,1} \, \leqslant \,n \\
m_{\,2} \, + \,m_{\,1} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} \right.} {\left( {n - v_{\,1} + 1} \right)^{\,m_{\,1} - 1} \left( {n - v_{\,2} + 1} \right)^{\,m_{\,2} - 1} } = \hfill \\
= \sum\limits_{\left\{ \begin{subarray}{l}
1\, \leqslant \,k_{\,1} \, < \,k_{\,2} \, \leqslant \,n\, \\
m_{\,2} \, + \,m_{\,1} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} \right.} {k_{\,1} ^{\,m_{\,1} - 1} \;k_{\,2} ^{\,m_{\,2} - 1} } \hfill \\
\end{gathered}
$$
to which corresponds a probability
$$
P_{\,2} (n,q) = \frac{{N_{\,2} (n,q)}}
{{n^{\,q} }} = \frac{1}
{{n^{\,2} }}\sum\limits_{\left\{ \begin{subarray}{l}
1\, \leqslant \,k_{\,1} \, < \,k_{\,2} \, \leqslant \,n\, \\
m_{\,2} \, + \,m_{\,1} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} \right.} {\left( {\frac{{k_{\,1} }}
{n}} \right)^{\,m_{\,1} - 1} \;\left( {\frac{{k_{\,2} }}
{n}} \right)^{\,m_{\,2} - 1} }
$$
and so forth.
I do not know whether the multiple summations can be reduced to a simpler form. (*)
However, if we increase speed granularity (i.e. $n$) till continuum then we can replace the summation
over the $k's$ in the expression for the probability into integrals
$$
\begin{gathered}
P_{\,g} (q) = \sum\limits_{\begin{subarray}{l}
\\
\\
\,m_{\,1} \, + \,m_{\,2} + \, \cdots \, + m_{\,g} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} } {\mathop {\int {} }\limits_{\begin{subarray}{l}
\\
0\, \leqslant \,x_{\,1} \, < \,x_{\,2} < \, \cdots \, < \,x_{\,g} \, \leqslant \,1
\end{subarray} } \prod\limits_{1\, \leqslant \,j\, \leqslant \,g\,} {x_{\,j} ^{\,m_{\,j} - 1} \;dx_{\,j} } } = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\\
\,m_{\,1} \, + \,m_{\,2} + \, \cdots \, + m_{\,g} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} } {\frac{1}
{{m_{\,1} }}\;\mathop {\int {} }\limits_{0\, \leqslant \,\,x_{\,2} < \, \cdots \, < \,x_{\,g} \, \leqslant \,1} x_{\,2} ^{\,m_{\,1} + m_{\,2} - 1} \;dx_{\,2} \prod\limits_{3\, \leqslant \,j\, \leqslant \,l\,} {x_{\,j} ^{\,m_{\,j} - 1} \;dx_{\,j} } } = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\\
\,m_{\,1} \, + \,m_{\,2} + \, \cdots \, + m_{\,g} \, = \,q\;\;\left| {\;1\, \leqslant \,m_{\,k} \,} \right.
\end{subarray} } {\frac{1}
{{m_{\,1} }}\frac{1}
{{m_{\,1} + m_{\,2} }}\, \cdots \,\frac{1}
{{m_{\,1} + m_{\,2} + \, \cdots \, + m_{\,g} }}} \hfill \\
\end{gathered}
$$
We can see that $P$ satisfies the following recursion
$$ \bbox[lightyellow] {
P_{\,g} (q) = \frac{1}
{q}\;\sum\limits_{\left( {0\, \leqslant \,g - 1 \leqslant } \right)\,k\, \leqslant \,q - 1} {P_{\,g - 1} (k)}
}$$
and for the initial conditions we can put that $0$ sheeps can only be arranged into a empty group and v.v.
that an empty group can gather only $0$ sheeps.
$$
\left\{ \begin{gathered}
P_{\,g} (q) = 0\quad \left| {\;g < 0\; \vee \;p < 0} \right. \hfill \\
P_{\,g} (0) = \delta _{\,g,\,0} \hfill \\
P_{\,0} (q) = \delta _{\,q,\,0} \hfill \\
\end{gathered} \right.
$$
Now, starting from a known identity about Stirling N. of the 1st kind, we have
$$
\begin{gathered}
\left[ \begin{gathered}
n + 1 \\
m + 1 \\
\end{gathered} \right] = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left[ \begin{gathered}
k \\
m \\
\end{gathered} \right]n^{\,\underline {\,n - k\,} } } = n!\sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left[ \begin{gathered}
k \\
m \\
\end{gathered} \right]\frac{1}
{{k!}}} \quad \Rightarrow \hfill \\
\Rightarrow \quad \left[ \begin{gathered}
n \\
m \\
\end{gathered} \right] = \left( {n - 1} \right)!\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\left[ \begin{gathered}
k \\
m - 1 \\
\end{gathered} \right]\frac{1}
{{k!}}} \quad \left| {\;1 \leqslant n,m} \right.\quad \Rightarrow \hfill \\
\Rightarrow \quad \left( {\frac{1}
{{n!}}\left[ \begin{gathered}
n \\
m \\
\end{gathered} \right]} \right) = \frac{1}
{n}\;\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\left( {\frac{1}
{{k!}}\left[ \begin{gathered}
k \\
m - 1 \\
\end{gathered} \right]} \right)} \hfill \\
\end{gathered}
$$
Finally, having the same recursion and same initial conditions, we reach to:
$$ \bbox[lightyellow] {
P_{\,g} (q) = \frac{1}
{{q!}}\left[ \begin{gathered}
q \\
g \\
\end{gathered} \right]
}$$
and then it is known (**) that
$$ \bbox[lightyellow] {
\overline g (q) = \sum\limits_{\left( {0\, \leqslant } \right)\,g\,\left( { \leqslant \,q} \right)} {\frac{g}
{{q!}}\left[ \begin{gathered}
q \\
g \\
\end{gathered} \right]} = H(q)
}$$
-----------
Note (*)
I have later found a closed form for $N_{g}(n,q)$, which is presented in this related post.
----------
Note (**)
because we have in fact
$$
\begin{gathered}
\frac{{\left( {x + 1} \right)^{\,\overline {\,n\,} } }}
{{n!}} = \prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{{x + 1 + k}}
{{1 + k}}} = \exp \left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\ln \left( {1 + \frac{x}
{{1 + k}}} \right)} } \right) = \sum\limits_{\left( {0\, \leqslant } \right)\,k} {\frac{1}
{{n!}}\left[ \begin{gathered}
n \\
k \\
\end{gathered} \right]\left( {x + 1} \right)^{\,k} } \hfill \\
\left( {x + 1} \right)\frac{d}
{{d\,x}}\frac{{\left( {x + 1} \right)^{\,\overline {\,n\,} } }}
{{n!}} = \exp \left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\ln } \left( {1 + \frac{x}
{{k + 1}}} \right)} \right)\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{{x + 1}}
{{x + k + 1}}} = \sum\limits_{\left( {0\, \leqslant } \right)\,k} {\frac{k}
{{n!}}\left[ \begin{gathered}
n \\
k \\
\end{gathered} \right]\left( {x + 1} \right)^{\,k} } \hfill \\
\left. {\left( {\left( {x + 1} \right)\frac{d}
{{d\,x}}\frac{{\left( {x + 1} \right)^{\,\overline {\,n\,} } }}
{{n!}} = \left( {x + 1} \right)\frac{d}
{{d\,x}}\frac{{\Gamma \left( {x + 1 + n} \right)}}
{{\Gamma \left( {x + 1} \right)\Gamma \left( {1 + n} \right)}}} \right)\;} \right|_{\;x\, = \,0} = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\frac{1}
{{k + 1}}} = \sum\limits_{\left( {0\, \leqslant } \right)\,k} {\frac{k}
{{n!}}\left[ \begin{gathered}
n \\
k \\
\end{gathered} \right]} \hfill \\
\end{gathered}
$$
