$N$ balls with label $1..N$. After the first $q$ balls, the number of sequences is $(N-q)!$

$N$ balls with different weight and same radius. Pick these balls one by one and put them in sequence. After picking of a ball, re-mark numbers on the already-picked balls and the number represent the rank by their weight. Two examples of the sequence of the ranks is $1 \rightarrow 21 \rightarrow 312 \rightarrow 4132, 1 \rightarrow 21 \rightarrow 321 \rightarrow 4312$. After $q$ balls, the number of possible sequences is $q!$ and the number of the future sequences after $q$ balls is $N! q!$  

The number of the sequences whose weightest ball among first $q$ balls remains the weightest ball in first $I$ balls is $q(I-1)!$ And the probability is $\frac{q(I-1)!}{I!}=\frac{q}{I}$ . After all $N$ balls are picked, denote $X$ the rank of the first $q$ balls and $I + 1$ the position of a ball of a higher rank $Y$ is located. The number of possible rank sequences is $q P_{I-1}^{X - 1}(N-1-I)!$ and the possibility is $\begin{aligned}\frac{q P_{I - 1}^{X - 1}(N - 1 - I)!}{N !} = \frac{q P_{I - 1}^{X - 1}}{P_{I + 1}^N}\end{aligned}$

For [katex=Y] labled balls, counting by partition of max labeled number and location of this ball, it follows $P_I^Y = I \sum_{X = I}^Y P_{I - 1}^{X - 1}$ . When $Y$ is set to $N$, the possibility of picking the weightest ball after all $N$ balls is

$\begin{aligned}\sum_{I = q}^{N - 1}{\sum_{X = I}^{N - 1}\frac{q P_{I - 1}^{X - 1}}{P_{I + 1}^N}}= \sum_{I = q}^{N - 1}\frac{q \frac{P_I^{N - 1}}{I}}{P_{I + 1}^N} = \sum_{I = q}^{N - 1} \frac{q}{N I}\end{aligned}$

Rescale $q$ and $I$ as $N q$ and $N I$ then it is the sum

$\begin{aligned}q \sum_{I = q, step \frac{1}{N}}^{1 - \frac{1}{N}} \frac{1}{I} \times \frac{1}{N}\end{aligned}$

which approaches to $q \int_q^1 \frac{d I}{I} = - q \ln{q}$

Similarly for picking the $y$-th weightest ball, the probability is

$\begin{aligned}\sum_{I = q, step \frac{1}{N}}^{y - \frac{1}{N}} {\frac{q}{I} \times \frac{1}{N}} \prod_{J = \frac{1}{N}, step \frac{1}{N}}^{I}\frac{y-J}{1 - J}\end{aligned}$

Because $\frac{y - J}{1 - J}< y$, this value goes to zero as $N$ goes to infinity. It makes more sense to see accumulated probability

$\begin{aligned}\sum_{y = q + \frac{1}{N}}^Y {\sum_{I = q, step \frac{1}{N}}^{y - \frac{1}{N}} {\frac{q}{I} \times \frac{1}{N}} \prod_{J = \frac{1}{N}, step \frac{1}{N}}^{I}\frac{y-J}{1 - J}}\end{aligned}$