Probability to seat at the same table in a poker tournament. Warning. This question is inspired by the ongoing Main Event of the Word Series of Poker, but it is a purely mathematical question and requires no knowledge of poker.
Consider a poker tournament with $N = 10000$ players. The $N$ players are ranked (randomly) from $1$ to $N$ and have uniform equity: for each $k \in \{1, \ldots, N\}$, each player has probability $1/N$ to finish the tournament in $k$-th position. The tournament goes as follows: players are gathered in $N/10 = 1000$ tables (numbered from $1$ to $N/10$) of $10$ players each: Table $1$ gathers players $1$ to $10$, Table $2$ gathers players $11$ to $20$ and so on. 
Each time $10$ players are eliminated, the tables are filled again according to the following process: if there were $K$ remaining tables (numbered from $1$ to $K$), the players of Table $K$ are used to fill the remaining $K - 1$ tables, starting from Table $1$, up to Table $K-1$. After this step, there are again $K-1$ full tables of $10$ players$^{(*)}$.  The tournament finishes when there is only one remaining player, the winner.
Question. Two friends register for this tournament. What is the probability that they seat at the same table at some point of the tournament?
$^{(*)} \scriptsize\text{If you need a more precise algorithm, let us say that the players fill the tables in decreasing rank order.}$
 A: Let's assume table refilling occurs randomly (which makes the problem simpler compared to refilling by rank!).
The probability of meeting immediately in a tournament with $T$ tables of size $S$ is $\frac {S-1}{ST-1}$. If this does not happen, the friends are at different tables $1\le a<b\le T$, and nothing of interest happens until the table count is about to decrease from $b$ to $b-1$.
Assuming that the two have survived so far (probability of this: $\frac{Sb\choose 2}{ST\choose 2}$), the friend from table $b$ is assigned to a random position at the other tables, so the friends meet with probability $\frac{S-1}{S(b-1)-1}$. In fact, we may consider this as a new tournament with $b-1$ tables.
Thus if $f(T)$ denotes the desired probability, we have
$$\begin{align}f(T)& =\frac{S-1}{ST-1}+\sum_{b=2}^T\frac{S\cdot (b-1)S}{ST\choose 2}f(b-1)\\&=\frac{S-1}{ST-1}+\sum_{k=1}^{T-1}\frac{kSf(k)}{T(ST-1)}.\end{align}$$
Numerically, with $S=10$, we find $f(1000)\approx 0.006896773971525475279786665979649$.
