The tea bag problem: probability of extracting a single bag of tea

Say you select tea bag $i$. Your question, then, is what is the probability that teabag $\hat i$ has previously been selected. (where $\forall i$, teabag $i$ was initially paired with $\hat i$). thus we are asked to compute the probability that a given teabag survived the selection process conditioned on the fact that we know some other specified teabag did survive. the conditioning just requires us to start with a sample of $2N_0-1$ and then the probability that a given teabag survived $k$ selections is $$\frac {2N_0-2}{2N_0-1}\times \frac {2N_0-3}{2N_0-2}\times \cdots\times \frac {2N_0-(k+1)}{2N_0-k}=\frac {2N_0-(k+1)}{2N_0-1}$$

The answer you seek is the compliment of this, hence $$1-\frac {2N_0-(k+1)}{2N_0-1}=\frac k{2N_0-1}$$