Start with $n$ elements (let's say $\{0,1,...,n\}$), then sample $m$ elements uniformly with $m<n$, then add another element ($n+1$), sample $m$ elements uniformly, and keep doing that for $T$ steps.
Let $n+j$ be the current step, then the probability of sampling $i$ with $i<n+j$ is: $$P(X=i) = \dfrac{m!(n+j-m)!}{(n+j)!}$$
Therefore, after $T$ steps, the probability of sampling element $i$ with $i>n$ is: $$P(X=i)= \dfrac{1}{K}\sum_{t=i}^T\dfrac{m!(n+t-m)!}{(n+t)!}$$
where $K$ is a constant to make the quantity a probability (make all the masses add up to 1).
Now, I am interested in the likelihood ratio of samples $i$ and $j$, $\dfrac{P(X=i)}{P(X=j)}$, but that is not easy to compute, especially for large values. This seems like a very generic situation, so I'd imagine there is a nice formula or some bounds, e.g., $d=j-i$, then the likelihood ratio is bounded below by a quantity dependent on $d$. I'd appreciate any help on how to compute or obtain some bounds for likelihood ratio.