I have a black-box discrete probability distribution $\mathbf{p}$ (I can only get samples from it), two items in the support: $\sigma_1, \sigma_2$, and a procedure which estimates the following quantity: $$ r = \frac{\mathbf{p}(\sigma_1)}{\mathbf{p}(\sigma_1)+\mathbf{p}(\sigma_2)} $$
The procedure runs for $T$ iterations ($T$ is not known in advance).
At each iteration $t$, I get a multiset $S_t$ of $n_t$ independent samples from $\mathbf{p}$ conditioned on the set $\{\sigma_1, \sigma_2\}$. Then, my estimate for $r$ is: $$ \hat{r}_t = \frac{1}{n_t} \cdot \sum_{s \in S_t} [s = \sigma_1] $$
(where $[P]$ is the Iverson bracket notation).
Think of this procedure as estimating the bias of a coin, with $\{\sigma_1, \sigma_2\} = \{H, T\}$.
Can I optimize the procedure by reusing the samples from the previous iteration when computing the estimate $\hat{r}_{t+1}$? That is, instead of drawing $n_{t+1}$ samples at the $(t+1)$-th iteration, can I just draw the difference $n_{t+1} - n_t$ and reuse the $n_t$ samples from the $t$-th iteration? Note that $n_1 < n_2 < \dots < n_T$.
Since samples are independent at each iteration, I suppose this is fine. And if this is the case, my next question is: how to formally prove that reusing samples is fine and does not affect the estimate?
Clarifications
The whole discussion stemmed from the following goal, that is, I want: $$ \Pr \bigg[ \hat{r} \in \big[r (1-\epsilon), \ r (1+\epsilon) \big] \bigg] \geq 1-\delta $$ Thus, I have derived a lower bound on the number of samples needed, using a multiplicative Chernoff bound, which yielded: $$ n = \Theta \left( \frac{1}{(\epsilon \cdot \hat{r})^2} \cdot \log \frac1\delta \right) $$ so the estimation is part of the bound. To this extent, we need to make reasonable guesses: for the $t$-th iteration, we guess that $r \geq 1/2^t$, and set $\hat{r} = 1/2^t$ in the bound for the number of samples, $n_t$. Then, after sampling $n_t$ values, if we get that the estimate is indeed $\geq 1/2^t$, we stop; otherwise $t \gets t + 1$ and we continue.
It's worth noting that, assuming $r \sim \mathsf{Uniform}(0,1)$, we have that $T \sim \mathsf{Geometric}(1/2)$.