If a treatment has constant success probability of $p$ per trial, the probability of success on trial $k$ is given by the First Success Distribution. $$P(K\!=\!k\,|\,p) = p(1-p)^{k-1} \qquad k = 1, 2, 3, ...$$ Let $q = 1-p$, $$P(K\!=\! k\,|\,p) = pq^{k-1}$$ $$P(K \le k \,|\, p) = 1 - q^k$$ $$P(K > k \,|\,p) = q^k$$ Let $S_k = P(K>k|p)$. $$S_k = q^k$$ My problem is that the probability of success $p$ is not constant. After each unsuccessful trial, the probability of success decreases by $r$, so on trial $k$, $$p_k = (1-r)^{k-1}p \qquad k = 1, 2, 3, ...$$ Let $u = 1-r$, $$p_k = u^{k-1}p \qquad k = 1, 2, 3, ...$$
$$P(K > k \,|\,p, u) = \prod_{i=1}^k (1 - u^{i-1}p) \qquad k = 1, 2, 3, .. $$ Again, let $P(K > k \,|\,p, u) = S_k$. $$S_k= \prod_{i=1}^k (1 - u^{i-1}p)$$ $$P(K <= k) = 1 - S_k$$ $$P(K = k) = u^{k-1}pS_{k-1}$$ Does this distribution have a name?
Is there a simpler analytic form for $S_k$ that does not require a product from $i = 1$ to $k$?
Context: This question came up in planning a trial for an infertility treatment.
We are using $p = 0.09$ and $r = 0.22$, so $u = 1- 0.22 = 0.78$.
$$S_5 = \prod_{i=1}^5 (1 - 0.78^{i-1} \times 0.09) = 0.7401$$.
$$P(K <= 5) = 1 - 0.7401 = 0.2599$$
$$S_4 = \prod_{i=1}^4 (1 - 0.78^{i-1} \times 0.09) = 0.7656$$.
$$P(K = 5) = 0.78^{4} \times 0.7656 \times 0.09 = 0.0255$$
This says that the probability of success in up to 5 treatments is only 0.26, and the probability of success on the fifth treatment is only 0.026.
This is (unfortunately) about right.
I found a related answer here.
