Expected Number of Trials to Achieve a Result With Dependent Probability I'm trying to find the expected number of trials, $x$, to achieve $n$ number of successes, when the probability of a success is dependent on the results of the last trial.
Specifically, I'm trying to calculate the effects of gacha pity on the expected number of trials to achieve $n$ number of successes. How it works is if a certain number of trials, $t$, have occurred without a success, the probability of success, $p$, will be increased by a set amount, $q$, for each failure. Once a success has been achieved, the number of trials in order to activate the probability increase is set back to $t$. I know that the expected number of trials for a number of successes is given by $x = n/p$, but this only works when $p$ is constant.
In the gacha system I'm examining:

$$p = 0.02,$$
$$t = 50,$$
$$q = +0.02$$

Any ideas or places you could point me to would be appreciated!
 A: The expected number of attempts for $n$ successes is $n$ times the expected number of attempts for one success, and it is easier to look at the expected number of attempts for one success.

*

*The probability of no success after $m$ attempts when $m\le t$ is $(1-p)^m$

*The probability of no success after $m$ attempts when $t \lt m\le t+\frac{1-p}{q}$ is $(1-p)^t \prod\limits_{k=1}^{m-t}(1-p-kq)$

*The probability of no success after $m$ attempts when $t+\frac{1-p}{q} \lt m$ is $0$
If you add these all up to give the expected number of attempts for one success and then multiply by $n$, then  you get $$n \left(\frac{1-p^{t+1}}{1-p} + \left(1-p^{t}\right)\sum\limits_{m=t+1}^{t+\lfloor(1-p)/q\rfloor} \prod\limits_{k=1}^{m-t}(1-p-kq)\right)$$
This will not have a closed form, but in the example in the question with $n=3, p=q=0.02$ and $t=50$ it seems to give about $3\times 34.594555=103.783665$ from the earlier result
It may be possible to give approximations, at least for large $t$ and small $p$ and $q$ in special circumstances.  For example if $c=\frac1p=\frac1q$ for some integer $c$ then I think you may be able to use, if I have done this correctly, an approximation like $$n\left(c-e^{-t/c}\left(c-\sqrt{\frac \pi 2}\sqrt{c}+\frac56-\frac7{12}\sqrt{\frac \pi 2}\sqrt{\frac{1}{c}} + \frac{617}{1080}\frac1c -\cdots\right) \right) $$
which with $n=3, t=50$ and $c=50$ gives about $103.7806$, which is not far away from the earlier result.
