Calculate number of successes after T trials with linearly increasing probability of success Lets say i have a set of 100 trials i want to test, each one has a probability of success of T / 100, where T = trial number (so first trial is 1 / 100, second trial is 2 / 100).
Also everytime a trial succeed, our trial number starts back to 1, but in total we still do 100 trials.
Now the question is, what is number of successes i should expect after 100 trials?
To give a better idea, i'm trying to do a point system, where everytime you do something you gain 1 point, and you repeat this for 100 times. Each time you do that "something" if a random generated number, 1 to 100, is lower or equal to the points you have, you have a success, so the points you have go to 0, otherwise you go on increasing points.
So again what i want to know is the formula to calculate how many times i'm expected to have a success with this system.
The "100 times/trial", and the probability increasing by 1% are just an example, as i said in the comments i actually have an old system that i want to somewhat reproduce with this new system.
 A: Let $N=100$. The number of steps since the last success occurred performs a Markov chain on $\{1,2,\ldots,N\}$ where each transition $k\to1$ has probability $k/N$ and each transition $k\to k+1$ has probability $1-k/N$. Successes are visits to the state $1$.
By the classical one-step analysis, the stationary measure $(\pi_k)$ of this Markov chain solves the system
$$\pi_{k+1}=(1-k/N)\pi_k\ (1\leqslant k\lt N),\qquad\pi_1=\sum\limits_{k=1}^N(k/N)\pi_k.$$ Thus,
$$
\pi_1=\frac{N^N}{N!}\,\left(\sum_{k=0}^{N-1}\frac{N^k}{k!}\right)^{-1}.
$$
Using Stirling's formula to estimate the prefactor and the central limit theorem to estimate the parenthesis, one sees that, when $N\to\infty$,
$$
\pi_1\sim\sqrt{\frac2{\pi N}},
$$
hence, for $N$ fixed, when $T\to\infty$, the number of visits to the state $1$ up to time $T$ is approximately
$$
\sqrt{\frac2{\pi N}}\,T.
$$
Assuming that $N$ is large enough to be considered as a large number of steps for the Markov chain on  $\{1,2,\ldots,N\}$ (a rather dubious hypothesis), the mean number of visits of state $1$ (aka, the mean number of successes) would scale as
$$
\sqrt{\frac{2N}\pi},
$$
which, for $N=100$, is about $7.98$.
