# Time to first success in case of increasing probability at each trial

It may be an obvious question, but how can we correctly evaluate, in general, the time to first success in a process in which the probability increases at each trial?

The particular example I have in mind is the event $L_k$ "to get, in $k$ independent trials (i.e. with replacement), at least one element of kind A and at least one element of kind B", taken from an urn containing $\alpha$ elements of kind A, $\beta$ elements of kind B and $\gamma$ elements of kind G, and $c=\alpha+\beta+\gamma$. Clearly, $$P(L_k)=1-\left(\frac{\alpha+\gamma}{c}\right)^k-\left(\frac{\beta+\gamma}{c}\right)^k+\left(\frac{\gamma}{c}\right)^k,$$

which (given $\alpha,\beta,\gamma>0$) is a strictly monotonically increasing function of $k$. Then, the question is: If we perform $k=n$ independent trials, what is the expected number of trials to first success for the event $L_n$?

This a useful trick for computing expected values of integer-valued random variables: $$E[X] = \sum_{i=0}^\infty P(X>i)$$ In your case, let $X$ be number of trials it takes to get at least one of A and at least one of B. Then$$P(X>i)=1-P(L_i)=\left(\frac{\alpha+\gamma}{c}\right)^i+\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\gamma}{c}\right)^i,$$ so using the formula for the sum of an infinite geometric series, $$E[X]=\sum_{i=0}^\infty\left(\frac{\alpha+\gamma}{c}\right)^i+\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\gamma}{c}\right)^i=\frac{c}{\beta}+\frac{c}{\alpha}-\frac{c}{\alpha+\beta}$$

• Thanks Mike, looks great. This is the expected value of the event $L_n$, right? But I don't understand in which sense it can be seen as a "time to first success", maybe I miss some knowledge here.
– user559615
Commented May 7, 2018 at 15:06
• I made some changes, it should be clearer now. You cannot take the expected value of an event, $L_n$, but the calculation you did for $L_n$ is useful for computing the expected value of the random variable. Commented May 7, 2018 at 15:11
• Thanks a lot, Mike. Yes, I understand better now. I will think more about it and, in case, I will come back to you. Many thanks again!
– user559615
Commented May 7, 2018 at 15:36
• Some observations: 1) I don't catch clearly why $P(X>i)=1-P(L_i)$. 2) The index $i$ spans from $0$ to $\infty$. However my question is the expected time to first success in $n$ trials. Should I simply use the formula for the geometric progression? It is important because the event $L_n$ can not occur at the first trial, therefore the index $i$ should span from 0 to $n-1$. 3) How do we take into account the fact that if the event $L_n$ occurs in one trial, then it occurs in all the following trials? Thanks a lot!
– user559615
Commented May 8, 2018 at 9:40