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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 $L_n$?

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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, $$ E[X]=\sum_{i=0}^\infty\left(\frac{\alpha+\gamma}{c}\right)^i+\left(\frac{\beta+\gamma}{c}\right)^i-\left(\frac{\gamma}{c}\right)^i=\frac{c}{\beta}+\frac{c}{\alpha}-\frac{c}{\alpha+\beta} $$

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  • $\begingroup$ 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. $\endgroup$
    – user559615
    Commented May 7, 2018 at 15:06
  • $\begingroup$ 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. $\endgroup$ Commented May 7, 2018 at 15:11
  • $\begingroup$ 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! $\endgroup$
    – user559615
    Commented May 7, 2018 at 15:36
  • $\begingroup$ 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! $\endgroup$
    – user559615
    Commented May 8, 2018 at 9:40

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