Given a Bernoulli trial with probability of success p and a number of trials x, what is the expected number of trials until success in a case success was achieved in x trials? It is assumed that after a successful attempt, the single experiment ends and no further trials are taken.

e.g. if p = 0.02 and x = 60, I would like to find the expected number of trials taken in case success was obtained 60 trials. Indeed there is a chance to fail all 60, but I'm interested, for the case of success within 60 trials, in the average/expected number of trials it takes. I understand it is lower than 1/0.02 and decreases with x.

  • $\begingroup$ So you want to find, given that there was at least one success, the expected number of trials taken to reach the first success? $\endgroup$ Jul 7, 2019 at 20:44
  • $\begingroup$ Well for the case of p = 0.02 and x = 60, I understand that 1-((1-0.02)^60) ~ 0.7 of experiments result in success. For those experiments, what is the expected number of trials to reach first (and only) success? $\endgroup$ Jul 7, 2019 at 20:54
  • $\begingroup$ Are you only considering experiments with exactly one success? $\endgroup$ Jul 7, 2019 at 20:56
  • $\begingroup$ Yes the experiment terminates after a successful attempt $\endgroup$ Jul 7, 2019 at 20:57

1 Answer 1


The waiting time to the first success is geometrically distributed: $$ P(T\le t) = 1-(1-p)^{t}.$$ You are interested in conditioning on $T\le 60,$ so for $t\le 60,$ we have $$ P(T\le t\mid T\le 60) = \frac{P(T\le t,T\le 60)}{P(T\le 60)} = \frac{P(T\le t)}{P(T\le 60)} =\frac{1-(1-p)^{t}}{1-(1-p)^{60}},$$ and differencing gives $$ P(T=t\mid T\le 60) = \frac{p(1-p)^{t-1}}{1-(1-p)^{60}}.$$ Thus, $$ E(T\mid T\le 60) = \frac{p}{1-(1-p)^{60}}\sum_{t=1}^{60}t(1-p)^{t-1}.$$

This sum can be done in closed form using a standard trick $$ \sum_{t=1}^{60}tz^{t-1}= \frac{d}{dz}\sum_{t=1}^{60} z^t = \frac{d}{dz}\frac{z-z^{61}}{1-z}=\frac{z-z^{61}}{(1-z)^2}+\frac{1-60z^{59}}{1-z}$$ so taking $z=1-p,$ we have $$ E(T\mid T\le 60) = \frac{1-z}{1-z^{60}}\left(\frac{z-z^{61}}{(1-z)^2}-\frac{1-61z^{60}}{1-z}\right) = \frac{1-p}{p} + \frac{1-61(1-p)^{60}}{1-(1-p)^{60}}\\= \frac{1}{p}- \frac{60(1-p)^{60}}{1-(1-p)^{60}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.