Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This article states that the formula for the average run lengths for large numbers of trials is:$$\frac{1}{1-Pr(event\ in\ one\ trial)}.$$

My questions

  • What is the intuition behind this formula?
  • Do you know an elementary proof for this result?
share|improve this question
2  
@Downvoter: Why the downvote: Please comment how to enhance the question at least - thank you! –  vonjd Jul 15 '12 at 12:49
add comment

2 Answers

up vote 4 down vote accepted

This is one of the workhorses of first courses in probability. Assume one repeats an experiment whose probability to happen in one trial is $p$, that the results of the successive trials are independent, and that one wishes to estimate the mean number $R$ of consecutive successes before the next failure (that is, a run of successes) once a success occurred.

For every $r\geqslant1$, the event $[R\geqslant r]$ corresponds to $r-1$ supplementary successes after the first one, hence $\mathrm P(R\geqslant r)=p^{r-1}$. Thus the distribution of $R$ is geometric with parameter $p$ and the mean length of a run is the expectation of $R$, that is, $$ \langle R\rangle=\sum_{r\geqslant1}r\cdot\mathrm P(R=r)=\sum_{r\geqslant1}\mathrm P(R\geqslant r)=\sum_{s\geqslant0}p^s=\frac1{1-p}. $$ Edit Alternatively, note that either a failure occurs immediately after the first success and then $R=1$, which happens with probability $1-p$, or a second success occurs immediately after the first success and then $R=1+R'$ where $R'$ is distributed like $R$, which happens with probability $p$. Taking expectations, one sees that $\langle R\rangle=(1-p)\cdot1+p\cdot(1+\langle R\rangle)$, that is, $\langle R\rangle=1/(1-p)$.

share|improve this answer
add comment

The run is ended by a failure, and the probability of failure is $1-p$, where $p$ is the probability of success. Since the density of failures in the sequence of results is $1-p$, the average length of a run ending in a failure must be the reciprocal of that. The failure itself isn't part of the run, so you might think you'd have to subtract $1$ from this result; however, if you're calculating the average length of a run, the fact that you're only looking at runs already guarantees that there's at least one success, and this success that you get "for free" offsets the failure to be subtracted.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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