Probability that multiple rare events happen in a time window over a long time span Yesterday at work I hit a small probability problem:
Once per minute, a Bad Thing might happen with probability 0.01.  If three or more Bad Things happen within 15 minutes of each other, an alarm will sound.  What is the probability that an alarm will sound sometime in the next 24 hours?
I can get the probability that an alarm will fire in a given 15-minute interval with a Poisson distribution, but after that I get stuck: modelling the day as a sequence of disjoint 15-minute intervals will undercount, but modelling it as a sequence of overlapping intervals will overcount (won't it?).  Nothing else I've been able to find in the Poisson family seems appropriate either.  Suggestions?
 A: It seems reasonable to assume that the time between events has an exponential distribution.  With that assumption, $P(X \le 1) = 0.01$ implies the distribution has rate $\lambda = -\ln(1 -0.01) = 0.0100503$ per minute. Suppose three successive events occur at times $T_1, T_2$ and $T_3$ with $T_2-T_1 = X_1$ and $T_3 - T_2 = X_2$, where $X_1$ an $X_2$ are independent events from an exponential distribution with rate $\lambda$. Then it is known that $T_3 - T_1 = X_1 + X_2$ has a gamma distribution with shape $2$ and rate $\lambda$, so $P(X_1 + X_2 \le 15) =  0.01028352$. (I used the R function pgamma for this computation.)
With these facts in hand, we can make a crude estimate of the probability that three or more events occur within $15$ minutes at least once in $1440$ minutes (one day), as follows. We know that the mean time between events is $1 / \lambda = 99.50$ minutes, so we expect around $1440 / 99.50 = 14.47$ events in a day.  Except for the last two, each event is the start of a three-event sequence.  So we expect around $12.47$ such sequences, each of which has a probability of $0.01028352$ of occurring in less than $15$ minutes. So if we assume independence, the probability of experiencing three events within $15$ minutes at least once in a day is about
$$1 - (1 - 0.01028352)^{12.47} \approx 0.121$$
The assumption of independence is false, because the three-event sequences overlap.  For comparison, I ran a Monte Carlo simulation of $10^6$ days and found the probability of having a bad day was about $0.134$. So it seems the approximate approach isn't too bad.
