What are the odds of a bad run?

When something happens at random at a steady average rate — an outage, a defect slipping through, a customer walking in — the count in any window follows the Poisson distributionThe distribution of how many independent, constant-rate events land in a fixed window. Its one parameter, λ, is both the mean and the variance.. "Two a month on average" doesn't mean two every month: some months you'll get none, some four. Feed in the average and read off the odds of each.

The rate
On averageper.
Over a window ofmonths, that's λ = 4.5 expected.
1.1%chance of none
98.9%at least one
4.5 ± 2.1expected ± spreadThe Poisson mean equals its variance, so the typical spread is √λ. Most windows land within one of these of the average.
The odds
The chance of at leastoutage in the window
98.9%P(≥ 1)

Over the window, the chance of at least 1 outage is 98.9%.

19%0 outages: 1.1%1 outage: 5%2 outages: 11.2%3 outages: 16.9%4 outages: 19%5 outages: 17.1%6 outages: 12.8%7 outages: 8.2%8 outages: 4.6%9 outages: 2.3%10 outages: 1%11 outages: 0.43%12 outages: 0.16%13 outages: 0.06%14 outages: 0.02%
0268101214λ = 4.5

number of outages in the window →

exactly 1: 5% · at most 1: 6.1% · at least 1: 98.9%

A low average doesn't make a quiet window a sure thing — and a bad run of several isn't a sign the rate jumped, just the tail of the same distribution. The bars highlighted are the ones your question sums; tap any bar to ask about that count.

the Poisson distribution — first famously fit to the number of Prussian cavalrymen kicked to death by their horses each year