At least once probability calculator

Imagine you are fare-dodging on the tram, and on each journey a ticket inspector checks you with a probability of 5%. On one journey, you are almost certain to get away with it. But over a month you take twenty journeys. What is the chance that you are caught at least once? You cannot simply add up the percentages: twenty journeys would give 100%, and thirty would give 150%. Instead, we use the complementary event: we work out the probability that the inspector never comes, and subtract it from one.

The attempts must be independent of each other and keep the same probability p.

How many attempts do you need?

You can also turn the question around: how many times would you have to ride without a ticket for the inspector to catch you almost certainly, say with a probability of 90%?

How it works

The probability that an event happens at least once in n attempts is

$$P=1-(1-p{)}^n$$

To find the number of attempts, we look for the smallest n for which 1 − (1 − p)ⁿ ≥ c, where c is the required probability. Taking logarithms of this inequality gives

$$n\ge \frac{\ln (1-c)}{\ln (1-p)}$$

We round the result up, because the number of attempts must be a whole number.

On one journey, the inspector does not come with a probability of 1 − 0.05 = 0.95. The journeys are independent, so for twenty journeys without a check we multiply the probabilities:

$${0.95}^{20}\approx 0.3585$$

Subtract from one:

$$P=1-0.3585=0.6415$$

Over twenty journeys, you are caught at least once with a probability of about 64.15%. How many journeys are needed for 90%? Substitute:

$$n\ge \frac{\ln (1-0.9)}{\ln (1-0.05)}=\frac{\ln 0.1}{\ln 0.95}\approx 44.89$$

Round up to 45 journeys. Over those, you are caught with a probability of about 90.06%. A ticket is the cheaper option.