If you have a given probability for a specific outcome in a specific event or period of time, how can you calculate the probability of it happening over a length time or number of events?

I've seen answers that might explain what I'm talking about, but I don't understand the notation.

If I have a 10% chance of something happening in a given day, it is NOT a 100% chance over 10 days, but how do you calculate the actual odds?

I thought the way to do it was subtracting the chance it didn't happen from one, for example:

If I have a .1 chance of an outcome on a given day, I think the likelihood of it happening at some point over 10 days is 1-.9^10 which is approximately .65

Did I do this right? Does a 10 percent chance of an outcome per event only have a 65% chance of happening over the course of 10 events?

For a 1% probability over 100 events I tried 1-.99^100 which approximated to 63.4%.

If this is wrong, how do you accrue the likelihood of an event over a number of periods when only the likelihood in a single period is given?

  • 2
    $\begingroup$ You are correct. If the probability of the event occurring on a given day is $p$, and different days are independent, then the probability that the event occurs at least once in $n$ days is $1-(1-p)^n$. $\endgroup$ – angryavian Jan 10 at 21:11

Yes, I agree to you and the comment of angyavian. Maybe a simplification of the term is useful. We have $1-\left(0.99\right)^{100}=1-\left(1-\frac{1}{100} \right)^{100}$

This term can be approximated. For large $n$

$\left(1-\frac{x}{n} \right)^{n}\approx e^{-x}$

In your case it is

$$1-\left(1-\frac{1}{100} \right)^{100}\approx 1-e^{-1}\approx63.2\%$$

In both cases the probability is approximately $63\%$.


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.