1
$\begingroup$

I have four people (n) and every time each is getting a random three digit combination (000-999). This means there are 1000 possible combinations (k).

I want to know how probable it is for two or more people to get the same three digits in idk let's say 10,000 trys.

I found this but I calculated it with n=4 and k=1000 and it doesn't seem to be the right formula for my use case. How would I calculate something like that?

$\endgroup$
1
$\begingroup$

The probability is $$ p=1-\left[\frac{(k-1)(k-2)(k-3)}{k^3}\right]^N, $$ where $N$ is the number of trials.

$\endgroup$
  • 1
    $\begingroup$ In 100 games there is a 50% chance? That's a lot! $\endgroup$ – MiXT4PE Jan 9 at 16:52
  • $\begingroup$ Yes, according to calculation it is about 45%. $\endgroup$ – user Jan 9 at 17:03

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.