1
$\begingroup$

I am trying to find the expected value of the number of days where exactly $k$ people have a birthday in a class which consists of $60$ independently chosen people.

For $k = 0, 1, 2, 3, 4$

I am confused if the answer here is similar to the answer to this post, as here we need the expected value of the days not people?

And does the term "independently chosen" imply that I also need to apply a probability of choosing a certain person from the 60 ppl or is it just an indication to get the product of the events?

Any hints greatly appreciated

$\endgroup$
2
$\begingroup$

With the usual assumptions (each person's birthday independently and uniformly distributed over $365$ days) the probability that exactly $k$ out of $60$ people share $1$ January as a birthday is binomial: ${60 \choose k}\frac{364^{60-k}}{365^{60}}$, and similarly for each of the $365$ days you might consider

So the expected number of days that have exactly $k$ people sharing that day as a birthday is $${60 \choose k}\frac{364^{60-k}}{365^{59}}$$

For small values of $k$ this gives the following (rounded) expectations

k     0       1       2        3         4           5            6
E  309.6022 51.03333 4.135943 0.2196746 0.008599899 0.0002646123 0.000006663771

and these add up to about $365$ as you might expect; unrounded and taking $k$ from $0$ through to $60$, the sum would be exactly $365$

$\endgroup$

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.