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The problem of the calculating the probability that there is a birthday shared by at least 2 people in a group of size n is well known. I am wondering if there is a way of finding the probability of there being a birthday shared by m people in a group of size n. I couldn't find any info about this online and was unable to solve it myself.

The particular question asked by a friend of mine was on the chances of there being some day which is at least 4 peoples birthday from a group of 50. I was able to get the answer through a monte-carlo simulation, but am still interested in an analytic solution.

Edit: I have since solved the problem, using Bernoulli trials, a technique I just learned in my discrete math class. The general formula is $1-\left(\sum_{i=0}^{m-1}\operatorname{nCr}\left(n,i\right)\cdot\frac{1}{365}^{i}\cdot\frac{364}{365}^{\left(50-i\right)}\right)^{365}$. The part inside the sum is the chance of, for a fixed day, that there are exactly i shared birthdays. The sum finds the chance that there are less than m shared birthdays. Raising this to the power of 365 finds the chance that there are less than m shared birthdays every day. Subtracting this from 1 gets the chance that there are m or more shared birthdays. The answer I got for the specific case agreed with my simulation within +-.000001.

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    $\begingroup$ Sometimes approximations / simulations can be a good thing and a time saver. Here is a Math.SE post in the case of $3$ people sharing a birthday out of $30$ people and you can see how complicated the answer gets for it. I would imagine it would be further complicated for your case but perhaps not impossible to use inspiration from that post to construct a solution that would work $\endgroup$
    – WaveX
    Commented Feb 18, 2020 at 0:53
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    $\begingroup$ Note that the linked question itself (and the accepted answer) only mention the case where $3$ people share a birthday, but other answers discuss various ways to compute exact or approximate probabilities for $M$ people sharing a birthday for some arbitrary positive integer $M$. $\endgroup$
    – David K
    Commented Feb 18, 2020 at 1:49
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    $\begingroup$ This is also relevant and has useful information in its answers: stats.stackexchange.com/questions/1308/… $\endgroup$
    – David K
    Commented Feb 18, 2020 at 1:56

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Take a look at Hildebrand's Short course on asymptotics. One of the problems for which he develops an asymptotic approximation is the birthday problem, you should be able to extend it to your case. So you get an exact expression (very messy) and a reasonably clean approximation.

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    $\begingroup$ Good materials! Thanks! $\endgroup$
    – alan23273850
    Commented May 17, 2023 at 16:37

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