How many people are needed for a 50% chance that 3 have the same birthday?

[wolf]

A really popular "birthday paradox" is "how many people are needed for a 50% chance that 2 have the same birthday"? The answer is 23 and the mathematics for that calculation can be found at my website: https://www.1728.org/birthday.htm and a great many other websites.
I am curious to calculate this for 3 people, which I understand is much more difficult than the 2 person calculation.
I have found many websites that discuss solutions for this problem but they really don't explain the exact steps to take.
For example, at this website (page 858) https://www.math.uchicago.edu/~fcale/CCC/DC.pdf
you can see the number of people you'd need to have 3 coincident birthdays (88 people), 4 coincident birthdays (187 people), and so on.
I would like to know precisely how to calculate this.
From what I've read, the best way is to use a Poisson distribution but no site explains exactly what to do.
I'd like to know how to calculate the probability.
Thank you.

 

[tkhunny]

What makes it a "paradox"?

 

[wolf]

The "paradox" refers to the case in which you have to calculate the probability of 2 people having the same birthday. As we know, it only requires 23 randomly-selected people to have a 50% chance that 2 have the same birthday. Intuitively, you would think it should be much higher. After all, there are 365 possible birthdays that each person may have and so you would think that it would probably take 182 randomly-chosen people.

 

[tkhunny]

So, not a paradox at all, just a result that doesn't match up with naive expectation. Fair enough.

 

[wolf]

It seems I might have found a formula here:

Birthday Paradox/General/3 - ProofWiki

proofwiki.org

According to the text, n should be 365.
I wonder where the 88 people and the 3 "variable" should be entered or
(to make it easier) where to insert 23 people and a variable of 2.

 

[Dr.Peterson]

That is only part of a formula for the probability; and the formula given does not directly give the answer to your question (how many so the probability is 50%).

The formula they show is [MATH]Pr(r)=1−\sum_{d=0}^{\lfloor r/2\rfloor}\frac{n!r!}{n^r2^dd!(r−2d)!(n+d−r)!}[/MATH] where [MATH]n = 365[/MATH].

This gives the probability of at least 3 of r people sharing a birthday, by repeating the calculation you showed (the summand) for every number d from 0 through r/2, adding them all up, and subtracting from 1.

If you want the probability for 88 people, you would put that for r and do the summation. For only two sharing a birthday, you would use a different formula; the page says this is for three.

To answer the original question, you have to do this whole calculation for each possible value of r (1, 2, ...) until you get the probability you are looking for, namely at least 0.5.

(I haven't checked whether everything is correct.)

To my knowledge, something of this sort is necessary. The only solution I recall seeing for the problem is similar, but with a slightly less imposing formula.

 

[wolf]

Dr. Peterson, thanks for the quick reply!
So, for that formula I would use n = 365, r = 88 and for d I would use 1, 2, 3?

 

[Dr.Peterson]

This gives the probability of at least 3 of r people sharing a birthday, by repeating the calculation you showed (the summand) for every number d from 0 through r/2, adding them all up, and subtracting from 1.

Dr. Peterson, thanks for the quick reply!
So, for that formula I would use n = 365, r = 88 and for d I would use 1, 2, 3?

You would add up the results for d = 0, 1, 2, 3, 4, 5, ... up to 44.

I'd use Excel, or something similar. (And, again, I don't guarantee it's right; I'm just telling you what they're saying.)

 

[wolf]

I found yet another equation here: https://www.math.uchicago.edu/~fcale/CCC/DC.pdf

That formula works okay but it starts failing when the number of people = 6
It calculates 459 people are needed when the correct answer is 460.
Does anyone have an equation or a procedure that works correctly all the time?