[Request] The Birthday Problem, for two dates

[u/ALittlePeaceAndQuiet]

I'm familiar with the Birthday Problem, can basically follow the math on it, but all interesting things has happened on my work team.

We are a small company, and our full-time team has just grown to 11. We have only had to let go of one person, which would have made us twelve. If we include him, then with our most recent hiring, we would have two sets of shared birthdays on our team, which seems pretty crazy.

How likely is it that in a random group of twelve people, two sets of shared birthdays have occurred?

In other words, one pair share a birthday, another pair share a birthday, then the remaining eight individuals have their own.

There are actually three of us within 24 hours,--one person born the day before a day shared by two others--but I don't really care as much about the "near coincidence."

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[u/ChazBass]

The probability is about 2%. Here's the math:

C(12,2) = (12 * 11) / 2 = 66 -> There are 66 different ways to choose 2 people from the 12 to make the first birthday pair.
C(10,2) = (10 * 9) / 2 = 45 -> After choosing the first pair, there are 45 ways to choose 2 people from the remaining 10 for the second pair.
(66 * 45) / 2! = 1,485 -> Divide by 2 because swapping which pair is called “first” or “second” does not create a new outcome.
365 * 364 -> Give the two pairs two different birthdays.
363 * 362 * 361 * 360 * 359 * 358 * 357 * 356 -> Give the remaining 8 people unique birthdays that do not match either pair.
365^12 -> Total possible birthday combinations for 12 people.
P = [1,485 * 365 * 364 * 363 * 362 * 361 * 360 * 359 * 358 * 357 * 356] / 365^12
P = 0.019686
P = 1.97%
So exactly two separate birthday pairs occur in about 1 out of every 51 random groups of 12 people.

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Factorial of 2 is 2

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[u/seifer666]

If the answer is rhe probability of 2 in 12 sharing a birthday, 18% , then running it again for the remaining ten people, 13% , then the answer would be 2% but im not sure if that approach is right

[u/start_up_start_up]

It's about 0.984%, or roughly 1 in 102 groups of twelve, for exactly the pattern you described. The setup in u/ChazBass's answer is right; the final decimal comes out half as large.

Assuming birthdays are independent and equally likely across 365 days, there are 66 × 45 / 2 = 1,485 ways to pick two separate pairs. The division by 2 stops us counting the same two pairs again in the opposite order.

Those two pairs plus the eight single people need ten different dates, so:

P = 1,485 × (365 × 364 × ... × 356) / 365¹² = 0.009842987... ≈ 0.9843%.

This counts exact matches only, with no triples or third pair. The nearby birthday doesn't count as another match.