why in a room with 23 people there’s a good chance 2 of them share a birthday
In: Mathematics
Because for any probability we might call “good”, there exists a number X of congregated people that will have a “good” chance of containing at least one pair sharing a birthday.
And since we’re calling 50% our “good” chance, that number is just 23.
Now, what is that probability exactly? It’s a sum of probabilities:
* that all 23 share, plus
* 22 sharing, plus
* 21 sharing, plus
* 21 sharing (and the other 2 form a pair), plus
* 20 sharing, plus
* 20 sharing (and another 2 form a pair), plus
* 20 sharing (and another 3 form a pair)
* …
* only 2 share
Are you tired yet? I am. A lot of times in statistics you might bruteforce a probability by listing every event that would meet the condition. There’s a lot of them. Instead, let’s find the counterpart: the probability that **none** will have the same birthday. This is easier.
The probability that *two* people have different birthdays is 364/365.
If it is given that two people have different birthdays, the probability that a third shares with neither is 363/365.
And the probability that adding up to 23 people, none will share a birthday with the previous, or anyone before that, is of 343, which is (365-(23-1)):
364/365×363/365×362/365×…×343/365
And that’s equal to:
364×363×362×…×343/365^22
Which is [0.4927](https://www.google.com/search?client=firefox-b-1-d&q=364*363*362*361*360*359*358*357*356*355*354*353*352*351*350*349*348*347*346*345*344*343%2F365%5E22).
That is the probability that none will have the same birthday. And to subtract that probability from one tells us the chance that at least two will.
0.50729723432
Or 50.7 percent.
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