Question

what is the probability of 2 out of 23 people having the same birthday? answer key

Answer 1

search it up on quora.

Answer 2

This is usually stated to be "what is the probability of AT LEAST 2 out of 23 people having the same birthday?"

= 1 - prob(all have different birthday dates)

Start with any student, the prob that he will have a birthday that year
= 365/365 = 1
then the prob that the first 2 students will have different birthdays
= (365/365)(364/365)
the prob that the first 3 students will have different birthdays
= (365/365)(364/365)(363/365)
....
the prob that the 23 students will have different birthdays
= (365/365)(364/365)(363/365)...(343/365)
= (365!/342!)/365^23

Your standard calculator cannot handle such large numbers, but if we
do 23 divisions, we get ...
= .4927

prob(that at least 2 of 23 will have the same birthday)
= 1 -.4927 = .5073

Very patiently check my calculations to get .4927
I only did it once, might have hit a wrong key somewhere.
The concept is correct.

we could try this using a high level "calculator" such as Wolfram
let k = (365! - 342!)/365^23
ln k = ln [ (365! - 342!)/365^23 ]
= ln 365! - ln 342! - 23ln 365

Wolfram gave me:
ln 365! = 1792.33
ln 343! = 1657.34
23ln365 = 135.698
ln k = ln 365! - ln 342! - 23ln 365 = 1792.33-1657.34-135.698 = -.708
k = .4926 , hey, almost dead-on both ways

prob(that at least 2 of 23 will have the same birthday)
= 1 -.4926 = .5074 , off by .0001 from the other answer

How about just say: 50.7%

Wow, that was fun!!!