How would I calculate the probability of atleast 2 people having the same birthday given a group of 3 using counting principles. I know that using P(same bday) = 1 - P(not same bday). I calculated this to be $1- \frac{365!}{(365-3)!365^3}$. Is there a way to find P(same bday) using combinatorics principles only?
2026-04-12 07:43:08.1775979788
Birthday Problem Probability
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yes of course it's possible!
HINT
start from $365^3$ overall possibilities and then evaluate all the possible favourable cases
EG
three distinct birthdays = ${365}\choose3$
same birthdays for three = ${365}\choose1$
etc.