Birthday Paradox

Question on the 3-person version.

Thanks for the note, great question. Exactly as you guessed, enumerating the probabilities for 3 people is trickier. In the 2-person case, we can get the probability that nobody matches, and subtract this from 100%. (The remainder is the chance of at least one match.)

For 3 people, there’s all sorts of ways two people can match but third doesn’t. I had to google around, there’s a solution here:

There is a probability distribution that can model the scenario in questions (independent birthdays appearing) and we can see the chances that 3 such events happen at the same time. (Another way to look at it… as we walk through the calendar day by day, there’s a random chance someone will stand up and say “It’s my birthday!”. We want to know the chances at least 3 people stand up. It’s similar to the chances 3 or more machines will break on the same day in a factory.)

Inline image 1

That page works it out to about 83 or 84 people to get a 50% chance of a triple. The actual number is closer to 87 it seems, likely due to the imprecision of the probability scenario (you have exactly one birthday, not a 1/365 “random chance” of raising your hand on any given day… in that viewpoint, it’d be possible for you to raise your hand twice!).