In a room with 96 people, there’s a one in a million chance you all have different birthdays. Why? 365!/((365-96)!365^96) = 1.06 x 10^(-6).— Brian Greene (@bgreene) January 15, 2014
To put it another way, if you have roughly 100 people in a room, it is virtually guaranteed that there will be two people in that room who share a birthday.
Greene's tweet reminded me of something daredevil Even Knievel once said in an interview:
One of Slim's favorite tricks was to bet that two of any 25 people chosen at random would have the same birthday. He always won that bet—the math was huge in his favor.
The "Slim" Knievel was talking about was legendary gambler Amarillo Slim. In his own book, Amarillo Slim in a World Full of Fat People, Slim said he tried for a room full of 30 people, not 25. That would definitely make the math even huger in his favor.
It's called the Birthday Paradox, and it's termed a paradox because it's counterintuitive and, on the surface, doesn't seem to make sense. Let's restate Greene's tweet in the manner that Knievel and Slim knew the Birthday Paradox: In any gathering of 30 random people, there is a 70.6-percent probability that two of them will have the same birthday.
With just 10 random people, the probability is only 11.7 percent that any two of them will share a birthday. At 23, it's 50-50. At 30 people, as noted, it's 70.6-percent. But at 50 people the odds go up to 97.0-percent that some two people in that room will share a birthday, and the odds reach close to certainty at 95 people.
Again: If there are 30 people in a room, there is a 70.6-percent likelihood that someone in that room with have the same birthday as someone else in that room.
If you really want to dive into the math, you can do so here. But more fun would be next time you're at a gathering of 30 more people, make a bet with someone that there are two people in that gathering who share a birthday. You'll win far, far more often than not.