How many people should be gathered in the same room before it is more likely?

Question

than not that two people share the same birthday?

Answer 1

23 is 50%

Answer 2

Not very many. To calculate this mathmatically, the probability that two people do not share the same birthday is 364/365 (let's ignore leap years), or 99.726%. However, every person in the room has a chance to pair with every other person, so the probability that nobody shares a birthday is going to be .99726 to the power of the number of connections in the room. The breakeven point will be between 252 and 253 connections. For n number of people, the total number of connections will be 1+2+3...+(n-1), or (n-1)*n/2. Once you have 23 people in the room there will be 253 connections between them, and you will be slightly more likely to have someone share a birthday than not. If you increase the number of people to 42, now you have 90% chance that two people share a birthday. Up it to 71 people, and now there's a 99.9% chance that two people share the same day.

Answer 3

Assuming a random distribution of birthdays, the answer is 23.

Let's call the people Andrew, Barbara, Charles, ..., Vanessa(22), and William(23).

Start with A by himself.
Add B. She has a 364/365 chance of mismatching A.
Add C. He has a 364*363/(365*365) of mismatching both.
Add D. She has a 364*363*362/(365^3) of mismatching all 3.
...
Add V. She has a 364*...*343/(365^22) of mismatching all 21.
Add W. He has a 364*...*342/(365^23) of mismatching all 22.

Excel makes it easy to set up the calculation:
1--: A1 = 1; B1 = 365; C1=1.
2--: A2 =A1+1; B2 =B1-1; C2 =C1*B2/$B$1.
3--: Drag row 2 downwards to populate two dozen more rows.

Results*:
|--: With just Andrew (1 person), it's 100% certain he matches nobody.
|--: With A and B (2 people), it's 99.73% likely they both mismatch.
|--: With A,B,C (3 people), it's 99.18% likely they all mismatch.
...
|--: With A-V (22 people), it's 52.43% likely they all mismatch.
|--: With A-W (23 people), it's 49.27% likely they all mismatch.
* - Assumes 365 birthdays, each 1/365 likely.


Ultraprecise results*:
|--: With A-V (22 people), it's 52.45% likely they all mismatch.
|--: With A-W (23 people), it's 49.29% likely they all mismatch.
* - Assumes 365 birthdays each 4/1461 likely and a 366th birthday that is 1/1461 likely.

Answer 4

23
To figure it out, compute the chances of NO ONE having the same birthday.
The chance that a second person does not have the same birthday as the first is 364365.
The chance that the third does not have either the same birthday as #1 or #2 is 363/365
for "N" people, it is
364*363*362*361.... *(365-(n-1))/(365^(n-1))
This, for N=21, is .52, and for N=22, .49

Answer 5

366

Answer 6

I know 3 people that share my birthday randomly..

Answer 7

28.