There are a number of people in a group. We will assign the variable “n” to the amount of people in that group.

How likely is it that two people will share the same birthday?

But more importantly...

How many people do you need in a group so that the probability that two people will have the same birthday is 50%?

The guided questions below helped me find the solution:

The formula I came up with was:

365! / 365n (365-n)

I used factorial to more easily multiply numbers with each other, to learn what will give us close to 50%.

Mr. Tejeras walked us through how he got the answer after we came up with our own answers. His solution for the problem was the same one I got. He has the same exact formula and calculated that at 23 people there is a 50% chance that two people will have the same birthday. For the extension I had to come up with how many people you needed in a group to have a 99% probability that two will have the same birthday. The solution I got for that one was 70 people, which blew my mind. I really enjoyed working on this problem because it had a real interesting solution. It’s still hard to wrap my head around the solutions and how low a number of people it is. With a group of 23 people 50% of the time two people in the group will have the same birthday and in a group of 70 people 99% will have the same birthday.

How likely is it that two people will share the same birthday?

But more importantly...

How many people do you need in a group so that the probability that two people will have the same birthday is 50%?

The guided questions below helped me find the solution:

- How many birthdates are there in a regular year (not including leap year)?
- In a regular year, there are 365 possible birthdates.

- In a regular year, there are 365 possible birthdates.
- If my birthday is April 19th, and I meet a new person without the same birthday, how many other days of the year can they possibly have their birthday?
- A new person without my exact same birthday must have their birthday on 364 of the other days in the year.

- A new person without my exact same birthday must have their birthday on 364 of the other days in the year.
- What about a third person? On what other possible days can this third person have their birthday so that it is not on the same as the me or the second person?
- Following the pattern, the third person can have their birthday on 363 of the 365 days in the year.

- Following the pattern, the third person can have their birthday on 363 of the 365 days in the year.

The formula I came up with was:

365! / 365n (365-n)

I used factorial to more easily multiply numbers with each other, to learn what will give us close to 50%.

Mr. Tejeras walked us through how he got the answer after we came up with our own answers. His solution for the problem was the same one I got. He has the same exact formula and calculated that at 23 people there is a 50% chance that two people will have the same birthday. For the extension I had to come up with how many people you needed in a group to have a 99% probability that two will have the same birthday. The solution I got for that one was 70 people, which blew my mind. I really enjoyed working on this problem because it had a real interesting solution. It’s still hard to wrap my head around the solutions and how low a number of people it is. With a group of 23 people 50% of the time two people in the group will have the same birthday and in a group of 70 people 99% will have the same birthday.