Confusion with two Birthday problems questions

I have been trying to understand the difference between Problem 1 and Problem 3 in the Four Birthday Problems in Week 1.

Problem 1 is worded as: given a pre-defined date, what is the value of `n` such that the probability of having a match is greater than or equal to 0.5?

Problem 3: given a classroom with students, what is the value of `n` such that the probability of having a match is greater than or equal to 0.5 for any two students?

I get the Problem 3 is including the ‘for any two students’ but isn’t that implied in Problem 1 as well? Since I am a not able to properly understand the distinction between the two it’s impeding my understanding of the analytical solutions as well.

I apologize if this is a poor question but would love to know about an approach that sheds more light into the distinction between the two

Thank you!

Hi @Ritesh1227,

Firstly, welcome to our community! Sor for the late response.

There is no need to apologize, both three problems look quite similar indeed.

The difference between problems 1 and 2 is essentialy what you’ve stated. In problem 1, you have fixed a date and you are comparing it with every student in the classroom.

Problem 3 however, there is no fixed date, but you are looking for (at least) two students that match their birthday.

Problem 1 is a bit simpler in the sense that you must compare all students with one fixed date, whereas in problem 3 the comparsion occurs within the classroom.

I hope that this makes everything clear. If not, please let me know.

Thanks,
Lucas

@Ritesh1227

Assumption: We will give names to the two Calsses as Class A and Class B.

In problem 4, First: we choose a student’s birthday independently from class A such that the probability of choosing one day is

``````                         (1-1/365)
``````

as in problem 3. Then we use that day as a pre-defined birthday D1. We compare the birthday D1 with student’s birthdays in class B, as in problem 1, such that the probability is

``````                          (1-1/365)^n for each birthday D1 as in problem 1
``````

Second: We choose another birthday from class A such that the probability of choosing another one day D2 is

``````                           (1-1/365)
``````

then we compare day D2 with the student’s birthdays from class B such that the probability is

``````                           (1-1/365)^n   for each birthday D2 as in problem 1
``````

We repeat this for all birthdays in class A and compare them with all birthdays from class B. The total probability is

``````                       (1-1/365)^n x (1-1/365)^n = (1-1/365)^(n^2)
``````