# Random Variable - Birthday Problem

1. How many people are needed so that the probability that at least two of them were born on the same day of the week is at least 1/2? (Assume that the days of the week are equally likely to be the birthday of each person.)
2. How many people are needed so that the probability that at least two of them were born in the same month of the year is at least 3/4? (Assume that the months are equally likely to be the birth month of each person.)
3. How many people, none of whom was born on February 29, are needed so that the probability that at least two of them were born on the same day of the year is at least 7/8? (Assume the days of the year are equally likely to be the birthday of each person.)

The steps to follow are :
Calculate the probability of $$P_n$$ for all different birthdays.
The book suggests that this is calculated by the formula $$\frac{365}{366}.\frac{364}{366}...\frac{367-n}{366}$$
This value is very hard to calculate.
So, the probability among $$n$$ people having the same birthday is : $$1 - P_n = 1 - \frac{365}{366}.\frac{364}{366}...\frac{367-n}{366}$$
Since I got all the answers wrong, I am trying to reverse engineer the solution to get the value of $$X$$ which is $$1 - P_n = X * \frac{367 - n}{366}$$

I searched this site and found that for question 1 (above) we have to consider 7 days instead of the 366 used in the bigger problem.
My question is how do we arrive at $$n = 4$$
$$1 - P_n = X * \frac{8 - n}{7}$$

I have tried $$1 - P_n = 1 - \frac{6}{7}*\frac{5}{7}*\frac{4}{7}*\frac{3}{7}*\frac{2}{7}*\frac{1}{7}*\frac{8-n}{7}$$

Please help me with one solution, for the first one, I will reverse engineer and understand this for the other two.

• The first thing to do is to actually think about what you are copying down from somewhere else. Note that February $29$ birthdays have been specifically excluded. If you do not have to account for leap year days, how many different birthdays are possible in a year? It is not $366$. Also, where did you get $367-n$? That would mean that the first person (where you have $365$ in the numerator) would correspond to $n=2$. Commented Dec 1, 2021 at 20:00

The argument goes that $$3$$ is not enough to have the probability that at least two of them were born on the same day of the week at least $$\frac12$$, but $$4$$ is, since
• $$1-\frac77\times\frac67\times\frac57\qquad \approx 0.39 <\frac12 \text{ when } n=3$$
• $$1-\frac77\times\frac67\times\frac57\times\frac47 \approx 0.65 \ge\frac12 \text{ when } n=4$$
Your $$P_n = \frac{7-(n-1)}{7} P_{n-1}= \frac{8-n}{7} P_{n-1}=\frac{7!}{(7-n)!\,7^n}$$ gives the probability that no two of them were born on the same day of the week, obviously starting with $$P_0=P_1=1$$, though the question asks for $$1-P_n$$