Other days of the week - Birthday Paradox

Question

I'm trying to solve a birthday paradox problem, I'm really close to the answer but I guess I'm doing something wrong. The problem states that the probability of being born on a friday is $\frac{1}{3}$ and is equally probable to any other day of the week. Assuming 4 (or any number n people) where selected, what is the probability to have for example two born on friday and two born on other days of the week?

I know that the probability of two people sharing the same birthday is: $$\frac{(365*364*363*362)}{365^n}$$ Another note I was able to come to is that the chance of last two not sharing birthdays with the first two is: $$1-\frac{5}{7}*\frac{4}{7} \approx 0.591$$

Assuming that a person can be born on a Friday with probability 1/3 and with equal probability any other day of the week. What is the probability that among 4 randomly selected people, two were born on the same day and the other two in two other days?

The correct answer according to my exercise sheet is about $0.52$. Can anyone point me in the right direction?

Answer 1

It is of course unrealistic that the chance of a person born on a Friday is $\frac{1}{3}$, but so be it. We are also told that for all other days of the week the chance is the same, so that would be

$$\frac{1-\frac{1}{3}}{6}=\frac{1}{9}$$ (e.g. the chance of being born on a Wednesday is $\frac{1}{9}$)

Now, there are various ways to have 2 people being born on the same day, and the other two on 2 other days:

$A$. Two born on a Friday, and the other two on two different days other than Friday

$B$. Two born on the same day but other than a Friday, one born on Friday, and the last on a different day yet.

$C$. Two born on the same day but other than a Friday, and the other two on two different days other than Friday

$$P(A) = {4 \choose 2} \cdot \big( \frac{1}{3} \big)^2\cdot \frac{6}{9} \cdot \frac{5}{9} = \frac{180}{729}$$

(the $4 \choose 2$ is the number of ways to pick the two people born on a Friday)

$$P(B) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2\cdot 2 \cdot \frac{1}{3} \cdot \frac{5}{9} = \frac{120}{729}$$

(the extra $2$ term is to distinguish the two people: one born on Friday, the other not. The $6$ is to pick one of the 6 days that the first two are born on)

$$P(C) = {4 \choose 2} \cdot 6 \cdot \big( \frac{1}{9} \big)^2 \cdot \frac{5}{9} \cdot \frac{4}{9} = \frac{80}{729}$$

Hence, the chance that 2 people being born on the same day, and the other two on 2 other days is:

$$P(A)+P(B)+P(C)=\frac{380}{729} \approx 0.52$$

Answer 2

First of all, you should not involve number of days in a year, because there is no assumption about uniform distribution on each days of a year.

In fact, since the assumption is evenly distributed across a week, it is slightly not evenly distributed across days in a year

$$\frac{\binom{4}{2} \times 1 \times 1 \times 6 \times 6}{7^4}$$

EDIT according to OP's edit

Notice that Friday has a different probability than other days.

$$\binom{4}{2}\times \frac{1}{3}\times \frac{1}{3} \times \frac{6}{9} \times \frac{5}{9} $$ $$+ \binom{4}{2} \times \binom{7-1}{1} \times \frac{1}{9} \times \frac{1}{9} \times \binom{2}{1} \times \frac{1}{3} \times \frac{5}{9}$$ $$+\binom{4}{2} \times \binom{7-1}{1} \times \frac{1}{9} \times \frac{1}{9} \times \frac{5}{9} \times \frac{4}{9}$$