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:
$$P(A)+P(B)+P(C)=\frac{380}{729} \approx 0.52$$
It certainly changes things. Say $n=5$, then we know at least one of the possible pairs have the same birthday. There are $10$ pairs, and each is equally likely to have the same birthday, so the probability that it is Adam and Eve who have the same birthday is at least $\frac1{10}$, much more than the $\frac1{365}$ it would be without the extra information.
To get the exact value, use
$$P(X\mid Y)=\frac{P(X\text{ and }Y)}{P(Y)}.$$
Here $X$ is Adam and Eve having the same birthday, and $Y$ is some two people having the same birthday, so $P(X\text{ and }Y)=P(X)=\frac1{365}$.
Now you just need to find $P(Y)$. It is easier to calculate the probability that no two people have the same birthday, and subtract from $1$. (Hint: what is the probability that the first two have different birthdays? If they do, what is the probability the third person has a different birthday from both of them?)
Best Answer
Here are the first several results for the same day of the week: $$ \begin{align} 1-\frac77&=0&\text{$1$ person}&(0\%)\\ 1-\frac77\frac67&=\frac17&\text{$2$ people}&(14.29\%)\\ 1-\frac77\frac67\frac57&=\frac{19}{49}&\text{$3$ people}&(38.78\%)\\ 1-\frac77\frac67\frac57\frac47&=\frac{223}{343}&\text{$4$ people}&(65.01\%)\\ \end{align} $$