Birthday problem- Adam and Eve

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?)

Hint: Let $B$ be the event that Adam and Eve have the same birthday. Let $S$ be the event that two people have the same birthday. You are trying to compute the conditional probability $$P(B|S) = \frac{P(B\cap S)}{P(S)}.$$

Yes it does. As noted above, we are computing a conditional probability. Let $P(A')$ be the probability that no two people have the same birthday. Therefore, $$ P(A') = \frac{365}{365} \frac{364}{365} ... \frac{365 - (n-1)}{365} $$

Let $P(A)$ be the probability that at least two people have the same birthday, that is $P(A) = 1- P(A')$. ($A$ and $A'$ are complementary events.)

Let $P(B)$ be the probability that Adam and Eve share the same birthday. That is, $$ P(B) = \frac{365}{365} \frac{1}{365}$$

As noted above by the other answers, we are trying to find $$P(B|A) = \frac{P(B \cap A)}{P(A)}$$

So, we must find $P(B \cap A)$. Suppose Adam and Eve have the same birthday. The event $B$ is fully contained in $A$, hence $P(B \cap A) = P(B)$.

The total number of combinations T is

$$T=365^n$$

The number of combinations where no pair has the same birthday, i.e. invalid combinations I is

$$I=365 \cdot 364 \cdot \ ... \ \cdot (365-(n-1))$$

So the number of valid (i.e. at least one pair with same birthday) combinations V is

$$V=T-I = 365^n - 365 \cdot 364 \cdot \ ... \ \cdot (365-(n-1))$$

The number of correct combinations C where Adam and Eve have the same birthday

$$C=365 * 365^{n-2} = 365^{n-1}$$

So now the probability is:

$$\frac{C}{V} = \frac{C}{T-I}$$

or

$$\frac {365^{n-1}} {365^n - 365 \cdot 364 \cdot \ ... \ \cdot (365-(n-1))}$$

If desired this can be rewritten to

$$\frac { \frac {1}{365}} {1 - \frac {365 \cdot 364 \cdot \ ... \ \cdot (365-(n-1))}{365^n}}$$

$$\frac { \frac {1}{365}} {1 - P(No \ one\ with \ same \ birthday)}$$

$$\frac { \frac {1}{365}} {P(At \ least \ two \ with \ same \ birthday)}$$