Question:
Adam and Eve are in a room with $n − 2$ other people. Suppose you know that at least two of the people in the room celebrate their birthday on the same day. What is the probability that Adam and Eve celebrate their birthday on the same day? (Assume that a year has 365 days and that the distribution of births over a year is uniform.)
MyApproach:
I think it is knowing that at least two people have the same birthday that is confusing to me. Does it change anything to have this info?
$\endgroup$ 24 Answers
$\begingroup$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?)
$\endgroup$ 1 $\begingroup$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 $$P(B|S) = \frac{P(B\cap S)}{P(S)}.$$
$\endgroup$ $\begingroup$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 $P(B \cap A) = P(B)$.
$\endgroup$ 3 $\begingroup$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
$$\frac { \frac {1}{365}} {1 - \frac {365 \cdot 364 \cdot \ ... \ \cdot (365-(n-1))}{365^n}}$$
$$\frac { \frac {1}{365}} {1 - P(No \ one\ with \ same \ birthday)}$$
$$\frac { \frac {1}{365}} {P(At \ least \ two \ with \ same \ birthday)}$$
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