Computing Conditional Probability on Birthday Problem

Question

I have a question concerning the birthday problem in conditional problem. Say there's a given group of $12$ people and a regular year with only $365$ days. Then the probability of no duplicate will be $\frac{365!}{365^{12}353!} = 0.833$. I am asked to compute given that the first twelve don't have duplicates, the probability that there will be a duplicate after the next $8$ people enter the group.

From conditional probability $P(A|B)=\frac{P(A \cap B)}{P(B)}$, I already computed $P(B) = 0.833$. Is $P(A \cap B)$ which is P(duplicate in next 8 and no duplicate in first 12) simply equal to $1 - \frac{353!}{365^{20}345!}$?

Answer 1

• The probability of no duplicate in the first twelve would be $\dfrac{365!}{365^{12} 353!}$
• The conditional probability of at least one duplicate in the next eight given no duplicate in the first twelve would be $1-\dfrac{353!}{365^{8} 345!}$
• The probability of no duplicate in the first twelve and at least one in the next eight would be $\dfrac{365!}{365^{12} 353!}\left(1-\dfrac{353!}{365^{8} 345!}\right)$

A slightly different approach with the same answer:

• The probability of no duplicate in the first twelve would be $\dfrac{365!}{365^{12} 353!}$
• The probability of no duplicate in the first twenty would be $\dfrac{365!}{365^{20} 345!}$
• The probability of no duplicate in the first twelve and at least one in the next eight would be $\dfrac{365!}{365^{12} 353!} - \dfrac{365!}{365^{20} 345!}$