-
- What is the probability that a randomly selected person shares a birthday with you? (Interesting read: The Birthday Problem)
- How many different arrangements could 9 pool balls be in?
- Find
$x$such that$x^2 − x − 6 < 0.$
-
Have you tried finding the probability that the
$k^{th}$person is the first birthday duplicate? Call this probability$P(A_k).$ -
… by first finding the probability that all the previous
$k-1$people have different birthdays. -
What is the probability that the third person shares a birthday with a candidate earlier in the queue?
-
Compare that with
$P(A_2)$-the probability that the second person is the first birthday duplicate. -
From the comparison and the question statement, what can you say about the value of
$P(A_k)$as$k$increases? -
… in particular, do you think
$P(A_k)$will increase indefinitely? -
Find the position
$k$that gives the highest probability of getting the free admission. -
… by finding
$k$such that$P(A_k)>P(A_{k+1}).$
-
Let
$A_k$be the event that the$k^{th}$person is the first birthday duplicate. The probability of$A_k$is the probability that all the previous$k-1$people have different birthdays and$k^{th}$is a duplicate, i.e.$P(A_k)=\frac{k-1}{N}\prod_{i=0}^{k-2} \frac{N-i}{N},$where$N=365$(or$366,$but as we’ll see it won’t change the result).It is easy to verify that
$P(A_2)<P(A_3)$and clearly$P(A_{366})>P(A_{367})=0$so there must exist a$k$for which$P(A_k)>P(A_{k+1})$has a positive real solution. Substituting, we get$\frac{k-1}{N}>\frac{N-k}{N}\cdot\frac{k}{N}$which gives$k^2>N.$Solving we get$k>\sqrt{365}$where we know$365=19^2+4$and hence$k>19,$so the smallest value is$k=20.$Note: The fully correct answer in fact is: At the end of the queue if the length of the queue is less than or equal to 20, or at position 20.