
 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
$k1$
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$k1$
people have different birthdays and$k^{th}$
is a duplicate, i.e.$P(A_k)=\frac{k1}{N}\prod_{i=0}^{k2} \frac{Ni}{N},$
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$\frac{k1}{N}>\frac{Nk}{N}\cdot\frac{k}{N}$
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.