The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # Practice Paper 4 Question 14

There is a queue for admissions to The Avengers team. You are offered free admission if you are the first candidate in the queue who shares a birthday with a candidate earlier in the queue. You have the power to sneak into any position in the queue undetected. Which position in the queue gives you the highest probability of getting the free admission?

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## Warm-up Questions

1. What is the probability that a randomly selected person shares a birthday with you? (Interesting read: The Birthday Problem)
2. How many different arrangements could 9 pool balls be in?
3. Find $$x$$ such that $$x^2 − x − 6 < 0.$$

## Hints

• Hint 1
Have you tried finding the probability that the $$k^{th}$$ person is the first birthday duplicate? Call this probability $$P(A_k).$$
• Hint 2
... by first finding the probability that all the previous $$k-1$$ people have different birthdays.
• Hint 3
What is the probability that the third person shares a birthday with a candidate earlier in the queue?
• Hint 4
Compare that with $$P(A_2)$$-the probability that the second person is the first birthday duplicate.
• Hint 5
From the comparison and the question statement, what can you say about the value of $$P(A_k)$$ as $$k$$ increases?
• Hint 6
... in particular, do you think $$P(A_k)$$ will increase indefinitely?
• Hint 7
Find the position $$k$$ that gives the highest probability of getting the free admission.
• Hint 8
... by finding $$k$$ such that $$P(A_k)>P(A_{k+1}).$$

## Solution

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.

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