In this intriguing probability problem, your goal is to maximize your chance of winning a free movie ticket based on birthdays.

The Setup

  • A line forms at a movie theater.
  • The first person whose birthday matches someone earlier in line wins a free ticket.
  • You can choose your position in the line.
  • Everyone’s birthday is uniformly and independently distributed over 365 days.
  • You don’t know anyone else’s birthday.

Question:

Which position in line should you choose to maximize your chance of being the winner?

Key Insight

You want to be the first person whose birthday matches someone earlier in line. This is related to the classic birthday paradox, but with a twist: you win only if your birthday matches someone before you, and you are the first such person.

Probabilistic Structure

Let’s define what needs to happen for the person in position \( i \) to win:

  1. No repeats among the first \( i-1 \) people.
    The probability that all \( i-1 \) people have distinct birthdays is: \[ P(\text{no repeats}) = \frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{365 - (i-2)}{365} \]

  2. Person \( i \) matches one of those \( i-1 \) distinct birthdays.
    Since there are exactly \( i-1 \) distinct birthdays before them, the probability that person \( i \)’s birthday is one of them is simply: \[ P(\text{match}) = \frac{i-1}{365} \]

So we compute the total probability that position \( i \) wins:

\[ P(i \text{ wins}) = P(\text{no repeats}) \times P(\text{match}) \]

To maximize this, we check where \( P(i+1) > P(i) \). After simplifying the ratio \( P(i+1)/P(i) \), the probability is maximized at position:

\[ \boxed{20} \]

Conclusion

If you can choose your place in line, position 20 gives you the highest chance of being the first person with a matching birthday and winning the free ticket.

Reference