Problem:
In a line to a movie theater, a free ticket is being offered. It is offered to the first person in line that has the same birthday as one of the members in front of him. What place in line maximizes your chances at getting the free ticket?
Solution:
The probability that person n gets the ticket is:
The probability that persons 1 through n - 1 don't have the same birthday times,
The probability that person n's birthday matches at least one person ahead of her.
We multiply both these numbers together since they both define the event in question and are independent, so we can use the "and" rule of probability.
To determine #1, start with n=3. The probability that the first two people don't have the same birthdays is just the probability that person 2 doesn't have person 1's birthday, which is 364/365.
If n=4, the probability that the first three people don't all have the same birthdays is the probability that person 2 doesn't have person 1's birthday and that person 3 doesn't have the same birthdays as person 1 or 2's. The first is just 364/365 by the above argument. The second is 363/365. Assuming person 2 and person 3 are independent from each other, the "and" part implies we can multiply the probabilities to get 364/365 x 363/365.
Continuing this logic, the probability that persons 1 through n -1 don't have the same birthdays is:
For #2, notice that the probability that j people have person n's birthday is a binomial random variable with success probability 1/365. So by the complement rule, the probability that person n's birthday matches at least one person ahead of her is 1 minus the probability that it doesn't match any of them, or:
which has a maximum when n=18.
This is an absolute classic problem testing whether you know how to deal with conditionally dependent sequences of random variables. If you're feeling rusty or need help developing strong fundamentals, learn more about our services and how we can help prepare you for a career in Quant Finance!
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