Section 12 An analytical solution
Returning to the case of 20 people in a non-leap year, we can compute the probability analytically via
\[ P(\mbox{at least two birthdays the same}) = 1 - P(\mbox{all 20 birthdays different}) \]
12.1 Total number of possibilities
To compute \(P(\mbox{all 20 birthdays different})\), first count the total number of possibilities (regardless of whether any birthdays are the same or not):
- there are 365 possible days for person 1
- for each possible day for person 1, there are 365 possible days for person 2…
- so there are \(365 \times 365\) possibilities for persons 1 and 2 combined.
Extending this argument, the total number of possibilities must be \(365 \times 365 \times 365 \times \ldots \times 365\), i.e. \(365^{20}\).
12.2 Number of possibilities where all birthdays are different
- person 1 can have a birthday on any day;
- person 2 must have a birthday on a different day: there are 364 possibilities for person 2, and \(365 \times 364\) possibilities for persons 1 and 2 combined;
- person 3 must have a birthday on different days to persons 1 and 2: there are 363 possibilities.
Extending this argument, the total number of possibilities must be \(365\times 364 \times 363 \times \ldots \times 346\).
Exercise 12.1 Using the analytical solution.
- Compute the probability that at least two birthdays are the same
- use the
:to create a sequence of integers - use the
prod()command to multiply the elements of a vector together
- use the
- Compare your answer with the Monte Carlo estimate.