The Birthday Problem
The Problem
• In a group of 50 students, what is the
probability that at least two students share
the same birthday?
Assumptions
• Only 365 days each year.
• Birthdays are evenly distributed throughout
the year, so that each day of the year has an
equal chance of being someone’s birthday.
Note: The accuracy of the probability calculations that we
do here depends on whether or not these
assumptions are valid.
Take group of 5 people….
Let A = event no one in group shares same birthday.
Then AC = event at least 2 people share same birthday.
Assuming independence and using classical approach:
P(A) = 365/365 × 364/365 × 363/365 × 362/365 × 361/365
= 0.973
So, then P(AC) = 1 - 0.973 = 0.027
That is, about a 3% chance that in a group of 5 people at
least two people share the same birthday.
Take group of 23 people….
Let A = event no one in group shares same birthday.
Then AC = event at least 2 people share same birthday.
Assuming independence and using classical approach:
P(A) = 365/365 × 364/365 × … × 343/365
= 0.493
So, then P(AC) = 1 - 0.493 = 0.507
That is, about a 51% chance that in a group of 23 people at
least two people share the same birthday.
Take group of 50 people….
Let A = event no one in group shares same birthday.
Then AC = event at least 2 people share same birthday.
Assuming independence and using classical approach:
P(A) = 365/365 × 364/365 × … × 316/365
= 0.03
P(AC) = 1 - 0.03 = 0.97
That is, about a 97% chance that in a group of 50 people at
least two people share the same birthday.