Birthday Probabilities

DCJ311 · #1 10-04-2005, 03:16 AM

Let's say you take 23 people and ask them for their birthdays (month and date). What is the probability that 2 of the people have the same birthday? Let's assume a person has a 1 in 365.25 chance of being born on any given date..

If you can do that, change the total # of people in the group to 30 and calculate the probability.

Luzion · #2 10-04-2005, 04:09 AM

[ QUOTE ]
Let's say you take 23 people and ask them for their birthdays (month and date). What is the probability that 2 of the people have the same birthday? Let's assume a person has a 1 in 365.25 chance of being born on any given date..

If you can do that, change the total # of people in the group to 30 and calculate the probability.

[/ QUOTE ]

This is a classic problem that they put on introductory courses in probability.

For 23 people, the chance that NO ONE shares a birthday is (365/365)(364/365)(363/365)(362/365)(361/365)......(343/365) or simply P(365,23)/365^23. I dont have a calculator around with me, but I recall the answer was approx 0.50. Thus the probability that at LEAST one person shares a birthday with another was also approximately 0.50.

If you want to find the probability of 30 people, do the same thing with P(365,30)/365^30 to find the probability NO ONE shares a birthday. To find the probability at least ONE PERSON shares a birthday, subtract your answer from 1. As in 1 - P(365,30)/365^30.

Little Fishy · #3 10-04-2005, 07:15 AM

this is such a common problem

for 23: 1-(364.25/365.25)(363.25/365.25)(362.25/365.25)*...((365.25-22)/365.25)

for 30 it's the same idea... there are a lot of differnt ways to do this problem but i think this is the most straight foreward and easiest to calcualte for some1 without a background in probability