Download the BIRTHDAY problem

http://people.bath.ac.uk/clp20/thebirthdayproblem.html
the BIRTHDAY problem
"How many people must be in a room before the probabilty that some share a
birthday, ignoring the year and ignoring leap days, becomes at least 50 percent?"
Above we have the Birthday Problem question. It was first proposed in 1939 by a
mathematician called Richard von Mises. Today, this is one of the most explored
probability problems in the classroom. The birthday problem is one of the most famous
problems in combinatorial probability. The problem is famous, in part, because the
answer is a bit suprising.
The classic statement
What is the probability that, in a group of n people, two people have the same
birthday?......
In a group of 400 people, you are guaranteed that two people will have the same birthday
(probability=1), because there are only 365 possible birthdays.
The phrase "have the same birthday" means celebrate their birthdays on the same day,
i.e. we ignore the year. So a person born on April 4th 1986 has the same birthday as a
person born April 4th 1966.
Assumptions
When we deal with this problem we make some assumptions:


only 365 days each year (ignoring leap years)
birthdays are evenly distributed throughout the year, so that each day of the year
has an equal chance of being someone's birthday.
Selected solutions
Below we have a table that shows the probability of a pair of matched birthdays for
different group sizes.
Group Size Probability
2
<1%
5
3%
10
12%
23
51%
30
71%
50
97%
http://people.bath.ac.uk/clp20/thebirthdayproblem.html
60
<99%
Converse probabilities are used to work out the infamous birthday problem.
Now I'm going to show how to work through this problem to get the solutions we have in
the table above.
Take group of 5 people ...
Let B = event no one in the group shares the same birthday
Let Bc = event at least 2 people share same birthday
P(B) = (365/365)(364/365)(363/365)(362/365)(361/365) = 0.973
P(Bc)= 1 - 0.973 = 0.027
That is, about a 3% chance that in a group of 5 people at least 2 people share the same
birthday.
Take group of 23 people ...
Let B = event no one in the group shares the same birthday
Let Bc = event at least 2 people share same birthday
P(B) = (365/365)(364/365)......(343/365) = 0.493
P(Bc)= 1 - 0.493 = 0.507
That is, about a 50% chance that in a group of 5 people at least 2 people share the same
birthday.
Take group of 50 people ...
Let B = event no one in the group shares the same birthday
Let Bc = event at least 2 people share same birthday
P(B) = (365/365)(364/365)......(316/365) = 0.03
P(Bc)= 1 - 0.03 = 0.97
That is, it's "virtually certain" that in a group of 5 people at least 2 people share the same
birthday.
http://people.bath.ac.uk/clp20/thebirthdayproblem.html
Many people refer to the birthday problem as the birthday paradox. We see that if there
are 23 people in the room then there is slghtly more than a 50% chance that at least 2
people share the same birthday. In a room with 60 or more people the probability that at
least 2 people share the same birthday is greater than 99%. It is a paradox in the sense
that it is a mathematical truth that contradicts common intuition (most people estimate the
chance of having 2 people with the same birthday in a group of 23 is much less than
50%).
General formula for calculating P(B)
To calculate the probability that in a a group of n people, at least 2 people share the same
birthday (P(Bc)), we will first calculate the probability that no one in the group shares the
same birthday P(B).
P(B) = (365/365)(364/365)(363/365)(362/365)... ((366-n)/365)
Then to calculate there is a match we subract the above calculation from 1.
The formula above can be written as the following:
The following graph shows the growth behaviour of this problem. We can see from the
graph that there must be at least 23 or more people in a room before the probability that
two people share a birthday becomes at least 50 percent.