Probability

Question

Probability that in a room of 400 people, at least 2 share a birthday.

Answer 1

You may either [A] Calculate this or [B] Think.

[A] Calculate:

(1) In the room, there are N people (no twins) who have independent birthdays.

 (2) Assume the 365 days in each year have an equal (uniform) probability that each day could be the birthdate for a person  (p=1/365 for each day).

 

(3) Now, restate the problem from the perspective of the calendar dates.  As you select people from the room, how likely is it that the remaining people in the room have unique birthdates (note: the probability that at least two of them share a birthdate is 100% minus this value).

 

 Solution:

 

For each person selected (and date eliminated), the number of open dates is one less.  So, for only two people in a room, the probability that they share a birthdate is (1 – probability of unique birthdays)

         P1 = 1 - (365/365)*(364/365).

 

For just three people in a room, the probability that at least two share a birthdate is

         P2 = 1 -  (365/365)*(364/365)*(363/365).

. . .

For 25 people in a room, the probability that at least two of them share a birthdate is

         P25 =  1  -  (365/365)*(364/365)*(363/365)*(362/365)* … *(341/365)    [note:  25 ratios]

         P25 = 0.5687

 

The 50% probability is first exceeded with only 23 people in the room (P=0.5073).  Perhaps surprisingly, 99% probability is achieved with only 57 people in the room.  When the room holds 366 people, there must be a duplicate birthday!

 

[B] Think

 

Wait a minute! I just said that "When there are at least 366 people, there must be a duplicate birthday!"

 

P=1