**Matching Birthdays**

(Adapted
from *Statistics: The Art and Science of Learning from *by Chris* Franklin’s Statistics. Section 5.4, page
229.*)

** **

** **

On average, there
are 365 days in a given year (excluding leap years). What is the probability that one of your classmates has the
same birthday? Since there are 365
days in a year, our intuition tells us that there is probably a small chance
that you share a birthday with another student in your class.

**Question for students:**

Let’s
say that there are 25 students total in your class. What is the probability that *at least* two students share the same birthday?

**Working through it:**

** **Keep in mind that the probability of *at least* one match includes 1 match, 2 matches, 3 matches,
etc. Evaluating each probability
would take a lot of work. To make
it simpler, we can find the probability of *at least* one match by finding the complement probability of *no* matches.
Note:

P(no matches) + P(at least one match) = 1

P(at least one match) = 1 – P(no matches)

·
Suppose the class only
has 2 students. The first
student’s birthday could be any of the 365 days in a year. So the chance that the second student
has a *different* birthday is
364/365. From our note above, the
probability that the two students have the same birthday is,

·
Suppose the class has 3
students. The probability that all
three have *different* birthdays is

P(no matches) = P(students 1, 2, and 3 have different
birthdays)

= P(students 1 and 2 are different) x P(student 3 is
different|students 1 and 2 are different)

The
probability that student 3 has a different birthday from students 1 and 2 is
363/365 since there are 363 days left that are different from student 1 and 2.

·
We can see that the
method follows when we add more students to our classroom. Now consider 25 students in your
class. Following the same method,
by the time you arrive at the 25^{th} student, to have a different
birthday he/she will have 24 less days to choose from out of 365. So the probability of the 25^{th}
student having a different birthday from students 1 through 24 is 341/365.

P(no matches) = P(students 1 and 2 and 3 and 4… and 25
have different birthdays)

The
product of these probabilities equals 0.43. Now, recall that we can easily find the probability of at
least one match by using the complement.

P(at least one match) = 1 – P(no matches)

P(at least one match) = 1 – 0.43

P(at least one match) = 0.57

**Conclusion:**

Wow! This probability is greater than
½! What do you think? Is this a higher chance than you
expected? It may help to consider
the idea of the complement: with a
class of 25 students, there are 300 possible pairs of students who can share
the same birthday. And we know
that if we have a large number of possible opportunities for something to
happen, then it is highly likely that it will happen.

**Extension for Teacher:**

Have
your students go around and interview each of their classmates to find
everyone’s birthday. Once everyone
is finished, ask the class if they found any students who had matching
birthdays. Or, announce the
concept of the activity to the class.
Then starting on one side of the room, ask each student to state their
birthday. See how far across the
room you can get until you (possibly) find a match. They may be surprised!
Take note of your students’ predictions and reactions.

*Note: This works best when you have a class
of at least 23 students. For a
class of at least 23 students, the probability that there is at least one match
is greater than ½.*

References

Agresti, Alan and Franklin, Christine A. (2007). *Statistics:
The art and science of learning*.
Upper Saddle River, NJ. Pearson Prentice Hall.