# Probability and the Birthday Paradox

A mysterious math problem from Science Buddies

Image: George Retseck

Key concepts
Mathematics
Probability
Statistics

Introduction
Have you ever noticed how sometimes what seems logical turns out to be proved false with a little math? For instance, how many people do you think it would take to survey, on average, to find two people who share the same birthday? Due to probability, sometimes an event is more likely to occur than we believe it to. In this case, if you survey a random group of just 23 people there is actually about a 50–50 chance that two of them will have the same birthday. This is known as the birthday paradox. Don't believe it's true? You can test it and see mathematical probability in action!

Background
The birthday paradox, also known as the birthday problem, states that in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday. Is this really true? There are multiple reasons why this seems like a paradox. One is that when in a room with 22 other people, if a person compares his or her birthday with the birthdays of the other people it would make for only 22 comparisons—only 22 chances for people to share the same birthday.

But when all 23 birthdays are compared against each other, it makes for much more than 22 comparisons. How much more? Well, the first person has 22 comparisons to make, but the second person was already compared to the first person, so there are only 21 comparisons to make. The third person then has 20 comparisons, the fourth person has 19 and so on. If you add up all possible comparisons (22 + 21 + 20 + 19 + … +1) the sum is 253 comparisons, or combinations. Consequently, each group of 23 people involves 253 comparisons, or 253 chances for matching birthdays.

Materials
•    Groups of 23 or more people (10 to 12 such groups) or a source with random birthdays (see Preparation below for tips)
•    Paper and pen or pencil
•    Calculator (optional)

Preparation
•    Collect birthdays for random groups of 23 or more people. Ideally you should get 10 to 12 groups of 23 or more people so you have enough different groups to compare. (You don't need the year for the birthdays, just the month and day.)
•    Tip: Here are a few ways that you can find a number of randomly grouped people: Ask school teachers to pass a list around each of their classes to collect the birthdays for students in the class (most schools have around 25 students in a class); use the birthdays of players on major league baseball teams (this information can easily be found on the Internet); or use the birthdays of other random people using online sources.

Procedure
•    For each group of 23 or more birthdays that you collected, sort through them to see if there are any birthday matches in each group.
•    How many of your groups have two or more people with the same birthday? Based on the birthday paradox, how many groups would you expect to find that have two people with the same birthday? Does the birthday paradox hold true?
•    Extra: In this activity you used a group of 23 or more people, but you could try it using bigger groups. If you use a group of 366 people—the greatest number of days a year can have—the odds that two people have the same birthday are 100 percent (excluding February 29 leap year birthdays), but what do you think the odds are in a group of 60 or 75 people?
•    Extra: Rolling dice is a great way to investigate probability. You could try rolling three 10-sided dice and five six-sided dice 100 times each and record the results of each roll. Calculate the mathematical probability of getting a sum higher than 18 for each combination of dice when rolling them 100 times. (This Web site can teach you how to calculate probability: Probability Central from Oracle ThinkQuest.) Which combination has a higher mathematical probability, and was this true when you rolled them?

View
1. 1. il--ya 02:05 PM 3/29/12

This statement is wrong:
"Every one of the 253 combinations has the same odds, 99.726027 percent, of not being a match."
Correct formula for probability of NOT having two persons in a group of 23 to share birthday (excluding 29 Feb):
P = 365! / ((365-23)! * 365^23)
Or, in other words:
P = (365/365)*(364/365)*(363/365)*...*(343*365) ≈ 0.492703 = 49.2703%, which is slightly less than 49.952%.
Simple explanation is that when we do a survey, first person can have any birthday (365/365); second person can have any birthday apart from first person's birthday, so 364 choices, hence (364/365); third person can have any birthday, apart from first and second persons' birthday, (363/365), etc.

2. 2. em_allways_right 07:01 PM 3/29/12

With your reasoning nobody would have the same birthday. You are assuming no replacement, like when you remove a card from a deck of cards. Birthdays have replacement.

3. 3. rafman 07:28 AM 3/30/12

Actually, il--ya is absolutely correct. The author's formula is close, but not technically correct. The problem is that birthdays are not like coin flips. The chances of individual pairs having the same birthday are not independent (if person 1 = person 3, and person 3= person 5, then person 1 = person 5). The practical difference is not very much as there is not much of a probability of such multiple matches among 23 people, but it exists.

Look at what happens if use 366 people and the author's formula. The result says there is a non-zero probability of them having different birthdays . . . a non-sensical result.

In problems of this sort, il--ya's approach is usually the correct: calculate the opposite and subtract from 1. This works because the answer you are trying to find and the opposite of that answer combine to equal 100% of all outcomes.

4. 4. kiteman 03:53 PM 3/31/12

I've never met anyone who has the same birthday as me. Is this unusual? My birthday is 31st. June.Just joking!

5. 5. kevin_neilson 05:51 PM 3/31/12

"If you use a group of 366 people—the greatest number of days a year can have—the odds that two people have the same birthday are 100 percent."

No comment.

6. 6. geojellyroll 02:08 AM 4/5/12

The author is correct.

Think of one person in a roonm and a second person entering...the odds are 365 to one that they have the same birthday. A third person enters and the odds halve...then a fourth...etc. Each added person is a variable for the next person to match. Odds are not one-on-one but collective.

7. 7. il--ya 08:50 PM 4/5/12

geojellyroll, OK, to keep it simple, let's imagine that we live on a planet where a year lasts for only 3 days. Now let's calculate the probability of the event that three random persons have all different birthdays, using authors method:
For 3 persons we need to make 1+2=3 comparisons.
"Every one of the 3 combinations has the same odds". For each person, probability of NOT having the same birthday is (3-1)/3 = 2/3. Probability of not having the same birthday for all three of them, following author's logic, is
(2/3)*(2/3)*(2/3)= 8/27

Now let's look at the sample space. I will list all possible combinations of birthdays for person 1, 2 and 3.
Day 1 - Day 1 - Day 1;
1-1-2
1-1-3
1-2-1
1-2-2
1-2-3 (all different)
1-3-1
1-3-2 (all different)
1-3-3
2-1-1
2-1-2
2-1-3 (all different)
2-2-1
2-2-2
2-2-3
2-3-1 (all different)
2-3-2
2-3-3
3-1-1
3-1-2 (all different)
3-1-3
3-2-1 (all different)
3-2-2
3-2-3
3-3-1
3-3-2
3-3-3
There is 27 possible outcomes in total. All outcomes are equally possible. Of these outcomes, only 6 satisfy our requirement. So the odds are 6/27. Which is different from what we've got following the author's method.
Using the method which I suggested, we get:
(3/3)*(2/3)*(1/3)=6/27, which is the correct answer.

8. 8. geojellyroll 10:04 PM 4/6/12

il--ya

No, you are still treating each sets of odds individually. The odds are cumulative. The addition of all possibilitiuers.

9. 9. Will_in_BC 07:34 PM 4/10/12

The suggestion of using MLB players is not a good one. MLB players birthdays are weighted to earlier in the year and are not a random sample.

See http://www.socialproblemindex.ualberta.ca/RelAgeMLB.pdf

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