University of Illinois at Urbana-Champaign
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Office for Mathematics, Science, and Technology Education
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The Birthday Problem by Becky Schmoyer

It's your birthday! Congratulations! But wait... doesn't the person sitting behind you have a birthday today also? What are the odds of that!?!

The Birthday Problem is a way of finding the probability of two people in the same room having the same birthday. To start, say there is one person. That person has a possibility of 365 days as being his birthday. (This is assuming there are 365 days in the year.) Now, if you have two people, the second person has 364 different days that could be his and not shared with the first person. If you take it up to three people, the third person has 363 different days that could be his and not shared with the other two. Therefore, the probability of a match being made between three people is this:

1 - (365) * (364) * (363) / (365) * (365) * (365) = .0082

To get this, first you must figure out the probability that no one in the group has a match. To do this, just find the product of each distinct day over the total days"

365/365 * 364/365 * 363/365 = .9918

Then, to find the probability that at least two people share the same birthday, subtract the previous answer from one.

1 - .9918 = .0082

From this, we can extract an equation that will aid in finding the probabilities in much larger groups.

P(A) = (365) * (364) * (363) * … (365 - N + 1) / 365^N

P(AC) 1 - (365) * (364) * (363) * … (365 - N + 1) / 365^N

WHEN

  • N = Number of people in the group
  • P(A) = Probability that no one shares a birthday
  • P(AC) = Probability that 2 people share the same birthday

Now, with these formulas and a basic understanding of The Birthday Problem, one can figure out the probability of similar birthdays in bigger groups, and also with different circumstances. For instance, how likely is it that in a group of six students there will be at least two birthdays in the same month?

To figure this out, we follow the same steps as above, but use 12 months in replace of the 365 days. The first person has a range of twelve. The next has 11 different months, the third, 10, and so on. So, using the equation above, we figure that

(12 * 11* 10 * 9 * 8 * 7) / 12^6 = .2228009

1 - .2228009 = .7772

So, the probability of 2 people having their birthday in the same month is greater than 2 students out of 6 having the same their birthday on the same day. (The probability of 2 out of 6 is around .04)

January, 1999

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