Same Birthday Odds: The Birthday Paradox

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It stands to reason that same birthday odds for one person meeting another are 1/365 (365 days in the year and your birthday is on one of them). But it may surprise you to know that with a class of 23 people, there is a 50% chance that two people will share a birthday. In a class of 57 people, it’s almost definite — with 99% probability — that two people will share the same birthday. This blew my mind when I was a student. There were 30 students in my undergrad statistics class and the professor said the odds of two of us having the same birthday were very high. In fact, two people in the class did have the same birthday. This didn’t seem to make sense to me, as there are 365 days in a year.

same birthday odds image showing probabilities for the borthday paradox
In probability theory, the birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are 366 possible birthdays, including February 29). However, 99% probability is reached with just 57 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday. [1]


My Initial (Incorrect) Reasoning of Same Birthday Odds

The odds are 1/365 that I will meet another person with the same birthday. But we’re not talking about just me in a class. We’re talking about every student having those odds. It’s like if I had a 1/10 chance of winning the lottery and I meet another person who also has a 1/10 chance of winning the lottery, then combined we have a 2/10 chance of winning the lottery. The odds of a “coincidence” increases with each person:

  • Me meeting a person with the same birthday: 1/365
  • Me and one other friend meeting someone with the same birthday: 1/(365/2) = 183
  • Three of us meeting someone with the same birthday: 1/(365/3) = 1/122
  • …Twenty nine of us meeting someone with the same birthday: 1/12.

Those are pretty good odds, but not high enough to account for all those coincidences. That left me with a peculiar puzzle. The odds are actually much higher (over 100 percent for a class of 30). The reason takes into account all of the possible combinations.

Why the Same Birthday Odds are Actually Much Higher

It isn’t quite as easy as just x/365:

  • One person has a 1/365 chance of meeting someone with the same birthday.
  • Two people have a 1/183 chance of meeting someone with the same birthday. But! Those two people might also have the same birthday, so you have to add odds of 1/365 for that. The odds become 1/365 + 1/182.5 = 0.008, or .8 percent.
  • Four people (lets call them ABCD) have a 1/91 chance, but there are 6 possible combinations (AB AC AD BD BC CD) so the probability becomes 1/91 + 6/365…and so on.

Another Way to Calculate Same Birthday Odds

If there are 30 students in a class, there are 435 ways two students can be paired. The odds of a “match” become 1/12 + 435/365…which is much greater than 100 percent. Seeing as the odds are 1/365 that any two students will match birthdays and there are 3 possible matches, it’s no surprise that two of those students share the same birthday. (Use the combinations calculator to figure the combinations out. It will also list all the possible name combinations if you really want to!).

Should I Perform This Experiment in Class?

I’ll be the first to admit I haven’t used this in class for the main reason that with 25 students in a class, the odds are a bit over 50/50 that this experiment will work. A second reason is that the above math is over simplified to be somewhat understandable. Even third or fourth year math majors will struggle a bit with the “true” probabilities behind why this works. Figuring out same birthday odds is very complex for many reasons including:

  • More people are born weekdays than weekends; mostly due to C-sections and induced births happening during the week, when doctors prefer to work.
  • Seasonal trends mean that more people are born in the summer than the winter.

Figuring out the true probabilities involves Bayesian logic; hop over to this Stanford University page for a more detailed explanation on Bayesian logic and same birthday odds.

Although it might not work in a small group, the birthday distribution has many applications, such as in cryptography, where it relates to the probability of collisions in hash functions, and in the design of computer networks, where it relates to the probability of congestion or packet collision. It is also used in social science research to study patterns of group behavior and interaction.

The Birthday Paradox

The birthday paradox is a veridical paradox — a situation that produces a solution that seems absurd, but is correct nonetheless: it seems wrong at first glance but is, in fact, true. The veridical paradox, also known as the veridicality problem, is a type of paradox that arises when a situation or observation challenges our expectations or beliefs about reality. In other words, the veridical paradox occurs when something is true, yet it goes against what we believed to be true.

Apart from the birthday paradox, there are many famous examples of the veridical paradox, which include:

    1. The McGurk Effect: This is an auditory-visual illusion that occurs when we see a person’s lips move in a way that contradicts the sound we hear. For example, if we see a person say “ga,” while we hear “ba,” our brain may perceive a third sound, such as “da.” This situation challenges our belief that our senses are reliable, and that what we see and hear should match.
    1. The Capgras Delusion: This is a psychological disorder in which a person believes that their loved ones have been replaced by identical imposters. This situation challenges our belief that our perception of people and objects is objective and accurate.
  1. The Monty Hall Problem: This is a probability puzzle that occurs in a game show where a contestant is given a choice between three doors. Behind one door is a prize, while the other two doors reveal nothing. After the contestant makes their initial choice, the host opens one of the other doors to reveal no prize. The contestant is then given the option to switch their choice. The veridical paradox arises because it seems counterintuitive that switching the choice would increase the probability of winning, even though it is mathematically true.


[1] Guillaume Jacquenot, CC BY-SA 3.0 <>, via Wikimedia Commons

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