It’s that day of the year again, when florists make half their yearly income from selling flora to those in, or freshly out of, love. I proud myself on having at least had email or some other form of internet conversation with my small contingent of FB friends. There’s only 150 odd of them in total. And yet, 2 of them have a birthday on Valentine’s Day, Feb 14 ! What are the odds ?
Well, the odds are actually pretty good, and I used that as a party trick on a quiet night shift once, we counted everyone, from ambulance officer, to patient, to ward clerk to policeman coming through the door to find 2 people with the same birthday, and it took only 36 or so to find them. How is this possible ? Well, it’s called the Birthday paradox, and it is based on the inability of our brains to think in exponentials, and to consider scenarios that we are not personally involved in:
It’s like asking “What’s the chance of getting one or more heads in 23 coin flips?” There are so many possibilities: heads on the first throw, or the 3rd, or the last, or the 1st and 3rd, the 2nd and 21st, and so on.
How do we solve the coin problem? Flip it around (Get it? Get it?). Rather than counting every way to get heads, find the chance of getting all tails, our “problem scenario”.
If there’s a 1% chance of getting all tails (more like .5^23 but work with me here), there’s a 99% chance of having at least one head. I don’t know if it’s 1 head, or 2, or 15 or 23: we got heads, and that’s what matters. If we subtract the chance of a problem scenario from 1 we are left with the probability of a good scenario.
The same principle applies for birthdays. Instead of finding all the ways we match, find the chance that everyone is different, the “problem scenario”. We then take the opposite probability and get the chance of a match. It may be 1 match, or 2, or 20, but somebody matched, which is what we need to find.
Fact of the matter is that you need just 23 people to have a 50:50 chance to find two with the same birthday. Which is why I now better go and send some virtual flowers to my two FB friends whose birthday it is today !
Correct on the odds of two people in a group sharing a birthday, Martin, but this does not apply when you select a specific date. Then the chance is 1/365 of any, particular, individual having that, particular, birthday. Because its a “special” date its one of these coincidences which religious types think must have “meaning”.
John McC , you’re quite sure the probability distribution is uniform?
The problem is to compute the approximate probability that in a room of n people, at least two have the same birthday. For simplicity, disregard variations in the distribution, such as leap years, twins, seasonal variations, etc., and assume that the 365 possible birthdays are equally likely. Real-life birthday distributions are not uniform since not all dates are equally likely.
If P(A) is the probability of at least two people in the room having the same birthday, it may be simpler to calculate P(A’), the probability of there not being any two people having the same birthday. Then, because P(A) and P(A’) are the only two possibilities and are also mutually exclusive, P(A’) = 1 − P(A).
In deference to widely published solutions concluding that 23 is the number of people necessary to have a P(A) that is greater than 50%, the following calculation of P(A) will use 23 people as an example.
When events are independent of each other, the probability of all of the events occurring is equal to a product of the probabilities of each of the events occurring. Therefore, if P(A’) can be described as 23 independent events, P(A’) could be calculated as P(1) × P(2) × P(3) × … × P(23).
The 23 independent events correspond to the 23 people, and can be defined in order. Each event can be defined as the corresponding person not sharing their birthday with any of the previously analyzed people. For Event 1, there are no previously analyzed people. Therefore, the probability, P(1), that person number 1 does not share his/her birthday with previously analyzed people is 1, or 100%. Ignoring leap years for this analysis, the probability of 1 can also be written as 365/365, for reasons that will become clear below.
For Event 2, the only previously analyzed people are Person 1. Assuming that birthdays are equally likely to happen on each of the 365 days of the year, the probability, P(2), that Person 2 has a different birthday than Person 1 is 364/365. This is because, if Person 2 was born on any of the other 364 days of the year, Persons 1 and 2 will not share the same birthday.
Similarly, if Person 3 is born on any of the 363 days of the year other than the birthdays of Persons 1 and 2, Person 3 will not share their birthday. This makes the probability P(3) = 363/365.
This analysis continues until Person 23 is reached, whose probability of not sharing their birthday with people analyzed before, P(23), is 343/365.
P(A’) is equal to the product of these individual probabilities:
(1) P(A’) = 365/365 × 364/365 × 363/365 × 362/365 × … × 343/365
The terms of equation (1) can be collected to arrive at:
(2) P(A’) = (1/365)²³ × (365 × 364 × 363 × … × 343)
Evaluating equation (2) gives P(A’) = 0.492703
Therefore, P(A) = 1 − 0.492703 = 0.507297 (50.7297%) or the chances are better than 1 in 2 that two people in a group of 23 will share a birthday.