Birthday Paradox

Why not delve into some serious maths? The 'Birthday Paradox' is actually not a paradox (a statement which is true and false at the same time) but it is a statement that is too interesting that it seems false to our human minds.

What is the probability that two persons will have the same birthday? There is a 1 in 365 chance for this to happen or a probability of 1/365. Then I can say the probability of 2 people not having the same birthday is

$\begin{aligned}1-\frac{1}{365}&=\frac{364}{365}\\\\&=0.99\end{aligned}$

Hence, there's a 99% chance of 2 people having different birthdays.

But if instead of 2 there were 23 people, then what would be the probability of 2 people having the same birthday. For this, we must consider all the possible pairings between these people and it is equal to

$\begin{aligned}C(23,2)&= \frac{23!}{21!\cdot 2!}\\\\&=\frac{23\cdot22}{2}\\\\&=253\end{aligned}$

There are a total of 253 possible pairings between a group of 23 people. Since each pair has a probability of 364/365 of having different birthdays, the probability of finding a pair among 253 possible pairs who have different birthdays will be equal to

$\begin{aligned}\left ( \frac{364}{365} \right )^{253}\end{aligned}$

Then the probability of finding a pair who has the same birthday is equal to

$\begin{aligned}1-\left ( \frac{364}{365} \right )^{253}&= 1-0.4995\\\\&=0.5\end{aligned}$

This means there is a 50% chance of two people having the same birthday in a group of 23 people. This is the birthday paradox. The probability just skyrocketed from 1% to 50% by just adding 21 people to the group.

To make this further interesting, let us consider a group of 50 people, then the probability of finding a pair with the same birthday will be equal to

$\begin{aligned}1-\left ( \frac{364}{365} \right )^{C(50,2)}&=1-\left ( \frac{364}{365} \right )^{1225}\\\\&=0.97\end{aligned}$

Imagine a room with 50 people, there is a 97% chance that two of them will have the same birthday, which is insane! Math never fails to surprise us.

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