Explain the Birthday Paradox

I recently read about the Birthday Paradox which states that in a group of 23 people, there's a probability of 50% that 2 people share their birthday, probability wise.

I calculated and don't think it's possible that it's true in any case (unless my math is wrong). So, can anyone please tell me how to prove or disprove it mathematically ?

Okay, here are my calculations.

Let us view the problem as this: Experiment: there are 23 people, each one is choosing 1 day for his birthday, and trying not to choose it so that it's same as others.

So the 1st person will easily choose any day according to his choice.

This leaves 364 days to the second person, so the second person will choose such day with probability 364/365.

Same with the third guy, but now he should not choose the day same as 1st as well as 2nd person and hence he has 363 days and probability= 363/365.

So the probability of the experiment is 1.(364/365).(363/365)....(343/365) which is approximately 50%.

I hope this helps.

For more discussions you can refer here.

Simple example with three balls: red, green and blue.

When we form a collection of two balls, we have

$$ 3^2 $$

possibilities.

But some do not contain the same color - and that is given by

$$ 3 \times 2 $$

So the number of collection such that two balls have the same color is given by

$$ 3^2 - 3 \times 2 = 3 $$

So the change of finding two balls with the same color in a collection of 2 balls is given by

$$ \frac{3^2 - 4 \times 3}{3^2} = \frac{3}{9} = \frac{1}{3} $$

We can do the same for 4 balls and a collection of 2 balls. the change of finding two balls with the same color in a collection of 2 balls is given by

$$ \frac{4^2 - 4 \times 3}{4^2} $$

The basic formula is then given by

$$ \frac{F^n - F \times (F-1) \times (F-2) \times \cdot (F-n)}{F^n} $$

where

$$ F $$

is the 'freedom' - the number of different colors for the balls, and

$$ n $$

is the number of balls in the collection.

Using some math we can write

$$ 1 - \frac{F!}{F^n \big(F-n\big)!} $$

Note that when $n>F$ we have

$$ k! $$

for a negative number.

But as

$$ \big(n-1\big)! = \frac{n!}{n} $$

we see that

$$ \big(-1\big)! = \frac{0!}{0} \rightarrow \infty $$

So in case $n > F$ the change becomes $1$

Instead of color - we can consider birthdays, so $F=365$ and we get

$$ 1 - \frac{365!}{365^n \big(365-n\big)!} $$

The case $n=23$ gives

$$ 1 - \frac{365!}{365^23 \big(365-23\big)!} = 50.7\% $$