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# The Birthday problem... amazing !!!

According to the "$$\color{Blue}{ \text{Pigeonhole}- \text{principle}}$$ " we know that if in a given set of people , you want to make sure there are 2 people with $$\color{Blue}{same}$$ $$\color{Blue}{ birthdays}$$ then you have to get 367 people in the set. (There are 366 choices for a birthday , therefore)

But do you know that $$\color{Red}{50}%$$ % probability of 2 people having same birthdays is actually achieved even in a set of 23 people !!!!!

And what's even more exciting is , $$\color{Red}{99.9 }%$$% probability is for the set of 70 people !!!!!

So hard to believe, but I have a mathematical function to show this, I saw it on the Wikipedia Page

Here, the probability of people having same birthday will be given by

$$\color{Green}{p(n)}$$ .. (probability in set of $$n$$ people) = $$\color{Green}{1\times \biggl(1-\dfrac{1}{365}\biggr) \times \biggl(1-\dfrac{2}{365}\biggr) \times \biggl(1-\dfrac{3}{365}\biggr)\times .... \times \biggl(1-\dfrac{n}{365}\biggr)}$$

That is $$\color{Purple}{p(n) = \dfrac{365!}{365^n \times (365-n)! }= \dfrac{n! \times \dbinom{365}{n}}{365^n}}$$

The graph of probability (of course it won't be a continuous graph because number of people has to be an integer) is as shown in attached image.

This shows how rapidly the graph grows in first few numbers.....

2 years, 9 months ago

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Actually if you think about it and calculate the numbers you can get this easily... it isn't rocket science. · 2 years, 9 months ago