• March 16th 2010, 08:51 PM
jrkevinfang1234
Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.

Here's the full solution on Wikipedia:
Birthday problem - Wikipedia, the free encyclopedia

I got lost at the transition from

to

What's happening here?
• March 16th 2010, 09:15 PM
Random Variable
let $x=-x$

then $e^{-x} \approx 1 -x$

so $1 - \frac{1}{365} \approx e^{-1/365}$

$1-\frac{2}{365} \approx e^{-2/365}$

and so on
• March 16th 2010, 09:16 PM
Drexel28
Quote:

Originally Posted by jrkevinfang1234
Apparently, there are two methods of solving this probability problem. The first, using p(n) = 1 - (365^n falling / 365^n) makes sense to me; however, there is an alternate solution using the Taylor series of e^x.

Here's the full solution on Wikipedia:
Birthday problem - Wikipedia, the free encyclopedia

I got lost at the transition from
$e^x\approx 1+x\implies e^-x\approx 1-x$ and so $\left(1-x\right)(1-y)(1-z)\cdots(1-t)\approx e^{-x}\cdot e^{-y}\cdot e^{-z}\cdots e^{-t}$