Birthday problem: What is the minimum number of people in a room such that at least two people have the same birthday has a probablity $\geq 0.5$ ( http://en.wikipedia.org/wiki/Birthday_problem )
I am trying to solve this by considering each day separately and then adding up the probabilities. This is equivalent to (birthday collision on a particular day) $\cdot \, 365$. I understand that I am doing something wrong as the probability goes more than $1$ after a few iterations. Can you please help me with identifying my mistake. Thanks in advance
$2$ persons: $A$ and $B$ $= 1/365 ( \text{prob of $A$ being born on day 1} ) \cdot 1/365 ( \text{prob of $B$ being born on day 1} ) + 1/365 ( \text{prob of $A$ being born on day 2}) \cdot 1/365 ( \text{prob of $B$ being born on day 2} ) + \dots = \frac{1}{(365 \cdot 365)} ( \text{probability on each day} ) \cdot 365 = 1/365$$= \frac{1}{365}\times ( \text{prob of $A$ being born on day 1} ) \cdot \frac{1}{365}\times ( \text{prob of $B$ being born on day 1} ) + \frac{1}{365}\times ( \text{prob of $A$ being born on day 2}) \cdot \frac{1}{365}\times ( \text{prob of $B$ being born on day 2} ) + \dots = \frac{1}{(365 \cdot 365)} ( \text{probability on each day} ) \cdot 365 = \frac{1}{365}$
$3$ persons: $A$, $B$ and $C$ $AB$ born on day 1 : $\frac{1}{(365 \cdot 365 )}$ $BC$ born on day 1 : $\frac{1}{(365 \cdot 365 )}$ $CA$ born on day 1 : $\frac{1}{(365 \cdot 365 )}$ for day 1 = $\frac{3}{(365\cdot 365)}$ for day 1 = number of pairs$/(365\cdot 365) = 3c2/(365\cdot 365)$ for all days $= 365 \cdot 3c2 / ( 365 \cdot 365 )$ $= 3c2 / 365$
4 persons: $= 4c2 / 365$
