Editing Birthday problem (section)

===Simple exponentiation===
The probability of any two people not having the same birthday is {{sfrac|364|365}}. In a room containing ''n'' people, there are {{math|{{pars|s=150%|{{su|p=''n''|b=2|a=c}}}} {{=}} {{sfrac|''n''(''n'' − 1)|2}}}} pairs of people, i.e. {{math|{{pars|s=150%|{{su|p=''n''|b=2|a=c}}}}}} events. The probability of no two people sharing the same birthday can be approximated by assuming that these events are independent and hence by multiplying their probability together. Being independent would be equivalent to picking [[Sampling (statistics)#Replacement of selected units|with replacement]], any pair of people in the world, not just in a room. In short {{sfrac|364|365}} can be multiplied by itself {{math|{{pars|s=150%|{{su|p=''n''|b=2|a=c}}}}}} times, which gives us

:<math>\bar p(n) \approx \left(\frac{364}{365}\right)^\binom{n}{2}.</math>

Since this is the probability of no one having the same birthday, then the probability of someone sharing a birthday is

:<math>p(n) \approx 1 - \left(\frac{364}{365}\right)^\binom{n}{2}.</math>

And for the group of 23 people, the probability of sharing is

:<math>p(23) \approx 1 - \left(\frac{364}{365}\right)^\binom{23}{2} = 1 - \left(\frac{364}{365}\right)^{253} \approx 0.500477 .</math>