Birthday

1. Mathematical Derivation

The birthday problem asks: what is the chance that at least two people in a group of \(n\) share a birthday?

It’s easier to compute the complement: the chance all birthdays are different. There are \(365 \times 364 \times \cdots \times (365-n+1)\) ways for this to happen, out of \(365^n\) possible outcomes:

\[ P(\text{no match}) = \frac{365 \times 364 \times \cdots \times (365-n+1)}{365^n}. \]

So the probability of a duplicate is:

\[ P(\text{duplicate}) = 1 - P(\text{no match}). \]

2. Reformulation with lfactorial and lchoose

The numerator can be written as \(\binom{365}{n} \times n!\), giving:

\[ P(\text{duplicate}) = 1 - \frac{\binom{365}{n} \times n!}{365^n}. \]

Using logarithms, we express this in R as:

\[ P(\text{duplicate}) = 1 - \exp\big( \log \binom{365}{n} + \log(n!) - n \log(365) \big). \]

3. The Function

This is implemented in R as:

MATH4753F25ianh::birthday
#> function (x) 
#> {
#>     1 - exp(lchoose(365, x) + lfactorial(x) - x * log(365))
#> }
#> <bytecode: 0x107f35168>
#> <environment: namespace:MATH4753F25ianh>