The problem

What is the probability that at least two people this room share the same birthday?

Is it something like \(\frac{27}{365} =\) 0.07

Analytical solution

\[\begin{align} 1 - \bar p(n) &= 1 \times \left(1-\frac{1}{365} \right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{n-1}{365}\right) \nonumber \\ &= \frac{ 365 \times 364 \times \cdots \times (365-n+1) }{ 365^n } \nonumber \\ &= \frac{ 365! }{ 365^n (365-n)!} = \frac{n!\cdot\binom{365}{n}}{365^n}\\ p(n= 27) &= 0.627 \nonumber \end{align}\]

Simulation

1 - Simulate 10,000 rooms with n = 27 random birthdays, and store the results in matrix where each row represents a room.
2 - For each room (row) compute the number of unique birthdays.

3 - Compute the average number of times a room has 40 unique birthdays, across 10,000 simulations, and report the complement.

birthday.prob = function(n.pers, n.sims) {
  # simulate birthdays
  birthdays = matrix(round(runif(n.pers * n.sims, 1, 
                                 365)), nrow = n.sims, 
                           ncol = n.pers)
  # for each room (row) get unique birthdays
  unique.birthdays = apply(birthdays, 1, unique)
  # Indicator with 1 if all are unique birthdays 
  all.different = (lapply(unique.birthdays, length) == n.pers) 
  # Compute average time all have different birthdays 
  result = 1 - mean(all.different)
return(result)
}
n.pers.param = n.pers
n.sims.param = 1e4 

res1 = birthday.prob(n.pers.param,n.sims.param)

The simulated probability is 0.64