If you put 23 people in a room, 2 of them probably have the same birthday

same_bday.rb

# inspired by http://www.smbc-comics.com/comic/monty-hall-problems
SAMPLES   = 50_000
DAYS      = 365
PRECISION = 4

def prob_of_same_bday(num_people)
  SAMPLES.times.count { share_bday_in_room_of_size? num_people } / SAMPLES.to_f
end

def share_bday_in_room_of_size?(num_people)
  bdays = Array.new DAYS, 0 # number of people born on the given day, index 0 ≈ 1 Jan,  DAYS-1 ≈ 31 Dec
  num_people.times.any? { (bdays[rand DAYS] += 1) == 2 }
end

def calculate_probability(num_people)
  1 - num_people.times.reduce(1.0) { |p, n| p * (DAYS-n) / DAYS }
end

def answer?(num_people)
  calculate_probability(num_people) > 0.5 &&
    calculate_probability(num_people-1) < 0.5
end

# Using 1+DAYS as the upper bound because: by the pigeonhole principle,
# we know that there is a 100% chance of two people sharing a birthday
# when there are more people than days. Eg: if there are 10 days in a year,
# and 11 people in a room, it is guaranteed that at least 2 of them share a birthday.
NUM_PEOPLE = 1.upto(1+DAYS).find do |n|
  p = prob_of_same_bday(n).round(PRECISION)

  n # => 1,   2,      3,      4,      5,      6,      7,      8,      9,      10,     11,     12,     13,     14,     15,     16,     17,     18,     19,     20,     21,     22,     23
  p # => 0.0, 0.0025, 0.0081, 0.0178, 0.0286, 0.0403, 0.0569, 0.0733, 0.0971, 0.1159, 0.1381, 0.1664, 0.1948, 0.2237, 0.2534, 0.2806, 0.3171, 0.3495, 0.3815, 0.4103, 0.4426, 0.4733, 0.5059

  p > 0.5
end

# Calculated results:
NUM_PEOPLE                       # => 23
prob_of_same_bday NUM_PEOPLE     # => 0.5097

# Validate results:
calculate_probability NUM_PEOPLE # => 0.5072972343239855
answer? NUM_PEOPLE               # => true
answer? NUM_PEOPLE-1             # => false
answer? NUM_PEOPLE+1             # => false