Birthday Paradox
October 12, 2012
In the study of probability, the Birthday Paradox states that in a group of 23 people, there is a 50% chance that two will have the same birthday; in a group of 57 people, the odds rise to 99%.
Your task is to simulate the birthday paradox over many trials and verify the odds. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
[…] today’s Programming Praxis exercise, our goal is to simulate the well-known birthday paradox. Let’s […]
My Haskell solution (see http://bonsaicode.wordpress.com/2012/10/12/programming-praxis-birthday-paradox/ for a version with comments):
import Control.Monad import Data.List import System.Random paradox :: Int -> IO Bool paradox n = fmap (\g -> or [elem h t | (h:t) <- tails . take n $ randomRs (1, 365 :: Int) g]) newStdGen trial :: Int -> IO Float trial = fmap ((/ 100) . genericLength . filter id) . replicateM 10000 . paradox main :: IO () main = do print =<< trial 23 print =<< trial 57Should one not allow for leap years?
Here is a video explaining birthday paradox: http://www.numberphile.com/videos/23birthday.html
A Python version
import random as RA
import itertools as IT
def analytic(N):
prob = 1
for i in range(0, N):
prob *= (1.0 – i / 365.0)
return 1 – prob
def gen_cases(N):
while 1:
yield [RA.randint(1, 365) for i in range(N)]
def monte_carlo(N, mcsamples = 1000):
“””calculate chance that 2 or more people have same birthday”””
cnt = sum(len(c) != len(set(c)) for c in IT.islice(gen_cases(N), mcsamples))
return cnt / float(mcsamples)
#[/sourcecode]
Repost. A python version.
import random as RA import itertools as IT def analytic(N): prob = 1 for i in range(0, N): prob *= (1.0 - i / 365.0) return 1 - prob def gen_cases(N): while 1: yield [RA.randint(1, 365) for i in range(N)] def monte_carlo(N, mcsamples = 1000): """calculate chance that 2 or more people have same birthday""" cnt = sum(len(c) != len(set(c)) for c in IT.islice(gen_cases(N), mcsamples)) return cnt / float(mcsamples)A Python version which takes into account leap years, 0-364 are the non-leap days of the year and 365 is Feb 29th:
import random def gen_day(): return random.randrange(366) if random.randrange(4) == 0 else random.randrange(365) def trial(people): birthdays = [gen_day() for p in range(people)] return len(birthdays) != len(set(birthdays)) def simulate(people, trials): return len([x for x in range(trials) if trial(people)])Hah, beat you to one for once. :)
Here’s a post that I wrote a week and a half ago on my own birthday that basically works out the math behind the problem and then lets you run a simulation build into the web page (Javascript w/ JQuery). You can choose any number of birthdays to simulate and it will generate that many for you over the next year, keeping statistics for each number, along with the expected value. I think it’s pretty neat at least. :)
Check it out here: The Birthday Paradox
Clojure solution.
(defn birthday? [n] (let [sample (map rand-int (repeat n 365))] (not= (count (distinct sample)) (count sample)))) (defn paradox [n, trials] (float (/ (count (filter birthday? (repeat trials n))) trials))) user=> (paradox 23 100000) 0.50874 user=> (paradox 33 100000) 0.7745 user=> (paradox 57 100000) 0.99029Solution written in Go (source code https://gist.github.com/3885112):
package main import ( "fmt" "math/rand" ) func main() { fmt.Printf("Chance of same birthday in group with 23 people: %.0f%%\n", getChanceOfSameBirthdayInGroup(23)*100) fmt.Printf("Chance of same birthday in group with 57 people: %.0f%%\n", getChanceOfSameBirthdayInGroup(57)*100) } func getChanceOfSameBirthdayInGroup(groupSize uint) float32 { n := 10000 numberOfSameBirthdays := 0 for i := 0; i < n; i++ { if isSameBirthdayInGroup(groupSize) { numberOfSameBirthdays++ } } return float32(numberOfSameBirthdays) / float32(n) } func isSameBirthdayInGroup(groupSize uint) bool { birthdays := make(map[int]uint8) for i := uint(0); i < groupSize; i++ { birthday := rand.Intn(365) birthdays[birthday]++ if birthdays[birthday] == 2 { return true } } return false }A year whose number is divisible by 100 is not a leap year, unless it is also divisible by 400. So the year 2000 was a leap year, but the year 2100 won’t be. A program trying to calculate birthday statistics for people currently alive can ignore this issue, but if it’s supposed to work across history it needs to take it into account. Of course, if one goes back before the Gregorian calendar was adopted, matters get extremely complicated, and figuring out how to deal with the eventual breakdown (at some not-yet-predictable time) of the Gregorian calendar is even trickier.
A Java Solution:
import java.util.HashSet; import java.util.Random; import java.util.Set; public class BirthdayParadox { private static final Random randGen = new Random(); static float simulate(int nPeople, int nTrials) { int n = 0; for(int i = 0; i < nTrials; i++) { if(trial(nPeople)) n++; } return (float)n/nTrials; } static boolean trial(int nPeople) { Set<Integer> birthdays = new HashSet<Integer>(); for (int i = 0; i < nPeople; i++) { int day = randGen.nextInt(365); if (birthdays.contains(day)) return true; birthdays.add(day); } return false; } public static void main(String[] args) { System.out.printf("%.4f\n", simulate(23, 1000)); # => 0.5220 System.out.printf("%.4f\n", simulate(57, 1000)); # => 0.9900 } }Leap years? Not sure what leap years have to do with the original problem stated in the main blog homepage. Here is my PHP solution for the first part.
for($i = 1; $i < 1000; $i++)
{
$arr = array();
for($q = 0; $q 1) { $total++; }
}
echo $total / 1000;
Not sure what’s wrong with your blog but it just can’t off half my code. Forget this. Moving on.
(defun sim-bday-paradox (size trials) (loop repeat trials counting (loop repeat size for r = (random 365) then (random 365) when (member r bday-lst) do (return t) collect r into bday-lst))) (sim-bday-paradox 23 10000) (sim-bday-paradox 57 10000)[…] Pages: 1 2 […]