# Birthday problem

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**Birthday problem****draft**programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

In probability theory, the **birthday problem**, or **birthday paradox** This is not a paradox in the sense of leading to a logical contradiction, but is called a paradox because the mathematical truth contradicts naïve intuition: most people estimate that the chance is much lower than 50%. pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 366 (by the pigeon hole principle, ignoring leap years). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack.

- Task

Using simulation, estimate the number of independent people required in a groups before we can expect a *better than even chance* that at least 2 independent people in a group share a common birthday. Furthermore: Simulate and thus estimate when we can expect a *better than even chance* that at least 3, 4 & 5 independent people of the group share a common birthday. For simplicity assume that all of the people are alive...

- Suggestions for improvement

- Estimating the error in the estimate to help ensure the estimate is accurate to 4 decimal places.
- Converging to the
^{th}solution using a root finding method, as opposed to using an extensive search. - Kudos (κῦδος) for finding the solution by proof (in a programming language) rather than by construction and simulation.

- See also

## Contents

## Ada[edit]

This solution assumes a 4-year cycle, with three 365-day years and one leap year.

with Ada.Command_Line, Ada.Text_IO, Ada.Numerics.Discrete_random;

procedure Birthday_Test is

Samples: constant Positive := Integer'Value(Ada.Command_Line.Argument(1));

-- our experiment: Generate a X (birth-)days and check for Y-collisions

-- the constant "Samples" is the number of repetitions of this experiment

subtype Day is integer range 0 .. 365; -- this includes leap_days

subtype Extended_Day is Integer range 0 .. 365*4; -- a four-year cycle

package ANDR is new Ada.Numerics.Discrete_Random(Extended_Day);

Random_Generator: ANDR.Generator;

function Random_Day return Day is (ANDR.Random(Random_Generator) / 4);

-- days 0 .. 364 are equally probable, leap-day 365 is 4* less probable

type Checkpoint is record

Multiplicity: Positive;

Person_Count: Positive;

end record;

Checkpoints: constant array(Positive range <>) of Checkpoint

:= ( (2, 22), (2, 23), (3, 86), (3, 87), (3, 88),

(4, 186), (4, 187), (5, 312), (5, 313), (5, 314) );

type Result_Type is array(Checkpoints'Range) of Natural;

Result: Result_Type := (others => 0);

-- how often is a 2-collision in a group of 22 or 23, ..., a 5-collision

-- in a group of 312 .. 314

procedure Experiment(Result: in out Result_Type) is

-- run the experiment once!

A_Year: array(Day) of Natural := (others => 0);

A_Day: Day;

Multiplicity: Natural := 0;

People: Positive := 1;

begin

for I in Checkpoints'Range loop

while People <= Checkpoints(I).Person_Count loop

A_Day := Random_Day;

A_Year(A_Day) := A_Year(A_Day)+1;

if A_Year(A_Day) > Multiplicity then

Multiplicity := Multiplicity + 1;

end if;

People := People + 1;

end loop;

if Multiplicity >= Checkpoints(I).Multiplicity then

Result(I) := Result(I) + 1;

-- found a Multipl.-collision in a group of Person_Cnt.

end if;

end loop;

end Experiment;

package TIO renames Ada.Text_IO;

package FIO is new TIO.Float_IO(Float);

begin

-- initialize the random generator

ANDR.Reset(Random_Generator);

-- repeat the experiment Samples times

for I in 1 .. Samples loop

Experiment(Result);

end loop;

-- print the results

TIO.Put_Line("Birthday-Test with" & Integer'Image(Samples) & " samples:");

for I in Result'Range loop

FIO.Put(Float(Result(I))/Float(Samples), Fore => 3, Aft => 6, Exp => 0);

TIO.Put_Line

("% of groups with" & Integer'Image(Checkpoints(I).Person_Count) &

" have" & Integer'Image(Checkpoints(I).Multiplicity) &

" persons sharing a common birthday.");

end loop;

end Birthday_Test;

- Output:

Running the program with a sample size 500_000_000 took about 25 minutes on a slow pc.

./birthday_test 500_000_000 Birthday-Test with 500000000 samples: 0.475292% of groups with 22 have 2 persons sharing a common birthday. 0.506882% of groups with 23 have 2 persons sharing a common birthday. 0.487155% of groups with 86 have 3 persons sharing a common birthday. 0.498788% of groups with 87 have 3 persons sharing a common birthday. 0.510391% of groups with 88 have 3 persons sharing a common birthday. 0.494970% of groups with 186 have 4 persons sharing a common birthday. 0.501825% of groups with 187 have 4 persons sharing a common birthday. 0.495137% of groups with 312 have 5 persons sharing a common birthday. 0.500010% of groups with 313 have 5 persons sharing a common birthday. 0.504888% of groups with 314 have 5 persons sharing a common birthday.

An interesting observation: The probability for groups of 313 persons having 5 persons sharing a common birthday is almost exactly 0.5. Note that a solution based on 365-day years, i.e., a solution ignoring leap days, would generate slightly but significantly larger probabilities.

## ALGOL 68[edit]

**File: Birthday_problem.a68**

#!/usr/bin/a68g --script #

# -*- coding: utf-8 -*- #

REAL desired probability := 0.5; # 50% #

REAL upb year = 365 + 1/4 # - 3/400 but alive, ignore those born prior to 1901 #,

INT upb sample size = 100 000,

upb common = 5 ;

FORMAT name int fmt = $g": "g(-0)"; "$,

name real fmt = $g": "g(-0,4)"; "$,

name percent fmt = $g": "g(-0,2)"%; "$;

printf((

name real fmt,

"upb year",upb year,

name int fmt,

"upb common",upb common,

"upb sample size",upb sample size,

$l$

));

INT required common := 1; # initial value #

FOR group size FROM required common WHILE required common <= upb common DO

INT sample with no required common := 0;

TO upb sample size DO

# generate sample #

[group size]INT sample;

FOR i TO UPB sample DO sample[i] := ENTIER(random * upb year) + 1 OD;

FOR birthday i TO UPB sample DO

INT birthday = sample[birthday i];

INT number in common := 1;

# special case = 1 #

IF number in common >= required common THEN

found required common

FI;

FOR birthday j FROM birthday i + 1 TO UPB sample DO

IF birthday = sample[birthday j] THEN

number in common +:= 1;

IF number in common >= required common THEN

found required common

FI

FI

OD

OD # days in year #;

sample with no required common +:= 1;

found required common: SKIP

OD # sample size #;

REAL portion of years with required common birthdays =

(upb sample size - sample with no required common) / upb sample size;

print(".");

IF portion of years with required common birthdays > desired probability THEN

printf((

$l$,

name int fmt,

"required common",required common,

"group size",group size,

# "sample with no required common",sample with no required common, #

name percent fmt,

"%age of years with required common birthdays",portion of years with required common birthdays*100,

$l$

));

required common +:= 1

FI

OD # group size #

**Output:**

upb year: 365.2500; upb common: 5; upb sample size: 100000; . required common: 1; group size: 1; %age of years with required common birthdays: 100.00%; ...................... required common: 2; group size: 23; %age of years with required common birthdays: 50.71%; ................................................................. required common: 3; group size: 88; %age of years with required common birthdays: 50.90%; ................................................................................................... required common: 4; group size: 187; %age of years with required common birthdays: 50.25%; ............................................................................................................................... required common: 5; group size: 314; %age of years with required common birthdays: 50.66%;

## C[edit]

Computing probabilities to 5 sigmas of confidence. It's very slow, chiefly because to make sure a probability like 0.5006 is indeed above .5 instead of just statistical fluctuation, you have to run the simulation millions of times.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define DEBUG 0 // set this to 2 for a lot of numbers on output

#define DAYS 365

#define EXCESS (RAND_MAX / DAYS * DAYS)

int days[DAYS];

inline int rand_day(void)

{

int n;

while ((n = rand()) >= EXCESS);

return n / (EXCESS / DAYS);

}

// given p people, if n of them have same birthday in one run

int simulate1(int p, int n)

{

memset(days, 0, sizeof(days));

while (p--)

if (++days[rand_day()] == n) return 1;

return 0;

}

// decide if the probablity of n out of np people sharing a birthday

// is above or below p_thresh, with n_sigmas sigmas confidence

// note that if p_thresh is very low or hi, minimum runs need to be much higher

double prob(int np, int n, double n_sigmas, double p_thresh, double *std_dev)

{

double p, d; // prob and std dev

int runs = 0, yes = 0;

do {

yes += simulate1(np, n);

p = (double) yes / ++runs;

d = sqrt(p * (1 - p) / runs);

if (DEBUG > 1)

printf("\t\t%d: %d %d %g %g \r", np, yes, runs, p, d);

} while (runs < 10 || fabs(p - p_thresh) < n_sigmas * d);

if (DEBUG > 1) putchar('\n');

*std_dev = d;

return p;

}

// bisect for truth

int find_half_chance(int n, double *p, double *dev)

{

int lo, hi, mid;

reset:

lo = 0;

hi = DAYS * (n - 1) + 1;

do {

mid = (hi + lo) / 2;

// 5 sigma confidence. Conventionally people think 3 sigmas are good

// enough, but for case of 5 people sharing birthday, 3 sigmas actually

// sometimes give a slightly wrong answer

*p = prob(mid, n, 5, .5, dev);

if (DEBUG)

printf("\t%d %d %d %g %g\n", lo, mid, hi, *p, *dev);

if (*p < .5) lo = mid + 1;

else hi = mid;

if (hi < lo) {

// this happens when previous precisions were too low;

// easiest fix: reset

if (DEBUG) puts("\tMade a mess, will redo.");

goto reset;

}

} while (lo < mid || *p < .5);

return mid;

}

int main(void)

{

int n, np;

double p, d;

srand(time(0));

for (n = 2; n <= 5; n++) {

np = find_half_chance(n, &p, &d);

printf("%d collision: %d people, P = %g +/- %g\n",

n, np, p, d);

}

return 0;

}

- Output:

2 collision: 23 people, P = 0.508741 +/- 0.00174794 3 collision: 88 people, P = 0.509034 +/- 0.00180628 4 collision: 187 people, P = 0.501812 +/- 0.000362394 5 collision: 313 people, P = 0.500641 +/- 0.000128174

## D[edit]

import std.stdio, std.random, std.algorithm, std.conv;

/// For sharing common birthday must all share same common day.

double equalBirthdays(in uint nSharers, in uint groupSize,

in uint nRepetitions, ref Xorshift rng) {

uint eq = 0;

foreach (immutable _; 0 .. nRepetitions) {

uint[365] group;

foreach (immutable __; 0 .. groupSize)

group[uniform(0, $, rng)]++;

eq += group[].any!(c => c >= nSharers);

}

return (eq * 100.0) / nRepetitions;

}

void main() {

auto rng = 1.Xorshift; // Fixed seed.

auto groupEst = 2;

foreach (immutable sharers; 2 .. 6) {

// Coarse.

auto groupSize = groupEst + 1;

while (equalBirthdays(sharers, groupSize, 100, rng) < 50.0)

groupSize++;

// Finer.

immutable inf = to!int(groupSize - (groupSize - groupEst) / 4.0);

foreach (immutable gs; inf .. groupSize + 999) {

immutable eq = equalBirthdays(sharers, groupSize, 250, rng);

if (eq > 50.0) {

groupSize = gs;

break;

}

}

// Finest.

foreach (immutable gs; groupSize - 1 .. groupSize + 999) {

immutable eq = equalBirthdays(sharers, gs, 50_000, rng);

if (eq > 50.0) {

groupEst = gs;

writefln("%d independent people in a group of %s share a common birthday. (%5.1f)",

sharers, gs, eq);

break;

}

}

}

}

- Output:

2 independent people in a group of 23 share a common birthday. ( 50.5) 3 independent people in a group of 87 share a common birthday. ( 50.1) 4 independent people in a group of 187 share a common birthday. ( 50.2) 5 independent people in a group of 313 share a common birthday. ( 50.3)

Run-time about 10.4 seconds with ldc2 compiler.

Alternative version:

import std.stdio, std.random, std.math;

enum nDays = 365;

// 5 sigma confidence. Conventionally people think 3 sigmas are good

// enough, but for case of 5 people sharing birthday, 3 sigmas

// actually sometimes give a slightly wrong answer.

enum double nSigmas = 3.0; // Currently 3 for smaller run time.

/// Given n people, if m of them have same birthday in one run.

bool simulate1(in uint nPeople, in uint nCollisions, ref Xorshift rng)

/*nothrow*/ @safe [email protected]*/ {

static uint[nDays] days;

days[] = 0;

foreach (immutable _; 0 .. nPeople) {

immutable day = uniform(0, days.length, rng);

days[day]++;

if (days[day] == nCollisions)

return true;

}

return false;

}

/** Decide if the probablity of n out of np people sharing a birthday

is above or below pThresh, with nSigmas sigmas confidence.

If pThresh is very low or hi, minimum runs need to be much higher. */

double prob(in uint np, in uint nCollisions, in double pThresh,

out double stdDev, ref Xorshift rng) {

double p, d; // Probablity and standard deviation.

uint nRuns = 0, yes = 0;

do {

yes += simulate1(np, nCollisions, rng);

nRuns++;

p = double(yes) / nRuns;

d = sqrt(p * (1 - p) / nRuns);

debug if (yes % 50_000 == 0)

printf("\t\t%d: %d %d %g %g \r", np, yes, nRuns, p, d);

} while (nRuns < 10 || abs(p - pThresh) < (nSigmas * d));

debug '\n'.putchar;

stdDev = d;

return p;

}

/// Bisect for truth.

uint findHalfChance(in uint nCollisions, out double p, out double dev, ref Xorshift rng) {

uint mid;

RESET:

uint lo = 0;

uint hi = nDays * (nCollisions - 1) + 1;

do {

mid = (hi + lo) / 2;

p = prob(mid, nCollisions, 0.5, dev, rng);

debug printf("\t%d %d %d %g %g\n", lo, mid, hi, p, dev);

if (p < 0.5)

lo = mid + 1;

else

hi = mid;

if (hi < lo) {

// This happens when previous precisions were too low;

// easiest fix: reset.

debug "\tMade a mess, will redo.".puts;

goto RESET;

}

} while (lo < mid || p < 0.5);

return mid;

}

void main() {

auto rng = Xorshift(unpredictableSeed);

foreach (immutable uint nCollisions; 2 .. 6) {

double p, d;

immutable np = findHalfChance(nCollisions, p, d, rng);

writefln("%d collision: %d people, P = %g +/- %g", nCollisions, np, p, d);

}

}

- Output:

2 collision: 23 people, P = 0.521934 +/- 0.00728933 3 collision: 88 people, P = 0.512367 +/- 0.00411469 4 collision: 187 people, P = 0.506974 +/- 0.00232306 5 collision: 313 people, P = 0.501588 +/- 0.000529277

Output with nSigmas = 5.0:

2 collision: 23 people, P = 0.508607 +/- 0.00172133 3 collision: 88 people, P = 0.511945 +/- 0.00238885 4 collision: 187 people, P = 0.503229 +/- 0.000645587 5 collision: 313 people, P = 0.501105 +/- 0.000221016

## Hy[edit]

We use a simple but not very accurate simulation method.

(import

[numpy :as np]

[random [randint]])

(defmacro incf (place)

`(+= ~place 1))

(defn birthday [required &optional [reps 20000] [ndays 365]]

(setv days (np.zeros (, reps ndays) np.int_))

(setv qualifying-reps (np.zeros reps np.bool_))

(setv group-size 1)

(setv count 0)

(while True

;(print group-size)

(for [r (range reps)]

(unless (get qualifying-reps r)

(setv day (randint 0 (dec ndays)))

(incf (get days (, r day)))

(when (= (get days (, r day)) required)

(setv (get qualifying-reps r) True)

(incf count))))

(when (> (/ (float count) reps) .5)

(break))

(incf group-size))

group-size)

(print (birthday 2))

(print (birthday 3))

(print (birthday 4))

(print (birthday 5))

## J[edit]

Quicky approach (use a population of 1e5 people to get a quick estimate and then refine against a population of 1e8 people):

PopSmall=: 1e5 ?@# 365

PopBig=: 1e8 ?@# 365

countShared=: [: >./ #/.~

avg=: +/ % #

probShared=: (1 :0)("0)

:

NB. y: shared birthday count

NB. m: population

NB. x: sample size

avg ,y <: (-x) countShared\ m

)

estGroupSz=: 3 :0

approx=. (PopSmall probShared&y i.365) I. 0.5

n=. approx-(2+y)

refine=. n+(PopBig probShared&y approx+i:2+y) I. 0.5

assert. (2+y) > |approx-refine

refine, refine PopBig probShared y

)

Task cases:

estGroupSz 2

23 0.507254

estGroupSz 3

88 0.510737

estGroupSz 4

187 0.502878

estGroupSz 5

313 0.500903

So, for example, we need a group of 88 to have at least a 50% chance of 3 people in the group having the same birthday in a year of 365 days. And, in that case, the simulated probability was 51.0737%

## Java[edit]

import static java.util.Arrays.stream;

import java.util.Random;

public class Test {

static double equalBirthdays(int nSharers, int groupSize, int nRepetitions) {

Random rand = new Random(1);

int eq = 0;

for (int i = 0; i < nRepetitions; i++) {

int[] group = new int[365];

for (int j = 0; j < groupSize; j++)

group[rand.nextInt(group.length)]++;

eq += stream(group).anyMatch(c -> c >= nSharers) ? 1 : 0;

}

return (eq * 100.0) / nRepetitions;

}

public static void main(String[] a) {

int groupEst = 2;

for (int sharers = 2; sharers < 6; sharers++) {

// Coarse.

int groupSize = groupEst + 1;

while (equalBirthdays(sharers, groupSize, 100) < 50.0)

groupSize++;

// Finer.

int inf = (int) (groupSize - (groupSize - groupEst) / 4.0);

for (int gs = inf; gs < groupSize + 999; gs++) {

double eq = equalBirthdays(sharers, groupSize, 250);

if (eq > 50.0) {

groupSize = gs;

break;

}

}

// Finest.

for (int gs = groupSize - 1; gs < groupSize + 999; gs++) {

double eq = equalBirthdays(sharers, gs, 50_000);

if (eq > 50.0) {

groupEst = gs;

System.out.printf("%d independent people in a group of "

+ "%s share a common birthday. (%5.1f)%n",

sharers, gs, eq);

break;

}

}

}

}

}

2 independent people in a group of 23 share a common birthday. ( 50,6) 3 independent people in a group of 87 share a common birthday. ( 50,4) 4 independent people in a group of 187 share a common birthday. ( 50,1) 5 independent people in a group of 314 share a common birthday. ( 50,2)

## Lasso[edit]

if(sys_listunboundmethods !>> 'randomgen') => {

define randomgen(len::integer,max::integer)::array => {

#len <= 0 ? return

local(out = array)

loop(#len) => { #out->insert(math_random(#max,1)) }

return #out

}

}

if(sys_listunboundmethods !>> 'hasdupe') => {

define hasdupe(a::array,threshold::integer) => {

with i in #a do => {

#a->find(#i)->size > #threshold-1 ? return true

}

return false

}

}

local(threshold = 2)

local(qty = 22, probability = 0.00, samplesize = 10000)

while(#probability < 50.00) => {^

local(dupeqty = 0)

loop(#samplesize) => {

local(x = randomgen(#qty,365))

hasdupe(#x,#threshold) ? #dupeqty++

}

#probability = (#dupeqty / decimal(#samplesize)) * 100

'Threshold: '+#threshold+', qty: '+#qty+' - probability: '+#probability+'\r'

#qty += 1

^}

- Output:

Threshold: 2, qty: 22 - probability: 47.810000 Threshold: 2, qty: 23 - probability: 51.070000 Threshold: 3, qty: 86 - probability: 48.400000 Threshold: 3, qty: 87 - probability: 49.200000 Threshold: 3, qty: 88 - probability: 52.900000 Threshold: 4, qty: 184 - probability: 48.000000 Threshold: 4, qty: 185 - probability: 49.800000 Threshold: 4, qty: 186 - probability: 49.600000 Threshold: 4, qty: 187 - probability: 48.900000 Threshold: 4, qty: 188 - probability: 50.700000 Threshold: 5, qty: 308 - probability: 48.130000 Threshold: 5, qty: 309 - probability: 48.430000 Threshold: 5, qty: 310 - probability: 48.640000 Threshold: 5, qty: 311 - probability: 49.370000 Threshold: 5, qty: 312 - probability: 49.180000 Threshold: 5, qty: 313 - probability: 49.540000 Threshold: 5, qty: 314 - probability: 50.000000

## PARI/GP[edit]

simulate(n)=my(v=vecsort(vector(n,i,random(365))),t,c=1); for(i=2,n,if(v[i]>v[i-1],t=max(t,c);c=1,c++)); t

find(n)=my(guess=365*n-342,t);while(1, t=sum(i=1,1e3,simulate(guess)>=n)/1e3; if(t>550, guess--); if(t<450, guess++); if(450<=t && t<=550, return(guess)))

find(2)

find(3)

find(4)

find(5)

## PL/I[edit]

*process source attributes xref;

bd: Proc Options(main);

/*--------------------------------------------------------------------

* 04.11.2013 Walter Pachl

* Take samp samples of groups with gs persons and check

*how many of the groups have at least match persons with same birthday

*-------------------------------------------------------------------*/

Dcl (float,random) Builtin;

Dcl samp Bin Fixed(31) Init(1000000);

Dcl arr(0:366) Bin Fixed(31);

Dcl r Bin fixed(31);

Dcl i Bin fixed(31);

Dcl ok Bin fixed(31);

Dcl g Bin fixed(31);

Dcl gs Bin fixed(31);

Dcl match Bin fixed(31);

Dcl cnt(0:1) Bin Fixed(31);

Dcl lo(6) Bin Fixed(31) Init(0,21,85,185,311,458);

Dcl hi(6) Bin Fixed(31) Init(0,25,89,189,315,462);

Dcl rf Bin Float(63);

Dcl hits Bin Float(63);

Dcl arrow Char(3);

Do match=2 To 6;

Put Edit(' ')(Skip,a);

Put Edit(samp,' samples. Percentage of groups with at least',

match,' matches')(Skip,f(8),a,f(2),a);

Put Edit('Group size')(Skip,a);

Do gs=lo(match) To hi(match);

cnt=0;

Do i=1 To samp;

ok=0;

arr=0;

Do g=1 To gs;

rf=random();

r=rf*365+1;

arr(r)+=1;

If arr(r)=match Then Do;

/* Put Edit(r)(Skip,f(4));*/

ok=1;

End;

End;

cnt(ok)+=1;

End;

hits=float(cnt(1))/samp;

If hits>=.5 Then arrow=' <-';

Else arrow='';

Put Edit(gs,cnt(0),cnt(1),100*hits,'%',arrow)

(Skip,f(10),2(f(7)),f(8,3),a,a);

End;

End;

End;

Output:

1000000 samples. Percentage of groups with at least 2 matches Group size 3000000 500000 samples 21 556903 443097 44.310% 44.343% 44.347% 22 524741 475259 47.526% 47.549% 47.521% 23 492034 507966 50.797% <- 50.735% <- 50.722% <- 24 462172 537828 53.783% <- 53.815% <- 53.838% <- 25 431507 568493 56.849% <- 56.849% <- 56.842% <- 1000000 samples. Percentage of groups with at least 3 matches Group size 85 523287 476713 47.671% 47.638% 47.631% 86 512219 487781 48.778% 48.776% 48.821% 87 499874 500126 50.013% <- 49.902% 49.903% 88 488197 511803 51.180% <- 51.127% <- 51.096% <- 89 478044 521956 52.196% <- 52.263% <- 52.290% <- 1000000 samples. Percentage of groups with at least 4 matches Group size 185 511352 488648 48.865% 48.868% 48.921% 186 503888 496112 49.611% 49.601% 49.568% 187 497844 502156 50.216% <- 50.258% <- 50.297% <- 188 490490 509510 50.951% <- 50.916% <- 50.946% <- 189 482893 517107 51.711% <- 51.645% <- 51.655% <- 1000000 samples. Percentage of groups with at least 5 matches Group size 311 508743 491257 49.126% 49.158% 49.164% 312 503524 496476 49.648% 49.631% 49.596% 313 498244 501756 50.176% <- 50.139% <- 50.095% <- 314 494032 505968 50.597% <- 50.636% <- 50.586% <- 315 489821 510179 51.018% <- 51.107% <- 51.114% <- 1000000 samples. Percentage of groups with at least 6 matches Group size 458 505225 494775 49.478% 49.498% 49.512% 459 501871 498129 49.813% 49.893% 49.885% 460 497719 502281 50.228% <- 50.278% <- 50.248% <- 461 493948 506052 50.605% <- 50.622% <- 50.626% <- 462 489416 510584 51.058% <- 51.029% <- 51.055% <-

extended to verify REXX results:

1000000 samples. Percentage of groups with at least 7 matches Group size 621 503758 496242 49.624% 622 500320 499680 49.968% 623 497047 502953 50.295% <- 624 493679 506321 50.632% <- 625 491240 508760 50.876% <- 1000000 samples. Percentage of groups with at least 8 matches Group size 796 504764 495236 49.524% 797 502537 497463 49.746% 798 499488 500512 50.051% <- 799 496658 503342 50.334% <- 800 494773 505227 50.523% <- 1000000 samples. Percentage of groups with at least 9 matches Group size 983 502613 497387 49.739% 984 501665 498335 49.834% 985 498606 501394 50.139% <- 986 497453 502547 50.255% <- 987 493816 506184 50.618% <- 1000000 samples. Percentage of groups with at least10 matches Group size 1179 502910 497090 49.709% 1180 500906 499094 49.909% 1181 499079 500921 50.092% <- 1182 496957 503043 50.304% <- 1183 494414 505586 50.559% <-

## Perl 6[edit]

For a start, we can show off how to get the exact solution. If we pick n people, the total number of possible arrangements of birthdays is `365 ^{n}`. Among those possibilities, there are

`C`where all birthdays are different. For each of these, there are

^{n}_{365}`n!`possible ways to arrange the n people. So the solution is

`1 - n!C`, which in Perl 6 can be written:

^{n}_{365}/365^{n}say "$_ :", 1 - combinations(365, $_)/365**$_ * [*] 1..$_ for ^Inf

- Output:

0 : 0 1 : 0 2 : 0.002740 3 : 0.0082042 4 : 0.016355912 5 : 0.027135573700 6 : 0.04046248364911 7 : 0.0562357030959754 8 : 0.0743352923516690285 9 : 0.0946238338891667 ^C

Now comparing with a simulation :

sub theory($n) { 1 - combinations(365, $n)/365**$n* [*] 1..$n }

sub simulation(:number-of-people($n), :sample-size($N) = 1_000) {

$N R/ grep ?*, ((^365).roll($n).unique !== $n) xx $N;

}

for 2 .. Inf -> $n {

printf "%3d people, theory: %.4f, simulation: %.4f\n",

$n, theory($n), simulation(number-of-people => $n);

}

- Output:

2 people, theory: 0.0027, simulation: 0.0020 3 people, theory: 0.0082, simulation: 0.0080 4 people, theory: 0.0164, simulation: 0.0130 5 people, theory: 0.0271, simulation: 0.0260 6 people, theory: 0.0405, simulation: 0.0340 7 people, theory: 0.0562, simulation: 0.0590 8 people, theory: 0.0743, simulation: 0.0730 9 people, theory: 0.0946, simulation: 0.1080 10 people, theory: 0.1169, simulation: 0.1120 11 people, theory: 0.1411, simulation: 0.1220 12 people, theory: 0.1670, simulation: 0.1740 13 people, theory: 0.1944, simulation: 0.2200 14 people, theory: 0.2231, simulation: 0.2290 15 people, theory: 0.2529, simulation: 0.2540 16 people, theory: 0.2836, simulation: 0.2820 17 people, theory: 0.3150, simulation: 0.3190 18 people, theory: 0.3469, simulation: 0.3740 19 people, theory: 0.3791, simulation: 0.3720 20 people, theory: 0.4114, simulation: 0.3810 21 people, theory: 0.4437, simulation: 0.4340 22 people, theory: 0.4757, simulation: 0.4700 23 people, theory: 0.5073, simulation: 0.4960 24 people, theory: 0.5383, simulation: 0.5200 25 people, theory: 0.5687, simulation: 0.5990 26 people, theory: 0.5982, simulation: 0.5980 27 people, theory: 0.6269, simulation: 0.6520 28 people, theory: 0.6545, simulation: 0.6430 29 people, theory: 0.6810, simulation: 0.6690 30 people, theory: 0.7063, simulation: 0.7190 31 people, theory: 0.7305, simulation: 0.7450 ^C

## Python[edit]

Note: the first (unused), version of function equal_birthdays() uses a different but equally valid interpretation of the phrase "common birthday".

from random import randint

def equal_birthdays(sharers=2, groupsize=23, rep=100000):

'Note: 4 sharing common birthday may have 2 dates shared between two people each'

g = range(groupsize)

sh = sharers - 1

eq = sum((groupsize - len(set(randint(1,365) for i in g)) >= sh)

for j in range(rep))

return (eq * 100.) / rep

def equal_birthdays(sharers=2, groupsize=23, rep=100000):

'Note: 4 sharing common birthday must all share same common day'

g = range(groupsize)

sh = sharers - 1

eq = 0

for j in range(rep):

group = [randint(1,365) for i in g]

if (groupsize - len(set(group)) >= sh and

any( group.count(member) >= sharers for member in set(group))):

eq += 1

return (eq * 100.) / rep

group_est = [2]

for sharers in (2, 3, 4, 5):

groupsize = group_est[-1]+1

while equal_birthdays(sharers, groupsize, 100) < 50.:

# Coarse

groupsize += 1

for groupsize in range(int(groupsize - (groupsize - group_est[-1])/4.), groupsize + 999):

# Finer

eq = equal_birthdays(sharers, groupsize, 250)

if eq > 50.:

break

for groupsize in range(groupsize - 1, groupsize +999):

# Finest

eq = equal_birthdays(sharers, groupsize, 50000)

if eq > 50.:

break

group_est.append(groupsize)

print("%i independent people in a group of %s share a common birthday. (%5.1f)" % (sharers, groupsize, eq))

- Output:

2 independent people in a group of 23 share a common birthday. ( 50.9) 3 independent people in a group of 87 share a common birthday. ( 50.0) 4 independent people in a group of 188 share a common birthday. ( 50.9) 5 independent people in a group of 314 share a common birthday. ( 50.6)

### Enumeration method[edit]

The following enumerates all birthday distributation of n people in a year. It's patentedly unscalable.

from collections import defaultdict

days = 365

def find_half(c):

# inc_people takes birthday combinations of n people and generates the

# new set for n+1

def inc_people(din, over):

# 'over' is the number of combinations that have at least c people

# sharing a birthday. These are not contained in the set.

dout,over = defaultdict(int), over * days

for k,s in din.items():

for i,v in enumerate(k):

if v + 1 >= c:

over += s

else:

dout[tuple(sorted(k[0:i] + (v + 1,) + k[i+1:]))] += s

dout[(1,) + k] += s * (days - len(k))

return dout, over

d, combos, good, n = {():1}, 1, 0, 0

# increase number of people until at least half of the cases have at

# at least c people sharing a birthday

while True:

n += 1

combos *= days # or, combos = sum(d.values()) + good

d,good = inc_people(d, good)

#!!! print d.items()

if good * 2 >= combos:

return n, good, combos

# In all fairness, I don't know if the code works for x >= 4: I probably don't

# have enough RAM for it, and certainly not enough patience. But it should.

# In theory.

for x in range(2, 5):

n, good, combos = find_half(x)

print "%d of %d people sharing birthday: %d out of %d combos"% (x, n, good, combos)

- Output:

2 of 23 people sharing birthday: 43450860051057961364418604769486195435604861663267741453125 out of 85651679353150321236814267844395152689354622364044189453125 combos 3 of 88 people sharing birthday: 1549702400401473425983277424737696914087385196361193892581987189461901608374448849589919219974092878625057027641693544686424625999709818279964664633586995549680467629183956971001416481439048256933422687688148710727691650390625 out of 3032299345394764867793392128292779133654078653518318790345269064871742118915665927782934165016667902517875712171754287171746462419635313222013443107339730598579399174951673950890087953259632858049599235528148710727691650390625 combos ...?

### Enumeration method #2[edit]

# ought to use a memoize class for all this

# factorial

def fact(n, cache={0:1}):

if not n in cache:

cache[n] = n * fact(n - 1)

return cache[n]

# permutations

def perm(n, k, cache={}):

if not (n,k) in cache:

cache[(n,k)] = fact(n) / fact(n - k)

return cache[(n,k)]

def choose(n, k, cache={}):

if not (n,k) in cache:

cache[(n,k)] = perm(n, k) / fact(k)

return cache[(n, k)]

# ways of distribute p people's birthdays into d days, with

# no more than m sharing any one day

def combos(d, p, m, cache={}):

if not p: return 1

if not m: return 0

if p <= m: return d**p # any combo would satisfy

k = (d, p, m)

if not k in cache:

result = 0

for x in range(0, p//m + 1):

c = combos(d - x, p - x * m, m - 1)

# ways to occupy x days with m people each

if c: result += c * choose(d, x) * perm(p, x * m) / fact(m)**x

cache[k] = result

return cache[k]

def find_half(m):

n = 0

while True:

n += 1

total = 365 ** n

c = total - combos(365, n, m - 1)

if c * 2 >= total:

print "%d of %d people: %d/%d combos" % (n, m, c, total)

return

for x in range(2, 6): find_half(x)

- Output:

23 of 2 people: 43450860....3125/85651679....3125 combos 88 of 3 people: 15497...50390625/30322...50390625 combos 187 of 4 people: 708046698...0703125/1408528546...0703125 combos 313 of 5 people: 498385488882289...2578125/99464149835930...2578125 combos

## Racket[edit]

Based on the Python task. For three digits precision use 250000 repetitions. For four digits precision use 25000000 repetitions, but it’s very slow. See discussion page.#lang racket

#;(define repetitions 25000000) ; for \sigma=1/10000

(define repetitions 250000) ; for \sigma=1/1000

(define coarse-repetitions 2500)

(define (vector-inc! v pos)

(vector-set! v pos (add1 (vector-ref v pos))))

(define (equal-birthdays sharers group-size repetitions)

(/ (for/sum ([j (in-range repetitions)])

(let ([days (make-vector 365 0)])

(for ([person (in-range group-size)])

(vector-inc! days (random 365)))

(if (>= (apply max (vector->list days)) sharers)

1 0)))

repetitions))

(define (search-coarse-group-size sharers)

(let loop ([coarse-group-size 2])

(let ([coarse-probability

(equal-birthdays sharers coarse-group-size coarse-repetitions)])

(if (> coarse-probability .5)

coarse-group-size

(loop (add1 coarse-group-size))))))

(define (search-upwards sharers group-size)

(let ([probability (equal-birthdays sharers group-size repetitions)])

(if (> probability .5)

(values group-size probability)

(search-upwards sharers (add1 group-size)))))

(define (search-downwards sharers group-size last-probability)

(let ([probability (equal-birthdays sharers group-size repetitions)])

(if (> probability .5)

(search-downwards sharers (sub1 group-size) probability)

(values (add1 group-size) last-probability))))

(define (search-from sharers group-size)

(let ([probability (equal-birthdays sharers group-size repetitions)])

(if (> probability .5)

(search-downwards sharers (sub1 group-size) probability)

(search-upwards sharers (add1 group-size)))))

(for ([sharers (in-range 2 6)])

(let-values ([(group-size probability)

(search-from sharers (search-coarse-group-size sharers))])

(printf "~a independent people in a group of ~a share a common birthday. (~a%)\n"

sharers group-size (~r (* probability 100) #:precision '(= 2)))))

**Output**

2 independent people in a group of 23 share a common birthday. (50.80%) 3 independent people in a group of 88 share a common birthday. (51.19%) 4 independent people in a group of 187 share a common birthday. (50.18%) 5 independent people in a group of 313 share a common birthday. (50.17%)

## REXX[edit]

### version 1[edit]

The method used is to find the average number of people to share a common birthday, and then use the **floor** of

that value (less the group size) as a starting point to find a new group size with an expected size that exceeds

50% common birthdays of the required size.

/*REXX program examines the birthday problem via random number simulation. */

parse arg grps samp seed . /*get optional arguments from the CL. */

if grps=='' | grps==',' then grps=5 /*Not specified? Then use the default.*/

if samp=='' | samp==',' then samp=100000 /* " " " " " " */

if datatype(seed,'W') then call random ,,seed /*RANDOM seed given for repeatability ?*/

diy =365 /*or: diy=365.25*/ /*the number of Days In a Year. */

diyM=diy*100 /*this expands the RANDOM (BIF) range.*/

/* [↓] get a rough estimate for %. */

do g=2 to grps; s=0 /*perform through 2 ──► group size. */

do samp; @.=0 /*perform some number of trials. */

do j=1 until @.day==g /*perform until G dup. birthdays found.*/

day=random(1,diyM) % 100 /*expand range RANDOM number generation*/

@.day=@.day+1 /*record the number of common birthdays*/

end /*j*/ /* [↓] adjust for the DO loop index.*/

s=s+j-1 /*add number of birthday hits to sum. */

end /*samp*/

start.g=s/samp%1-g /*define where the try-outs start. */

end /*g*/

say

say right('sample size is ' samp,40); say /*display this run's sample size. */

say ' required group % with required'

say ' common size common birthdays'

say ' ──────── ───── ────────────────'

/* [↓] where the try-outs happen. */

do g=2 to grps /*perform through 2 ──► group size. */

do try=start.g; s=0 /*perform try-outs until average > 50%.*/

do samp; @.=0 /*perform some number of trials. */

do try /*perform until G dup. birthdays found.*/

day=random(1,diyM) % 100 /*expand range RANDOM number generation*/

@.day=@.day+1 /*record the number of common birthdays*/

if @.day\==g then iterate /*not enough G (birthday) hits found ? */

s=s+1 /*another common birthday found. */

leave /* ··· and stop looking for more. */

end /*try;*/ /* [↓] bump the counter for Bday hits.*/

end /*samp*/

if s/samp>.5 then leave /*if the average is > 50%, then stop. */

end /*try=start.g*/

say right(g, 15) right(try, 15) center( format(s/samp*100,,5)'%', 30)

end /*g*/ /*stick a fork in it, we're all done. */

**output** when using the input of: ` 10 `

sample size is 100000 required group % with required common size common birthdays ──────── ───── ──────────────── 2 23 50.70900% 3 88 51.23200% 4 187 50.15100% 5 314 50.77800% 6 460 50.00600% 7 623 50.64800% 8 798 50.00700% 9 985 50.13400% 10 1181 50.22200%

### version 2[edit]

/*--------------------------------------------------------------------

* 04.11.2013 Walter Pachl translated from PL/I

* Take samp samples of groups with gs persons and check

*how many of the groups have at least match persons with same birthday

*-------------------------------------------------------------------*/

samp=100000

lo='0 21 85 185 311 458'

hi='0 25 89 189 315 462'

Do match=2 To 6

Say ' '

Say samp' samples . Percentage of groups with at least',

match ' matches'

Say 'Group size'

Do gs=word(lo,match) To word(hi,match)

cnt.=0

Do i=1 To samp

ok=0

arr.=0

Do g=1 To gs

r=random(1,365)

arr.r=arr.r+1

If arr.r=match Then

ok=1

End

cnt.ok=cnt.ok+1

End

hits=cnt.1/samp

If hits>=.5 Then arrow=' <-'

Else arrow=''

Say format(gs,10) cnt.0 cnt.1 100*hits||'%'||arrow

End

End

Output:

100000 samples . Percentage of groups with at least 2 matches Group size 21 55737 44263 44.26300% 22 52158 47842 47.84200% 23 49141 50859 50.85900% <- 24 46227 53773 53.77300% <- 25 43091 56909 56.90900% <- 100000 samples . Percentage of groups with at least 3 matches Group size 85 52193 47807 47.80700% 86 51489 48511 48.51100% 87 50146 49854 49.85400% 88 48790 51210 51.2100% <- 89 47771 52229 52.22900% <- 100000 samples . Percentage of groups with at least 4 matches Group size 185 50930 49070 49.0700% 186 50506 49494 49.49400% 187 49739 50261 50.26100% <- 188 49024 50976 50.97600% <- 189 48283 51717 51.71700% <- 100000 samples . Percentage of groups with at least 5 matches Group size 311 50909 49091 49.09100% 312 50441 49559 49.55900% 313 49912 50088 50.08800% <- 314 49425 50575 50.57500% <- 315 48930 51070 51.0700% <- 100000 samples . Percentage of groups with at least 6 matches Group size 458 50580 49420 49.4200% 459 49848 50152 50.15200% <- 460 49975 50025 50.02500% <- 461 49316 50684 50.68400% <- 462 49121 50879 50.87900% <-

## Tcl[edit]

proc birthdays {num {same 2}} {

for {set i 0} {$i < $num} {incr i} {

set b [expr {int(rand() * 365)}]

if {[incr bs($b)] >= $same} {

return 1

}

}

return 0

}

proc estimateBirthdayChance {num same} {

# Gives a reasonably close estimate with minimal execution time; the idea

# is to keep the amount that one random value may influence the result

# fairly constant.

set count [expr {$num * 100 / $same}]

set x 0

for {set i 0} {$i < $count} {incr i} {

incr x [birthdays $num $same]

}

return [expr {double($x) / $count}]

}

foreach {count from to} {2 20 25 3 85 90 4 183 190 5 310 315} {

puts "identifying level for $count people with same birthday"

for {set i $from} {$i <= $to} {incr i} {

set chance [estimateBirthdayChance $i $count]

puts [format "%d people => %%%.2f chance of %d people with same birthday" \

$i [expr {$chance * 100}] $count]

if {$chance >= 0.5} {

puts "level found: $i people"

break

}

}

}

- Output:

identifying level for 2 people with same birthday 20 people => %43.40 chance of 2 people with same birthday 21 people => %44.00 chance of 2 people with same birthday 22 people => %46.91 chance of 2 people with same birthday 23 people => %53.48 chance of 2 people with same birthday level found: 23 people identifying level for 3 people with same birthday 85 people => %47.97 chance of 3 people with same birthday 86 people => %48.46 chance of 3 people with same birthday 87 people => %49.55 chance of 3 people with same birthday 88 people => %50.66 chance of 3 people with same birthday level found: 88 people identifying level for 4 people with same birthday 183 people => %48.02 chance of 4 people with same birthday 184 people => %47.67 chance of 4 people with same birthday 185 people => %48.89 chance of 4 people with same birthday 186 people => %49.98 chance of 4 people with same birthday 187 people => %50.99 chance of 4 people with same birthday level found: 187 people identifying level for 5 people with same birthday 310 people => %48.52 chance of 5 people with same birthday 311 people => %48.14 chance of 5 people with same birthday 312 people => %49.07 chance of 5 people with same birthday 313 people => %49.63 chance of 5 people with same birthday 314 people => %49.59 chance of 5 people with same birthday 315 people => %51.79 chance of 5 people with same birthday level found: 315 people

## zkl[edit]

Pure simulation; adding a person to a population until there are the required number of collisions, then repeating that a bunch of times to get an average.

fcn bdays(N){ // N is shared birthdays in a population

year:=(0).pump(365,List.createLong(365).write,0); // 365 days == one year

shared:=people:=0; do{ // add a new person to population

bday:=(0).random(365); // with this birthday [0..364]

shared=shared.max(year[bday]+=1); people+=1;

}while(shared<N);

people // size of simulated population that contains N shared birthdays

}

fcn simulate(N,T){ avg:=0.0; do(T){ avg+=bdays(N) } avg/=T; } // N shared, T trials

foreach n in ([1..5]){

println("Average of %d people in a populatation of %s share birthdays"

.fmt(n,simulate(n,0d10_000)));

}

- Output:

Average of 1 people in a populatation of 1 share birthdays Average of 2 people in a populatation of 24.7199 share birthdays Average of 3 people in a populatation of 88.6416 share birthdays Average of 4 people in a populatation of 186.849 share birthdays Average of 5 people in a populatation of 312.399 share birthdays