# 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 group 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

- Wolfram entry: Birthday Problem

## 11l

```
F equal_birthdays(sharers, groupsize, rep)
V eq = 0
L 0 .< rep
V group = [0] * 365
L 0 .< groupsize
group[random:(group.len)]++
I any(group.map(c -> c >= @sharers))
eq++
R (eq * 100.) / rep
V group_est = 2
L(sharers) 2..5
V groupsize = group_est + 1
L equal_birthdays(sharers, groupsize, 100) < 50.
groupsize++
L(gs) Int(groupsize - (groupsize - group_est) / 4.) .< groupsize + 999
V eq = equal_birthdays(sharers, gs, 250)
I eq > 50.
groupsize = gs
L.break
L(gs) groupsize - 1 .< groupsize + 999
V eq = equal_birthdays(sharers, gs, 50'000)
I eq > 50.
group_est = gs
print(‘#. independent people in a group of #. share a common birthday. (#3.1)’.format(sharers, gs, eq))
L.break
```

- Output:

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. ( 50.1) 4 independent people in a group of 187 share a common birthday. ( 50.5) 5 independent people in a group of 313 share a common birthday. ( 50.4)

## Ada

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

**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

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 probability 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

## C++

```
#include <iostream>
#include <random>
#include <vector>
double equalBirthdays(int nSharers, int groupSize, int nRepetitions) {
std::default_random_engine generator;
std::uniform_int_distribution<int> distribution(0, 364);
std::vector<int> group(365);
int eq = 0;
for (int i = 0; i < nRepetitions; i++) {
std::fill(group.begin(), group.end(), 0);
for (int j = 0; j < groupSize; j++) {
int day = distribution(generator);
group[day]++;
}
if (std::any_of(group.cbegin(), group.cend(), [nSharers](int c) { return c >= nSharers; })) {
eq++;
}
}
return (100.0 * eq) / nRepetitions;
}
int main() {
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) / 4.0f);
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, 50000);
if (eq > 50.0) {
groupEst = gs;
printf("%d independant people in a group of %d share a common birthday. (%5.1f)\n", sharers, gs, eq);
break;
}
}
}
return 0;
}
```

- Output:

2 independant people in a group of 23 share a common birthday. ( 50.6) 3 independant people in a group of 87 share a common birthday. ( 50.2) 4 independant people in a group of 186 share a common birthday. ( 50.0) 5 independant people in a group of 313 share a common birthday. ( 50.5)

## D

```
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 /*@nogc*/ {
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

## Delphi

```
program Birthday_problem;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
const
_DAYS = 365;
var
days: array[0..(_DAYS) - 1] of Integer;
runs: Integer;
function rand_day: Integer; inline;
begin
Result := random(_DAYS);
end;
/// <summary>
/// given p people, if n of them have same birthday in one run
/// </summary>
function simulate1(p, n: Integer): Integer;
var
index, i: Integer;
begin
for i := 0 to High(days) do
begin
days[i] := 0;
end;
for i := 0 to p - 1 do
begin
index := rand_day();
inc(days[index]);
if days[index] = n then
Exit(1);
end;
Exit(0);
end;
/// <summary>
/// decide if the probability 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
/// </summary>
function prob(np, n: Integer; n_sigmas, p_thresh: Double; var d: Double): Double;
var
p: Double;
runs, yes: Integer;
begin
runs := 0;
yes := 0;
repeat
yes := yes + (simulate1(np, n));
inc(runs);
p := yes / runs;
d := sqrt(p * (1 - p) / runs);
until ((runs >= 10) and (abs(p - p_thresh) >= (n_sigmas * d)));
Exit(p);
end;
/// <summary>
/// bisect for truth
/// </summary>
function find_half_chance(n: Integer; var p: Double; var d: Double): Integer;
var
lo, hi, mid: Integer;
label
reset;
begin
reset:
lo := 0;
hi := _DAYS * (n - 1) + 1;
repeat
mid := (hi + lo) div 2;
p := prob(mid, n, 3, 0.5, d);
if p < 0.5 then
lo := mid + 1
else
hi := mid;
if hi < lo then
goto reset;
until ((lo >= mid) and (p >= 0.5));
Exit(mid);
end;
var
n, np: Integer;
p, d: Double;
begin
Randomize;
writeln('Wait for calculate');
for n := 2 to 5 do
begin
np := find_half_chance(n, p, d);
writeln(format('%d collision: %d people, P = %.8f +/- %.8f', [n, np, p, d]));
end;
writeln('Press enter to exit');
readln;
end.
```

- Output:

Wait for calculate 2 collision: 23 people, P = 0,50411600 +/- 0,00137151 3 collision: 88 people, P = 0,51060651 +/- 0,00352751 4 collision: 188 people, P = 0,50856493 +/- 0,00285022 5 collision: 313 people, P = 0,50093354 +/- 0,00031116 Press enter to exit

## FreeBASIC

```
Function Simulacion(N As Integer) As Integer
Dim As Integer i, dias(365)
For i = 0 To 365-1
dias(i) = 0
Next i
Dim As Integer R, personas = 0
Do
R = Rnd * 365
dias(R) += 1
personas += 1
If dias(R) = N Then Return personas
Loop
End Function
Dim As Integer N, grupo, t
For N = 2 To 5
grupo = 0
For t = 1 To 10000
grupo += Simulacion(N)
Next t
Print Using "Average of # people in a population of ### share birthdays"; N; Int(grupo/10000)
Next N
Sleep
```

## Go

```
package main
import (
"fmt"
"math"
"math/rand"
"time"
)
const (
DEBUG = 0
DAYS = 365
n_sigmas = 5.
WORKERS = 16 // concurrent worker processes
RUNS = 1000 // runs per flight
)
func simulate1(p, n int, r *rand.Rand) int {
var days [DAYS]int
for i := 0; i < p; i++ {
days[r.Intn(DAYS)]++
}
for _, d := range days {
if d >= n {
return 1
}
}
return 0
}
// send yes's per fixed number of simulate1 runs until canceled
func work(p, n int, ych chan int, cancel chan bool) {
r := rand.New(rand.NewSource(time.Now().Unix() + rand.Int63()))
for {
select {
case <-cancel:
return
default:
}
y := 0
for i := 0; i < RUNS; i++ {
y += simulate1(p, n, r)
}
ych <- y
}
}
func prob(np, n int) (p, d float64) {
ych := make(chan int, WORKERS)
cancel := make(chan bool)
for i := 0; i < WORKERS; i++ {
go work(np, n, ych, cancel)
}
var runs, yes int
for {
yes += <-ych
runs += RUNS
fr := float64(runs)
p = float64(yes) / fr
d = math.Sqrt(p * (1 - p) / fr)
if DEBUG > 1 {
fmt.Println("\t\t", np, yes, runs, p, d)
}
// .5 here is the "even chance" threshold
if !(math.Abs(p-.5) < n_sigmas*d) {
close(cancel)
break
}
}
if DEBUG > 1 {
fmt.Println()
}
return
}
func find_half_chance(n int) (mid int, p, dev float64) {
reset:
lo := 0
hi := DAYS*(n-1) + 1
for {
mid = (hi + lo) / 2
p, dev = prob(mid, n)
if DEBUG > 0 {
fmt.Println("\t", lo, mid, hi, p, dev)
}
if p < .5 {
lo = mid + 1
} else {
hi = mid
}
if hi < lo {
if DEBUG > 0 {
fmt.Println("\tMade a mess, will redo.")
}
goto reset
}
if !(lo < mid || p < .5) {
break
}
}
return
}
func main() {
for n := 2; n <= 5; n++ {
np, p, d := find_half_chance(n)
fmt.Printf("%d collision: %d people, P = %.4f ± %.4f\n",
n, np, p, d)
}
}
```

2 collision: 23 people, P = 0.5081 ± 0.0016 3 collision: 88 people, P = 0.5155 ± 0.0029 4 collision: 187 people, P = 0.5041 ± 0.0008 5 collision: 313 people, P = 0.5015 ± 0.0003

**Also based on the C version:**

```
package main
import (
"fmt"
"math"
"math/rand"
"runtime"
"time"
)
type ProbeRes struct {
np int
p, d float64
}
type Frac struct {
n int
d int
}
var DaysInYear int = 365
func main() {
sigma := 5.0
for i := 2; i <= 5; i++ {
res := GetNP(i, sigma, 0.5)
fmt.Printf("%d collision: %d people, P = %.4f ± %.4f\n",
i, res.np, res.p, res.d)
}
}
func GetNP(n int, n_sigmas, p_thresh float64) (res ProbeRes) {
res.np = DaysInYear * (n - 1)
for i := 0; i < DaysInYear*(n-1); i++ {
tmp := probe(i, n, n_sigmas, p_thresh)
if tmp.p > p_thresh && tmp.np < res.np {
res = tmp
}
}
return
}
var numCPU = runtime.NumCPU()
func probe(np, n int, n_sigmas, p_thresh float64) ProbeRes {
var p, d float64
var runs, yes int
cRes := make(chan Frac, numCPU)
for i := 0; i < numCPU; i++ {
go SimN(np, n, 25, cRes)
}
for math.Abs(p-p_thresh) < n_sigmas*d || runs < 100 {
f := <-cRes
yes += f.n
runs += f.d
p = float64(yes) / float64(runs)
d = math.Sqrt(p * (1 - p) / float64(runs))
go SimN(np, n, runs/3, cRes)
}
return ProbeRes{np, p, d}
}
func SimN(np, n, ssize int, c chan Frac) {
r := rand.New(rand.NewSource(time.Now().UnixNano() + rand.Int63()))
yes := 0
for i := 0; i < ssize; i++ {
if Sim(np, n, r) {
yes++
}
}
c <- Frac{yes, ssize}
}
func Sim(p, n int, r *rand.Rand) (res bool) {
Cal := make([]int, DaysInYear)
for i := 0; i < p; i++ {
Cal[r.Intn(DaysInYear)]++
}
for _, v := range Cal {
if v >= n {
res = true
}
}
return
}
```

- Output:

2 collision: 23 people, P = 0.5068 ± 0.0013 3 collision: 88 people, P = 0.5148 ± 0.0028 4 collision: 187 people, P = 0.5020 ± 0.0004 5 collision: 313 people, P = 0.5011 ± 0.0002

## Hy

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

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

```
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)

## Julia

```
function equalbirthdays(sharers::Int, groupsize::Int; nrep::Int = 10000)
eq = 0
for _ in 1:nrep
group = rand(1:365, groupsize)
grset = Set(group)
if groupsize - length(grset) ≥ sharers - 1 &&
any(count(x -> x == d, group) ≥ sharers for d in grset)
eq += 1
end
end
return eq / nrep
end
gsizes = [2]
for sh in (2, 3, 4, 5)
local gsize = gsizes[end]
local freq
# Coarse
while equalbirthdays(sh, gsize; nrep = 100) < .5
gsize += 1
end
# Finer
for gsize in trunc(Int, gsize - (gsize - gsizes[end]) / 4):(gsize + 999)
if equalbirthdays(sh, gsize; nrep = 250) > 0.5
break
end
end
# Finest
for gsize in (gsize - 1):(gsize + 999)
freq = equalbirthdays(sh, gsize; nrep = 50000)
if freq > 0.5
break
end
end
push!(gsizes, gsize)
@printf("%i independent people in a group of %s share a common birthday. (%5.3f)\n", sh, gsize, freq)
end
```

- Output:

2 independent people in a group of 23 share a common birthday. (0.506) 3 independent people in a group of 88 share a common birthday. (0.510) 4 independent people in a group of 187 share a common birthday. (0.500) 5 independent people in a group of 314 share a common birthday. (0.507)

## Kotlin

```
// version 1.1.3
import java.util.Random
fun equalBirthdays(nSharers: Int, groupSize: Int, nRepetitions: Int): Double {
val rand = Random(1L)
var eq = 0
for (i in 0 until nRepetitions) {
val group = IntArray(365)
for (j in 0 until groupSize) {
group[rand.nextInt(group.size)]++
}
eq += if (group.any { it >= nSharers}) 1 else 0
}
return eq * 100.0 / nRepetitions
}
fun main(args: Array<String>) {
var groupEst = 2
for (sharers in 2..5) {
// Coarse
var groupSize = groupEst + 1
while (equalBirthdays(sharers, groupSize, 100) < 50.0) groupSize++
// Finer
val inf = (groupSize - (groupSize - groupEst) / 4.0).toInt()
for (gs in inf until groupSize + 999) {
val eq = equalBirthdays(sharers, groupSize, 250)
if (eq > 50.0) {
groupSize = gs
break
}
}
// Finest
for (gs in groupSize - 1 until groupSize + 999) {
val eq = equalBirthdays(sharers, gs, 50_000)
if (eq > 50.0) {
groupEst = gs
print("$sharers independent people in a group of ${"%3d".format(gs)} ")
println("share a common birthday (${"%2.1f%%".format(eq)})")
break
}
}
}
}
```

- Output:

Expect runtime of about 15 seconds on a modest laptop:

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

```
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

## Nim

```
import random, sequtils, strformat
proc equalBirthdays(nSharers, groupSize, nRepetitions: int): float =
randomize(1)
var eq = 0
for _ in 1..nRepetitions:
var group: array[1..365, int]
for _ in 1..groupSize:
inc group[rand(1..group.len)]
eq += ord(group.anyIt(it >= nSharers))
result = eq * 100 / nRepetitions
proc main() =
var groupEst = 2
for sharers in 2..5:
# Coarse.
var groupSize = groupEst + 1
while equalBirthdays(sharers, groupSize, 100) < 50:
inc groupSize
# Finer.
let inf = (groupSize.toFloat - (groupSize - groupEst) / 4).toInt()
for gs in inf..(groupSize+998):
let eq = equalBirthdays(sharers, groupSize, 250)
if eq > 50:
groupSize = gs
break
# Finest.
for gs in (groupSize-1)..(groupSize+998):
let eq = equalBirthdays(sharers, gs, 50_000)
if eq > 50:
groupEst = gs
echo &"{sharers} independent people in a group of {gs:3} ",
&"share a common birthday ({eq:4.1f}%)"
break
main()
```

- Output:

2 independent people in a group of 23 share a common birthday (50.7%) 3 independent people in a group of 87 share a common birthday (50.0%) 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 (50.2%)

## PARI/GP

```
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)
```

## Perl

```
use strict;
use warnings;
use List::AllUtils qw(max min uniqnum count_by any);
use Math::Random qw(random_uniform_integer);
sub simulation {
my($c) = shift;
my $max_trials = 1_000_000;
my $min_trials = 10_000;
my $n = int 47 * ($c-1.5)**1.5; # OEIS/A050256: 16 86 185 307
my $N = min $max_trials, max $min_trials, 1000 * sqrt $n;
while (1) {
my $yes = 0;
for (1..$N) {
my %birthday_freq = count_by { $_ } random_uniform_integer($n, 1, 365);
$yes++ if any { $birthday_freq{$_} >= $c } keys %birthday_freq;
}
my $p = $yes/$N;
return($n, $p) if $p > 0.5;
$N = min $max_trials, max $min_trials, int 1000/(0.5-$p)**1.75;
$n++;
}
}
printf "$_ people in a group of %s share a common birthday. (%.4f)\n", simulation($_) for 2..5
```

- Output:

2 people in a group of 23 share a common birthday. (0.5083) 3 people in a group of 88 share a common birthday. (0.5120) 4 people in a group of 187 share a common birthday. (0.5034) 5 people in a group of 313 share a common birthday. (0.5008)

## Phix

constant nDays = 365 -- 5 sigma confidence. Conventionally people think 3 sigmas are -- good enough, but for the case of 5 people sharing a birthday, -- 3 sigmas actually sometimes gives a slightly wrong answer. constant nSigmas = 5.0; -- Change to 3 for smaller run time. function simulate1(integer nPeople, nCollisions) -- -- Given n people, if m of them have same birthday in one run. -- sequence days = repeat(0,nDays) for p=1 to nPeople do integer day = rand(nDays) days[day] += 1 if days[day] == nCollisions then return true end if end for return false; end function function prob(integer np, nCollisions, atom pThresh) -- -- 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. -- atom p, d; -- Probablity and standard deviation. integer nRuns = 0, yes = 0; while nRuns<10 or (abs(p - pThresh) < (nSigmas * d)) do yes += simulate1(np, nCollisions) nRuns += 1 p = yes/nRuns d = sqrt(p * (1 - p) / nRuns); end while return {p,d} end function function findHalfChance(integer nCollisions) -- Bisect for truth. atom p, dev integer mid = 1, lo = 0, hi = nDays * (nCollisions - 1) + 1; while lo < mid or p < 0.5 do mid = floor((hi + lo) / 2) {p,dev} = prob(mid, nCollisions, 0.5) if (p < 0.5) then lo = mid + 1; else hi = mid; end if if (hi < lo) then return findHalfChance(nCollisions) -- reset end if end while return {p,dev,mid} end function for nCollisions=2 to 6 do atom {p,d,np} = findHalfChance(nCollisions) printf(1,"%d collision: %d people, P = %g +/- %g\n", {nCollisions, np, p, d}) end for

- Output:

2 collision: 23 people, P = 0.520699 +/- 0.00688426 3 collision: 88 people, P = 0.507159 +/- 0.00238534 4 collision: 187 people, P = 0.504129 +/- 0.00137625 5 collision: 313 people, P = 0.501219 +/- 0.000406284 6 collision: 460 people, P = 0.502131 +/- 0.000710091

Output with nSigmas = 5.0:

2 collision: 23 people, P = 0.507817 +/- 0.00156278 3 collision: 88 people, P = 0.512042 +/- 0.00240772 4 collision: 187 people, P = 0.502546 +/- 0.000509275 5 collision: 313 people, P = 0.501218 +/- 0.000243516 6 collision: 460 people, P = 0.502901 +/- 0.000580137

## PL/I

```
*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% <-

## Python

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

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

```
# 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

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%)

## Raku

(formerly Perl 6) Gives correct answers, but more of a proof-of-concept at this point, even with max-trials at 250K it is too slow to be practical.

```
sub simulation ($c) {
my $max-trials = 250_000;
my $min-trials = 5_000;
my $n = floor 47 * ($c-1.5)**1.5; # OEIS/A050256: 16 86 185 307
my $N = min $max-trials, max $min-trials, 1000 * sqrt $n;
loop {
my $p = $N R/ elems grep { .elems > 0 }, ((grep { $_>=$c }, values bag (^365).roll($n)) xx $N);
return($n, $p) if $p > 0.5;
$N = min $max-trials, max $min-trials, floor 1000/(0.5-$p);
$n++;
}
}
printf "$_ people in a group of %s share a common birthday. (%.3f)\n", simulation($_) for 2..5;
```

- Output:

2 people in a group of 23 share a common birthday. (0.506) 3 people in a group of 88 share a common birthday. (0.511) 4 people in a group of 187 share a common birthday. (0.500) 5 people in a group of 313 share a common birthday. (0.507)

## REXX

### version 1

The root finding method used is to find the average number of people to share a 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% duplicate birthdays of the required size.

This REXX version doesn't need a precalculated group size to find the percentage required to exceed 50%.

```
/*REXX pgm examines the birthday problem via random# simulation (with specifiable parms)*/
parse arg dups samp seed . /*get optional arguments from the CL. */
if dups=='' | dups=="," then dups= 10 /*Not specified? Then use the default.*/
if samp=='' | samp=="," then samp= 10000 /* " " " " " " */
if datatype(seed, 'W') then call random ,,seed /*RANDOM seed given for repeatability ?*/
diy = 365 /*alternative: diy=365.25*/ /*the number of Days In a Year. */
diyM= diy*100 /*this expands the RANDOM (BIF) range.*/
do g=2 to dups; s= 0 /*perform through 2 ──► duplicate size*/
do samp; @.= 0 /*perform some number of trials. */
do j=0 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 /*add number of birthday hits to sum. */
end /*samp*/ /* [↓] % 1 rounds down the division.*/
start.g= s/samp % 1 - g /*define where the try─outs start. */
end /*g*/ /* [↑] get a rough estimate for %. */
say right('sample size is ' samp, 40); say /*display this run's sample size. */
say ' required trial % with required'
say ' duplicates size common birthdays'
say ' ──────────── ─────── ──────────────────'
do g=2 to dups /*perform through 2 ──► duplicate size*/
do try=start.g until s/samp>=.5; s= 0 /* " try─outs until average ≥ 50%.*/
do samp; @.= 0 /* " some number of trials. */
do try; day= random(1, diyM) % 100 /* " until G dup. birthdays found.*/
@.day= @.day + 1 /*record the number of common birthdays*/
if @.day==g then do; s=s+1; leave; end /*found enough G (birthday) hits ? */
end /*try;*/
end /*samp*/
end /*try=start.g*/ /* [↑] where the try─outs happen. */
say right(g, 15) right(try, 15) center( format( s / samp * 100, , 4)'%', 30)
end /*g*/ /*stick a fork in it, we're all done. */
```

- output when using the default inputs:

sample size is 10000 required trial % with required duplicates size common birthdays ──────────── ─────── ────────────────── 2 23 50.2300% 3 87 50.2400% 4 187 50.3800% 5 312 50.0100% 6 458 50.5200% 7 622 50.3900% 8 798 50.1700% 9 984 50.5700% 10 1182 51.4000%

### version 2

```
/*--------------------------------------------------------------------
* 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% <-

## Ruby

```
def equalBirthdays(nSharers, groupSize, nRepetitions)
eq = 0
for i in 1 .. nRepetitions
group = [0] * 365
for j in 1 .. groupSize
group[rand(group.length)] += 1
end
eq += group.any? { |n| n >= nSharers } ? 1 : 0
end
return (eq * 100.0) / nRepetitions
end
def main
groupEst = 2
for sharers in 2 .. 5
# Coarse
groupSize = groupEst + 1
while equalBirthdays(sharers, groupSize, 100) < 50.0
groupSize += 1
end
# Finer
inf = (groupSize - (groupSize - groupEst) / 4.0).floor
for gs in inf .. groupSize + 999
eq = equalBirthdays(sharers, groupSize, 250)
if eq > 50.0 then
groupSize = gs
break
end
end
# Finest
for gs in groupSize - 1 .. groupSize + 999
eq = equalBirthdays(sharers, gs, 50000)
if eq > 50.0 then
groupEst = gs
print "%d independant people in a group of %s share a common birthday. (%5.1f)\n" % [sharers, gs, eq]
break
end
end
end
end
main()
```

- Output:

2 independant people in a group of 23 share a common birthday. ( 50.5) 3 independant people in a group of 90 share a common birthday. ( 53.3) 4 independant people in a group of 187 share a common birthday. ( 50.3) 5 independant people in a group of 313 share a common birthday. ( 50.3)

## Scala

```
import scala.util.Random
import scala.util.control.Breaks._
object Test {
def equalBirthdays(
nSharers: Int,
groupSize: Int,
nRepetitions: Int
): Double = {
val rand = new Random(1)
var eq = 0
for (_ <- 1 to nRepetitions) {
val group = new Array[Int](365)
for (_ <- 1 to groupSize) {
val birthday = rand.nextInt(group.length)
group(birthday) += 1
}
if (group.exists(_ >= nSharers)) eq += 1
}
(eq * 100.0) / nRepetitions
}
def main(args: Array[String]): Unit = {
var groupEst = 2
for (sharers <- 2 until 6) {
// Coarse.
var groupSize = groupEst + 1
while (equalBirthdays(sharers, groupSize, 100) < 50.0)
groupSize += 1
// Finer.
val inf = (groupSize - (groupSize - groupEst) / 4.0).toInt
breakable({
for (gs <- inf until groupSize + 999) {
val eq = equalBirthdays(sharers, groupSize, 250)
if (eq > 50.0) {
groupSize = gs
break
}
}
})
// Finest.
breakable({
for (gs <- (groupSize - 1) until groupSize + 999) {
val eq = equalBirthdays(sharers, gs, 50000)
if (eq > 50.0) {
groupEst = gs
println(
f"$sharers independent people in a group of $gs share a common birthday. ($eq%5.1f)"
)
break
}
}
})
}
}
}
```

- Output:

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)

## SQL

birthday.sql

```
with
c as
(
select
500 nrep,
50 maxgsiz
from dual
),
reps as
(
select level rep
from dual
cross join c
connect by level <= c.nrep
),
pers as
(
select
round(sqrt(2*level)) npers
from dual
cross join c
connect by level <= c.maxgsiz*(c.maxgsiz+1)/2
),
bds as
(
select
reps.rep,
pers.npers,
floor(dbms_random.value(1,366)) bd
from
reps
cross join pers
),
mtch as
(
select
bds.npers,
case count(distinct bds.bd ) when bds.npers then 0 else 1 end match
from bds
group by
bds.rep,
bds.npers,
null
order by
bds.npers
),
nm as
(
select mtch.npers, sum (mtch.match) nmatch
from mtch
group by mtch.npers
),
sol as
(
select first_value ( nm.npers ) over ( order by abs ( nm.nmatch - c.nrep / 2 ) ) npers
from
nm
cross join c
)
select npers
from sol where rownum = 1
;
```

SQL> @ birthday.sql Connected.

NPERS

23

## Tcl

```
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

## Wren

```
import "random" for Random
import "./fmt" for Fmt
var equalBirthdays = Fn.new { |nSharers, groupSize, nRepetitions|
var rand = Random.new(12345)
var eq = 0
for (i in 0...nRepetitions) {
var group = List.filled(365, 0)
for (j in 0...groupSize) {
var r = rand.int(group.count)
group[r] = group[r] + 1
}
eq = eq + (group.any { |i| i >= nSharers } ? 1 : 0)
}
return eq * 100 / nRepetitions
}
var groupEst = 2
for (sharers in 2..5) {
// Coarse
var groupSize = groupEst + 1
while (equalBirthdays.call(sharers, groupSize, 100) < 50) groupSize = groupSize + 1
// Finer
var inf = (groupSize - (groupSize - groupEst) / 4).floor
for (gs in inf...groupSize + 999) {
var eq = equalBirthdays.call(sharers, groupSize, 250)
if (eq > 50) {
groupSize = gs
break
}
}
// Finest
for (gs in groupSize - 1...groupSize + 999) {
var eq = equalBirthdays.call(sharers, gs, 50000)
if (eq > 50) {
groupEst = gs
Fmt.write("$d independent people in a group of $3d ", sharers, gs)
Fmt.print("share a common birthday $2.1f\%", eq)
break
}
}
}
```

- Output:

2 independent people in a group of 23 share a common birthday 51.0% 3 independent people in a group of 88 share a common birthday 51.0% 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 50.7%

## XPL0

```
func Sim(N);
\Simulate birthdays and return number of people to have N same days
int N, I, People, R;
char Days(365);
[for I:= 0 to 365-1 do Days(I):= 0;
People:= 0;
loop [R:= Ran(365);
Days(R):= Days(R)+1;
People:= People+1;
if Days(R) = N then return People;
];
];
int N, Sum, Trial;
[for N:= 2 to 5 do
[Sum:= 0;
for Trial:= 1 to 10000 do
Sum:= Sum + Sim(N);
IntOut(0, N); Text(0, ": "); IntOut(0, Sum/10000); CrLf(0);
];
]
```

- Output:

2: 24 3: 88 4: 187 5: 311

## zkl

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