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{{Wikipedia pre 15 June 2009|pagename=Birthday Problem|lang=en|oldid=296054030|timedate=21:44, 12 June 2009}} {{draft task}} [[Category:Probability and statistics]] [[Category:Discrete math]]

In [[wp:probability theory|probability theory]], the '''birthday problem''', or '''birthday [[wp:paradox|paradox]]''' This is not a paradox in the sense of leading to a [[wp:logic|logic]]al contradiction, but is called a paradox because the mathematical truth contradicts naïve [[wp:intuition (knowledge)|intuition]]: most people estimate that the chance is much lower than 50%. pertains to the [[wp:probability|probability]] that in a set of [[wp:random|random]]ly chosen people some pair of them will have the same [[wp:birthday|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 [[wp:pigeon hole principle|pigeon hole principle]], ignoring leap years). The mathematics behind this problem leads to a well-known cryptographic attack called the [[wp:birthday attack|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 $n$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.

• Wolfram entry: {{Wolfram|Birthday|Problem}}

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

-- 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
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 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;


{{out}}

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

{{works with|ALGOL 68|Revision 1}} {{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-2.6 algol68g-2.6].}} {{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of '''format'''[ted] ''transput''.}} '''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;
}


{{out}}


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

{{trans|Python}}

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 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; } } } }  {{out}} 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: {{trans|C}} 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); } }  {{out}} 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  ## Go {{trans|C}} 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 }  {{Out}}  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 Translation of [[Birthday_problem#Python|Python]] via [[Birthday_problem#D|D]] {{works with|Java|8}} 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; 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 {{works with|Julia|0.6}} {{trans|Python}} 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 =  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  {{out}} 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 {{trans|Java}} // 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
}
}
}
}


{{out}} 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
^}


{{out}}

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

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

*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

{{trans|Perl 6}}

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


{{out}}

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)


## 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;  {{out}} 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)  ## Phix {{trans|D}} 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; -- Currently 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  {{out}}  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  ## 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 =  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))  {{out}} 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)  {{out}} 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)  {{out}}  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 {{trans|Python}}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 ### version 1 The 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. /*REXX pgm examines the birthday problem via random # simulation (with specifable 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 /*or: 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. */  {{out|output|text= 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% <-  ## 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
}
}
}


{{out}}


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

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)));
}


{{out}}


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

`