Birthday problem

 This page uses content from Wikipedia. The current wikipedia article is at Birthday Problem. The original RosettaCode article was extracted from the wikipedia article № 296054030 of 21:44, 12 June 2009 . The list of authors can be seen in the page history. As with Rosetta Code, the pre 5 June 2009 text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Birthday problem is a 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.

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 ${\displaystyle n}$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.

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

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;
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

Works with: ALGOL 68 version Revision 1
Works with: ALGOL 68G version Any - tested with release algol68g-2.6.
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 probablity of n out of np people sharing a birthday
// is above or below p_thresh, with n_sigmas sigmas confidence
// note that if p_thresh is very low or hi, minimum runs need to be much higher
double prob(int np, int n, double n_sigmas, double p_thresh, double *std_dev)
{
double p, d; // prob and std dev
int runs = 0, yes = 0;
do {
yes += simulate1(np, n);
p = (double) yes / ++runs;
d = sqrt(p * (1 - p) / runs);
if (DEBUG > 1)
printf("\t\t%d: %d %d %g %g \r", np, yes, runs, p, d);
} while (runs < 10 || fabs(p - p_thresh) < n_sigmas * d);
if (DEBUG > 1) putchar('\n');

*std_dev = d;
return p;
}

// bisect for truth
int find_half_chance(int n, double *p, double *dev)
{
int lo, hi, mid;

reset:
lo = 0;
hi = DAYS * (n - 1) + 1;
do {
mid = (hi + lo) / 2;

// 5 sigma confidence. Conventionally people think 3 sigmas are good
// enough, but for case of 5 people sharing birthday, 3 sigmas actually
// sometimes give a slightly wrong answer
*p = prob(mid, n, 5, .5, dev);

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

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

if (hi < lo) {
// this happens when previous precisions were too low;
// easiest fix: reset
if (DEBUG) puts("\tMade a mess, will redo.");
goto reset;
}
} while (lo < mid || *p < .5);

return mid;
}

int main(void)
{
int n, np;
double p, d;
srand(time(0));

for (n = 2; n <= 5; n++) {
np = find_half_chance(n, &p, &d);
printf("%d collision: %d people, P = %g +/- %g\n",
n, np, p, d);
}

return 0;
}
Output:
2 collision: 23 people, P = 0.508741 +/- 0.00174794
3 collision: 88 people, P = 0.509034 +/- 0.00180628
4 collision: 187 people, P = 0.501812 +/- 0.000362394
5 collision: 313 people, P = 0.500641 +/- 0.000128174

D

Translation of: 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[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: Translation of: 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 [email protected]*/ { static uint[nDays] days; days[] = 0; foreach (immutable _; 0 .. nPeople) { immutable day = uniform(0, days.length, rng); days[day]++; if (days[day] == nCollisions) return true; } return false; } /** Decide if the probablity of n out of np people sharing a birthday is above or below pThresh, with nSigmas sigmas confidence. If pThresh is very low or hi, minimum runs need to be much higher. */ double prob(in uint np, in uint nCollisions, in double pThresh, out double stdDev, ref Xorshift rng) { double p, d; // Probablity and standard deviation. uint nRuns = 0, yes = 0; do { yes += simulate1(np, nCollisions, rng); nRuns++; p = double(yes) / nRuns; d = sqrt(p * (1 - p) / nRuns); debug if (yes % 50_000 == 0) printf("\t\t%d: %d %d %g %g \r", np, yes, nRuns, p, d); } while (nRuns < 10 || abs(p - pThresh) < (nSigmas * d)); debug '\n'.putchar; stdDev = d; return p; } /// Bisect for truth. uint findHalfChance(in uint nCollisions, out double p, out double dev, ref Xorshift rng) { uint mid; RESET: uint lo = 0; uint hi = nDays * (nCollisions - 1) + 1; do { mid = (hi + lo) / 2; p = prob(mid, nCollisions, 0.5, dev, rng); debug printf("\t%d %d %d %g %g\n", lo, mid, hi, p, dev); if (p < 0.5) lo = mid + 1; else hi = mid; if (hi < lo) { // This happens when previous precisions were too low; // easiest fix: reset. debug "\tMade a mess, will redo.".puts; goto RESET; } } while (lo < mid || p < 0.5); return mid; } void main() { auto rng = Xorshift(unpredictableSeed); foreach (immutable uint nCollisions; 2 .. 6) { double p, d; immutable np = findHalfChance(nCollisions, p, d, rng); writefln("%d collision: %d people, P = %g +/- %g", nCollisions, np, p, d); } } Output: 2 collision: 23 people, P = 0.521934 +/- 0.00728933 3 collision: 88 people, P = 0.512367 +/- 0.00411469 4 collision: 187 people, P = 0.506974 +/- 0.00232306 5 collision: 313 people, P = 0.501588 +/- 0.000529277 Output with nSigmas = 5.0: 2 collision: 23 people, P = 0.508607 +/- 0.00172133 3 collision: 88 people, P = 0.511945 +/- 0.00238885 4 collision: 187 people, P = 0.503229 +/- 0.000645587 5 collision: 313 people, P = 0.501105 +/- 0.000221016 Go Based on the C version but multithreaded using Go channels package main import ( "math/rand" "fmt" "time" "math" "runtime" ) 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, dur := GetNP(i,sigma,0.5) fmt.Printf("%d collision: %d people, P = %f +/- %f, took %s\n",i,res.np,res.p,res.d,dur) } } func GetNP(n int, n_sigmas, p_thresh float64) (res ProbeRes, dur time.Duration){ start := time.Now() 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 } } dur = time.Since(start) return } func probe(np,n int, n_sigmas, p_thresh float64) ProbeRes{ var p,d float64 var runs, yes int cRes := make(chan Frac,runtime.NumCPU()) for i:=0; i < runtime.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){ yes := 0 for i := 0;i < ssize;i++ { if Sim(np,n) { yes++ } } c <- Frac{yes,ssize} } func Sim(p,n int) ( res bool){ Cal := make([]int,DaysInYear) for i := 0;i < p ;i++{ Cal[rand.Intn(DaysInYear)]++ } for _,v := range Cal{ if v >= n { res = true } } return } Output: 2 collision: 23 people, P = 0.506375 +/- 0.001197, took 1.5777555s 3 collision: 88 people, P = 0.516045 +/- 0.002945, took 2m17.5473911s 4 collision: 187 people, P = 0.502643 +/- 0.000507, took 1m11.1573736s 5 collision: 313 people, P = 0.501907 +/- 0.000367, took 1m49.7901966s 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 Python via D Works with: Java version 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[365]; for (int j = 0; j < groupSize; j++) group[rand.nextInt(group.length)]++; eq += stream(group).anyMatch(c -> c >= nSharers) ? 1 : 0; } return (eq * 100.0) / nRepetitions; } public static void main(String[] a) { int groupEst = 2; for (int sharers = 2; sharers < 6; sharers++) { // Coarse. int groupSize = groupEst + 1; while (equalBirthdays(sharers, groupSize, 100) < 50.0) groupSize++; // Finer. int inf = (int) (groupSize - (groupSize - groupEst) / 4.0); for (int gs = inf; gs < groupSize + 999; gs++) { double eq = equalBirthdays(sharers, groupSize, 250); if (eq > 50.0) { groupSize = gs; break; } } // Finest. for (int gs = groupSize - 1; gs < groupSize + 999; gs++) { double eq = equalBirthdays(sharers, gs, 50_000); if (eq > 50.0) { groupEst = gs; System.out.printf("%d independent people in a group of " + "%s share a common birthday. (%5.1f)%n", sharers, gs, eq); break; } } } } } 2 independent people in a group of 23 share a common birthday. ( 50,6) 3 independent people in a group of 87 share a common birthday. ( 50,4) 4 independent people in a group of 187 share a common birthday. ( 50,1) 5 independent people in a group of 314 share a common birthday. ( 50,2) Lasso if(sys_listunboundmethods !>> 'randomgen') => { define randomgen(len::integer,max::integer)::array => { #len <= 0 ? return local(out = array) loop(#len) => { #out->insert(math_random(#max,1)) } return #out } } if(sys_listunboundmethods !>> 'hasdupe') => { define hasdupe(a::array,threshold::integer) => { with i in #a do => { #a->find(#i)->size > #threshold-1 ? return true } return false } } local(threshold = 2) local(qty = 22, probability = 0.00, samplesize = 10000) while(#probability < 50.00) => {^ local(dupeqty = 0) loop(#samplesize) => { local(x = randomgen(#qty,365)) hasdupe(#x,#threshold) ? #dupeqty++ } #probability = (#dupeqty / decimal(#samplesize)) * 100 'Threshold: '+#threshold+', qty: '+#qty+' - probability: '+#probability+'\r' #qty += 1 ^} Output: Threshold: 2, qty: 22 - probability: 47.810000 Threshold: 2, qty: 23 - probability: 51.070000 Threshold: 3, qty: 86 - probability: 48.400000 Threshold: 3, qty: 87 - probability: 49.200000 Threshold: 3, qty: 88 - probability: 52.900000 Threshold: 4, qty: 184 - probability: 48.000000 Threshold: 4, qty: 185 - probability: 49.800000 Threshold: 4, qty: 186 - probability: 49.600000 Threshold: 4, qty: 187 - probability: 48.900000 Threshold: 4, qty: 188 - probability: 50.700000 Threshold: 5, qty: 308 - probability: 48.130000 Threshold: 5, qty: 309 - probability: 48.430000 Threshold: 5, qty: 310 - probability: 48.640000 Threshold: 5, qty: 311 - probability: 49.370000 Threshold: 5, qty: 312 - probability: 49.180000 Threshold: 5, qty: 313 - probability: 49.540000 Threshold: 5, qty: 314 - probability: 50.000000 PARI/GP 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 6  This example is incomplete. Please ensure that it meets all task requirements and remove this message. For a start, we can show off how to get the exact solution. If we pick n people, the total number of possible arrangements of birthdays is 365n. Among those possibilities, there are Cn365 where all birthdays are different. For each of these, there are n! possible ways to arrange the n people. So the solution is 1 - n!Cn365/365n, which in Perl 6 can be written: say "$_ :", 1 - combinations(365, $_)/365**$_ * [*] 1..$_ for ^Inf Output: 0 : 0 1 : 0 2 : 0.002740 3 : 0.0082042 4 : 0.016355912 5 : 0.027135573700 6 : 0.04046248364911 7 : 0.0562357030959754 8 : 0.0743352923516690285 9 : 0.0946238338891667 ^C Now comparing with a simulation : sub theory($n) { 1 - combinations(365, $n)/365**$n* [*] 1..$n } sub simulation(:number-of-people($n), :sample-size($N) = 1_000) {$N R/ grep ?*, ((^365).roll($n).unique !==$n) xx $N; } for 2 .. Inf ->$n {
printf "%3d people, theory: %.4f, simulation: %.4f\n",
$n, theory($n), simulation(number-of-people => $n); } Output: 2 people, theory: 0.0027, simulation: 0.0020 3 people, theory: 0.0082, simulation: 0.0080 4 people, theory: 0.0164, simulation: 0.0130 5 people, theory: 0.0271, simulation: 0.0260 6 people, theory: 0.0405, simulation: 0.0340 7 people, theory: 0.0562, simulation: 0.0590 8 people, theory: 0.0743, simulation: 0.0730 9 people, theory: 0.0946, simulation: 0.1080 10 people, theory: 0.1169, simulation: 0.1120 11 people, theory: 0.1411, simulation: 0.1220 12 people, theory: 0.1670, simulation: 0.1740 13 people, theory: 0.1944, simulation: 0.2200 14 people, theory: 0.2231, simulation: 0.2290 15 people, theory: 0.2529, simulation: 0.2540 16 people, theory: 0.2836, simulation: 0.2820 17 people, theory: 0.3150, simulation: 0.3190 18 people, theory: 0.3469, simulation: 0.3740 19 people, theory: 0.3791, simulation: 0.3720 20 people, theory: 0.4114, simulation: 0.3810 21 people, theory: 0.4437, simulation: 0.4340 22 people, theory: 0.4757, simulation: 0.4700 23 people, theory: 0.5073, simulation: 0.4960 24 people, theory: 0.5383, simulation: 0.5200 25 people, theory: 0.5687, simulation: 0.5990 26 people, theory: 0.5982, simulation: 0.5980 27 people, theory: 0.6269, simulation: 0.6520 28 people, theory: 0.6545, simulation: 0.6430 29 people, theory: 0.6810, simulation: 0.6690 30 people, theory: 0.7063, simulation: 0.7190 31 people, theory: 0.7305, simulation: 0.7450 ^C Python 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 Translation of: 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 common birthday, and then use the floor of that value (less the group size) as a starting point to find a new group size with an expected size that exceeds 50% common birthdays of the required size. /*REXX program examines the birthday problem via random number simulation. */ parse arg grps samp seed . /*get optional arguments from the CL. */ if grps=='' | grps==',' then grps=5 /*Not specified? Then use the default.*/ if samp=='' | samp==',' then samp=100000 /* " " " " " " */ if datatype(seed,'W') then call random ,,seed /*RANDOM seed given for repeatability ?*/ diy =365 /*or: diy=365.25*/ /*the number of Days In a Year. */ diyM=diy*100 /*this expands the RANDOM (BIF) range.*/ /* [↓] get a rough estimate for %. */ do g=2 to grps; s=0 /*perform through 2 ──► group size. */ do samp; @.=0 /*perform some number of trials. */ do j=1 until @.day==g /*perform until G dup. birthdays found.*/ day=random(1,diyM) % 100 /*expand range RANDOM number generation*/ @.day=@.day+1 /*record the number of common birthdays*/ end /*j*/ /* [↓] adjust for the DO loop index.*/ s=s+j-1 /*add number of birthday hits to sum. */ end /*samp*/ start.g=s/samp%1-g /*define where the try-outs start. */ end /*g*/ say say right('sample size is ' samp,40); say /*display this run's sample size. */ say ' required group % with required' say ' common size common birthdays' say ' ──────── ───── ────────────────' /* [↓] where the try-outs happen. */ do g=2 to grps /*perform through 2 ──► group size. */ do try=start.g; s=0 /*perform try-outs until average > 50%.*/ do samp; @.=0 /*perform some number of trials. */ do try /*perform until G dup. birthdays found.*/ day=random(1,diyM) % 100 /*expand range RANDOM number generation*/ @.day=@.day+1 /*record the number of common birthdays*/ if @.day\==g then iterate /*not enough G (birthday) hits found ? */ s=s+1 /*another common birthday found. */ leave /* ··· and stop looking for more. */ end /*try;*/ /* [↓] bump the counter for Bday hits.*/ end /*samp*/ if s/samp>.5 then leave /*if the average is > 50%, then stop. */ end /*try=start.g*/ say right(g, 15) right(try, 15) center( format(s/samp*100,,5)'%', 30) end /*g*/ /*stick a fork in it, we're all done. */ output when using the input of: 10 sample size is 100000 required group % with required common size common birthdays ──────── ───── ──────────────── 2 23 50.70900% 3 88 51.23200% 4 187 50.15100% 5 314 50.77800% 6 460 50.00600% 7 623 50.64800% 8 798 50.00700% 9 985 50.13400% 10 1181 50.22200% version 2 /*-------------------------------------------------------------------- * 04.11.2013 Walter Pachl translated from PL/I * Take samp samples of groups with gs persons and check *how many of the groups have at least match persons with same birthday *-------------------------------------------------------------------*/ samp=100000 lo='0 21 85 185 311 458' hi='0 25 89 189 315 462' Do match=2 To 6 Say ' ' Say samp' samples . Percentage of groups with at least', match ' matches' Say 'Group size' Do gs=word(lo,match) To word(hi,match) cnt.=0 Do i=1 To samp ok=0 arr.=0 Do g=1 To gs r=random(1,365) arr.r=arr.r+1 If arr.r=match Then ok=1 End cnt.ok=cnt.ok+1 End hits=cnt.1/samp If hits>=.5 Then arrow=' <-' Else arrow='' Say format(gs,10) cnt.0 cnt.1 100*hits||'%'||arrow End End Output: 100000 samples . Percentage of groups with at least 2 matches Group size 21 55737 44263 44.26300% 22 52158 47842 47.84200% 23 49141 50859 50.85900% <- 24 46227 53773 53.77300% <- 25 43091 56909 56.90900% <- 100000 samples . Percentage of groups with at least 3 matches Group size 85 52193 47807 47.80700% 86 51489 48511 48.51100% 87 50146 49854 49.85400% 88 48790 51210 51.2100% <- 89 47771 52229 52.22900% <- 100000 samples . Percentage of groups with at least 4 matches Group size 185 50930 49070 49.0700% 186 50506 49494 49.49400% 187 49739 50261 50.26100% <- 188 49024 50976 50.97600% <- 189 48283 51717 51.71700% <- 100000 samples . Percentage of groups with at least 5 matches Group size 311 50909 49091 49.09100% 312 50441 49559 49.55900% 313 49912 50088 50.08800% <- 314 49425 50575 50.57500% <- 315 48930 51070 51.0700% <- 100000 samples . Percentage of groups with at least 6 matches Group size 458 50580 49420 49.4200% 459 49848 50152 50.15200% <- 460 49975 50025 50.02500% <- 461 49316 50684 50.68400% <- 462 49121 50879 50.87900% <- Tcl proc birthdays {num {same 2}} { for {set i 0} {$i < $num} {incr i} { set b [expr {int(rand() * 365)}] if {[incr bs($b)] >= $same} { return 1 } } return 0 } proc estimateBirthdayChance {num same} { # Gives a reasonably close estimate with minimal execution time; the idea # is to keep the amount that one random value may influence the result # fairly constant. set count [expr {$num * 100 / $same}] set x 0 for {set i 0} {$i < $count} {incr i} { incr x [birthdays$num $same] } return [expr {double($x) / $count}] } foreach {count from to} {2 20 25 3 85 90 4 183 190 5 310 315} { puts "identifying level for$count people with same birthday"
for {set i $from} {$i <= $to} {incr i} { set chance [estimateBirthdayChance$i $count] puts [format "%d people => %%%.2f chance of %d people with same birthday" \$i [expr {$chance * 100}]$count]
if {$chance >= 0.5} { puts "level found:$i people"
break
}
}
}
Output:
identifying level for 2 people with same birthday
20 people => %43.40 chance of 2 people with same birthday
21 people => %44.00 chance of 2 people with same birthday
22 people => %46.91 chance of 2 people with same birthday
23 people => %53.48 chance of 2 people with same birthday
level found: 23 people
identifying level for 3 people with same birthday
85 people => %47.97 chance of 3 people with same birthday
86 people => %48.46 chance of 3 people with same birthday
87 people => %49.55 chance of 3 people with same birthday
88 people => %50.66 chance of 3 people with same birthday
level found: 88 people
identifying level for 4 people with same birthday
183 people => %48.02 chance of 4 people with same birthday
184 people => %47.67 chance of 4 people with same birthday
185 people => %48.89 chance of 4 people with same birthday
186 people => %49.98 chance of 4 people with same birthday
187 people => %50.99 chance of 4 people with same birthday
level found: 187 people
identifying level for 5 people with same birthday
310 people => %48.52 chance of 5 people with same birthday
311 people => %48.14 chance of 5 people with same birthday
312 people => %49.07 chance of 5 people with same birthday
313 people => %49.63 chance of 5 people with same birthday
314 people => %49.59 chance of 5 people with same birthday
315 people => %51.79 chance of 5 people with same birthday
level found: 315 people

zkl

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