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) |
In probability theory, the birthday problem, or birthday paradox This is not a paradox in the sense of leading to a logical contradiction, but is called a paradox because the mathematical truth contradicts naïve intuition: most people estimate that the chance is much lower than 50%. pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 366 (by the pigeon hole principle, ignoring leap years). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack.
- Task
Using simulation, estimate the number of independent people required in a groups before we can expect a better than even chance that at least 2 independent people in a group share a common birthday. Furthermore: Simulate and thus estimate when we can expect a better than even chance that at least 3, 4 & 5 independent people of the group share a common birthday. For simplicity assume that all of the people are alive...
- Suggestions for improvement
- Estimating the error in the estimate to help ensure the estimate is accurate to 4 decimal places.
- Converging to the th solution using a root finding method, as opposed to using an extensive search.
- Kudos (κῦδος) for finding the solution by proof (in a programming language) rather than by construction and simulation.
- See also
ALGOL 68
File: Birthday_problem.a68<lang algol68>#!/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 #</lang>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. <lang c>#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; }</lang>
- 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
<lang d>import std.stdio, std.random, std.algorithm;
/// For sharing common birthday must all share same common day. double equalBirthdays(ref Xorshift rng, in uint sharers=2,
in uint groupSize=23, in uint rep=100_000) { uint eq = 0; foreach (immutable j; 0 .. rep) { uint[365] group; foreach (immutable i; 0 .. groupSize) group[uniform(0, $, rng)]++; eq += group[].any!(c => c >= sharers); } return (eq * 100.0) / rep;
}
void main() {
auto rng = 1.Xorshift; auto groupEst = 2;
foreach (immutable sharers; 2 .. 6) { // Coarse. auto groupSize = groupEst + 1; while (equalBirthdays(rng, sharers, groupSize, 100) < 50.0) groupSize++;
// Finer. foreach (immutable gs; cast(int)(groupSize - (groupSize - groupEst) / 4.0) .. groupSize + 999) { immutable eq = equalBirthdays(rng, sharers, groupSize, 250); if (eq > 50.0) { groupSize = gs; break; } }
// Finest. foreach (immutable gs; groupSize - 1 .. groupSize + 999) { immutable eq = equalBirthdays(rng, sharers, gs, 50_000); 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; } } }
}</lang>
- 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.
Hy
We use a simple but not very accurate simulation method.
<lang lisp>(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))</lang>
J
Quicky approach (use a population of 1e5 people to get a quick estimate and then refine against a population of 1e8 people):
<lang J>bd5=: ?1e5#365 bd8=: ?1e8#365
cnt=: [: >./ #/.~ avg=: +/ % #
Prb=: (1 :0)("0)
NB. y: shared birthday count NB. m: population NB. x: sample size avg ,y <: (-x) cnt\ m
)
est=:3 :0
approx=. (bd5 Prb&y i.365) I. 0.5 n=. approx-(2+y) refine=. n+(bd8 Prb&y approx+i:2+y) I. 0.5 assert. (2+y) > |approx-refine refine, refine bd8 Prb y
)</lang>
Task cases:
<lang J> est 2 23 0.507254
est 3
88 0.510737
est 4
187 0.502878
est 5
313 0.500903</lang>
So, for example, we need a population of 88 to have at least a 50% chance of 3 people having the same birthday in a year of 365 days. And, in that case, the simulated probability was 51.0737%
Lasso
<lang 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 ^}</lang>
- 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
<lang parigp>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)</lang>
PL/I
<lang 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;</lang>
Output:
1000000 samples. Percentage of groups with at least 2 matches Group size 3000000 500000 samples 21 556903 443097 44.310% 44.343% 44.347% 22 524741 475259 47.526% 47.549% 47.521% 23 492034 507966 50.797% <- 50.735% <- 50.722% <- 24 462172 537828 53.783% <- 53.815% <- 53.838% <- 25 431507 568493 56.849% <- 56.849% <- 56.842% <- 1000000 samples. Percentage of groups with at least 3 matches Group size 85 523287 476713 47.671% 47.638% 47.631% 86 512219 487781 48.778% 48.776% 48.821% 87 499874 500126 50.013% <- 49.902% 49.903% 88 488197 511803 51.180% <- 51.127% <- 51.096% <- 89 478044 521956 52.196% <- 52.263% <- 52.290% <- 1000000 samples. Percentage of groups with at least 4 matches Group size 185 511352 488648 48.865% 48.868% 48.921% 186 503888 496112 49.611% 49.601% 49.568% 187 497844 502156 50.216% <- 50.258% <- 50.297% <- 188 490490 509510 50.951% <- 50.916% <- 50.946% <- 189 482893 517107 51.711% <- 51.645% <- 51.655% <- 1000000 samples. Percentage of groups with at least 5 matches Group size 311 508743 491257 49.126% 49.158% 49.164% 312 503524 496476 49.648% 49.631% 49.596% 313 498244 501756 50.176% <- 50.139% <- 50.095% <- 314 494032 505968 50.597% <- 50.636% <- 50.586% <- 315 489821 510179 51.018% <- 51.107% <- 51.114% <- 1000000 samples. Percentage of groups with at least 6 matches Group size 458 505225 494775 49.478% 49.498% 49.512% 459 501871 498129 49.813% 49.893% 49.885% 460 497719 502281 50.228% <- 50.278% <- 50.248% <- 461 493948 506052 50.605% <- 50.622% <- 50.626% <- 462 489416 510584 51.058% <- 51.029% <- 51.055% <-
extended to verify REXX results:
1000000 samples. Percentage of groups with at least 7 matches Group size 621 503758 496242 49.624% 622 500320 499680 49.968% 623 497047 502953 50.295% <- 624 493679 506321 50.632% <- 625 491240 508760 50.876% <- 1000000 samples. Percentage of groups with at least 8 matches Group size 796 504764 495236 49.524% 797 502537 497463 49.746% 798 499488 500512 50.051% <- 799 496658 503342 50.334% <- 800 494773 505227 50.523% <- 1000000 samples. Percentage of groups with at least 9 matches Group size 983 502613 497387 49.739% 984 501665 498335 49.834% 985 498606 501394 50.139% <- 986 497453 502547 50.255% <- 987 493816 506184 50.618% <- 1000000 samples. Percentage of groups with at least10 matches Group size 1179 502910 497090 49.709% 1180 500906 499094 49.909% 1181 499079 500921 50.092% <- 1182 496957 503043 50.304% <- 1183 494414 505586 50.559% <-
Python
Note: the first (unused), version of function equal_birthdays() uses a different but equally valid interpretation of the phrase "common birthday". <lang python> 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))</lang>
- 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. <lang python>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)
</lang>
- 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
<lang python># 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)</lang>
- Output:
23 of 2 people: 43450860....3125/85651679....3125 combos 88 of 3 people: 15497...50390625/30322...50390625 combos 187 of 4 people: 708046698...0703125/1408528546...0703125 combos 313 of 5 people: 498385488882289...2578125/99464149835930...2578125 combos
Racket
Based on the Python task. For three digits precision use 250000 repetitions. For four digits precision use 25000000 repetitions, but it’s very slow. See discussion page.
<lang Racket>#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)))))</lang>
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
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.
version 1
<lang rexx>/*REXX program solves the birthday problem via random number simulation.*/ parse arg grps samp seed . /*get optional arguments from CL.*/ if grps== | grps==',' then grps=5 /*Not specified? Use the default*/ if samp== | samp==',' then samp=100000 /*" " " " " */ if seed\==',' & seed \== then call random ,,seed /*repeatability?*/ diy =365 /*or: diy=365.25*/ /*the number of Days In a Year. */ diyM=diy*100 /*this expands the RANDOM 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 /*do until G dup birthdays found.*/ day=random(1,diyM) % 100 /*expand random number generation*/ @.day=@.day+1 /*record the # common birthdays.*/ end /*j*/ /* [↓] adjust for DO loop index.*/ s=s+j-1 /*add # 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 /*show 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 avg>50%.*/ do samp; @.=0 /*perform some number of trials. */ do try /*do until G dup birthdays found.*/ day=random(1,diyM) % 100 /*expand random number generation*/ @.day=@.day+1 /*record the # common birthdays.*/ if @.day\==g then iterate /*not enough G hits occurred ? */ s=s+1 /*another common birthday found. */ leave /* ··· and stop looking for more.*/ end /*do try;*/ /* [↓] bump counter for Bday hits*/ end /*samp*/ if s/samp>.5 then leave /*if the average is > 50%, stop. */ end /*do 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 done.*/</lang>
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
<lang rexx> /*--------------------------------------------------------------------
* 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</lang>
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
<lang 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 }
}
}</lang>
- 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