Birthday problem
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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%;
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
21 556903 443097 44.310%
22 524741 475259 47.526%
23 492034 507966 50.797% <-
24 462172 537828 53.783% <-
25 431507 568493 56.849% <-
1000000 samples. Percentage of groups with at least 3 matches
Group size
85 523287 476713 47.671%
86 512219 487781 48.778%
87 499874 500126 50.013% <-
88 488197 511803 51.180% <-
89 478044 521956 52.196% <-
1000000 samples. Percentage of groups with at least 4 matches
Group size
185 511352 488648 48.865%
186 503888 496112 49.611%
187 497844 502156 50.216% <-
188 490490 509510 50.951% <-
189 482893 517107 51.711% <-
1000000 samples. Percentage of groups with at least 5 matches
Group size
311 508743 491257 49.126%
312 503524 496476 49.648%
313 498244 501756 50.176% <-
314 494032 505968 50.597% <-
315 489821 510179 51.018% <-
1000000 samples. Percentage of groups with at least 6 matches
Group size
458 505225 494775 49.478%
459 501871 498129 49.813%
460 497719 502281 50.228% <-
461 493948 506052 50.605% <-
462 489416 510584 51.058% <-
REXX
<lang rexx>/*REXX programs solves "birthday problem" via random number simulations.*/ 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 /*do until g dup birthdays found.*/
day=random(1,diyM) % 100 /*expand random number generation*/
@.day=@.day+1 /*record the of a particular Bday*/
if @.day==g then leave /*when G hits have occurred ···*/
end /*j*/
s=s+j /*add number of people to sum. */
end /*samp*/
start.g=s/samp%1-g /*define where the try-outs start*/
end /*g*/
say ' sample size is ' samp /*show sample size of this run. */ say ___=' ' /*padding for easier eyeballing. */ 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 j=1 for try /*do until K dup birthdays found.*/
day=random(1,diyM) % 100 /*expand random number generation*/
@.day=@.day+1 /*record the of a particular Bday*/
if @.day\==g then iterate /*when G hits have occurred ···*/
s=s+1 /*another common birthday found. */
leave /* ... and stop looking for more.*/
end /*j*/
end /*samp*/
if s/samp>.5 then leave /*if the average is > 50%, stop. */
end /*try*/
say ___ center(g,8) ___ right(try,5) ___ center((s/samp*100)'%',16)
end /*g*/
/*stick a fork in it, we're done.*/</lang>
output
sample size is 100000
required group % with required
common size common birthdays
──────── ───── ────────────────
2 23 50.75100%
3 88 51.14700%
4 187 50.30400%
5 313 50.14100%