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.

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.

Task:

Using simulation, estimate the number of independent people required in a groups before we can expect a better then 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 then 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 nᵗʰ solution using a root finding method, as opposed to an using an extensive search.
• Kudos (κῦδος) for finding the solution by proof (in a programming language) rather then by construction and simulation.

See also:

• Birthday Problem

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<lang algol68>#!/usr/bin/a68g --script #

1. -*- coding: utf-8 -*- #

REAL desired probability := 0.5; # 50% #

REAL upb year = 365 + 1/4 # - 3/400 but alive, ignore >113yo #, INT upb sample size = 10000,

   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: 10000; 
.
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: 50.09%; 
................................................................
required repeat: 3; group size: 87; %age of years with required repeat birthdays: 50.37%; 
...................................................................................................
required repeat: 4; group size: 186; %age of years with required repeat birthdays: 50.70%; 
...............................................................................................................................
required repeat: 5; group size: 313; %age of years with required repeat birthdays: 50.92%;