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### Re: Re: Birthday Chances

by thraxil (Prior)
 on Feb 15, 2002 at 15:38 UTC ( #145698=note: print w/replies, xml ) Need Help??

in reply to Re: Birthday Chances

you're close on the math.

(ignoring leap-years) the probability that n people all have different birthdays is: ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365) = 365! / ((365-n)! * 365^n)

setting that equal to 50% and solving is probably harder than just plugging in numbers on a calculator till you get it. 23 is the magic number.

anders pearson

Replies are listed 'Best First'.
Re: Re: Re: Birthday Chances
by ckohl1 (Hermit) on Feb 15, 2002 at 22:56 UTC

To expand on what thraxil has said...

This probability and permutations exercise could be written as:

```#!/usr/bin/perl -w
use strict;

my \$DAYS_IN_YEAR = 365;

print "\n# of People \t Birthday Match Likelihood\n" . '-'x80 . "\n";

for my \$people ( 2 .. 40 ){
my \$tmp_days = \$DAYS_IN_YEAR;
my \$probability=1;
for ( 1 .. \$people ){
\$probability *= \$tmp_days--;
}
\$probability /= \$DAYS_IN_YEAR ** \$people;
\$probability = ( 1 - \$probability ) * 100;
print "\$people \t\t \$probability %\n";
}

exit;

Chris

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