Appoligies if this is out of place but this question really looks like a homework assignment.

Your information is out of date. Currently, Lotto Texas is six numbers from 1 to 44. From 7/19/2000 through 5/3/2003, it was six numbers from 1 to 54. Prior to 7/19/2000, it was six numbers from 1 to 50.

If I were to code up a Lotto script, I'd use WWW::Mechanize to pull the number frequency (and possibly all of the previous winning numbers) that the lottery commission provides.

my @numbers;
$numbers[ int(rand(54)) + 1 ]++ for 1 .. 10_000;
print join ', ', ( map { "$_ ".($numbers[$_]||'0') } reverse sort { $n
+umbers[ $a ] <=> $numbers[ $b ] } (1 .. 54) )[0..5];
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No, it's a command-line one-liner.... perl -e 'srand;push(@c,0)foreach(0..53);foreach(1..1e4){@d=(); foreach(0..53){$e{$_}=1;}foreach(0..5){$i=int(rand(54));while (!$e{$i}){$i=int(rand(54));} push(@d,$i);$e{$i}=0;} foreach(@d){$c[$_]++}}foreach(sort{$c[$b]<=>$c[$a]}0..$#c) {print $_+1," ",$c[$_],", ";}'

It also prevents duplicates per drawing. Explaining how it works and where they got it is left as an exercise to the student (as this smells a bit like homework)....

Update 22 Feb 2005: Fixed small typos, added a few spaces into code.

Update 22 Feb 2005: The code above appears to run in just under 1.0s, according to time(1), and weighs in around 267c total (256c for the code).

int(rand(53)) + 1 gives a number from 1 to 53. Did you mean int(rand(54)) + 1?

Yep, you got me :D I have a version here which doesn't pick numbers twice, but it's still running.. I guess this is not for production,

---

Ordinary morality is for ordinary people. -- Aleister Crowley

The code for your algorithm is easy:

my @random_numbers;
push(@random_numbers, int(rand(54)) + 1)
   foreach 1..10000;

my %counts;
$counts{$_}++
   foreach @random_numbers;

printf("%2d: %d\n", @$_)
   foreach
      sort { $a->[1] <=> $b[1] || $a->[0] <=> $b[0] }
         map { [ $_, $counts{$_} ] }
            keys(%counts);
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He generated 10000 random numbers between 1 and 54 and stored the frequency distribution in his array.

Can the same number be picked twice? If not, your algo isn't representative. here's a fixedup version:

# Picks 6 unique numbers uniformly between 1 and 54 inclusive.
sub pick6 {
   my @nums;
   push(@nums, int(rand(54 - $_)) + 1)
      foreach 0..5;

   for my $i (1..5) {
     for my $j (0..$i-1) {
        $num[$i]++ if $num[$i] >= $num[$j];
     }
   }

   return @nums;
}

my @random_numbers;
push(@random_numbers, pick6())
   foreach 1..10000;

my %counts;
$counts{$_}++
   foreach @random_numbers;

printf("%2d: %d\n", @$_)
   foreach
      sort { $a->[1] <=> $b[1] || $a->[0] <=> $b[0] }
         map { [ $_, $counts{$_} ] }
            keys(%counts);
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Hint: imbedded FOR LOOPS

"embedded" is spelled with an "e", and "nested" is the proper term when talking about loops within loops.

my @rand = 1..54;
my %rand = map { $_=int rand($rand[$#rand])+1, $rand{$_}++ } 1..10000;
print "$_\t$rand{$_}$/" for sort { $a <=> $b } keys %rand
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I'd do it with the statistics of numbers actually drewn in the lottery. Those stats are available from newspapers, or even from the internet. From these statistics of our national lottery, you get the following (The game is 5 numbers out of 1..90, and the statistics is from 2503 games.)

Update: was broken link.

To get the data, POST jatek=otos to www.szerencsejatek.hu/index.php?action=szamstat

my ( %c, @r );
$r[ int( rand(54) ) + 1 ]++ for ( 1 .. 10_000 );
printf "%2s $r[$_]\n", $_ for (
  sort { $r[$b] <=> $r[$a] || $a <=> $b } map {
    my $x;
    1 while ( $x = int( rand(54) ) + 1 and $c{$x}++ );
    $x
  } ( 1 .. 6 )
);
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