in reply to Rolling DND Dice.

Some possible improvements in efficiency:
Assume your dice are ordered highest to lowest (that is, 6-5-5-4 is the same result as 4-5-6-5).
You don't have to roll the lowest die, just note how many possible values there are for it.

Then the code looks like this: 

my ($sum, $rolls) = (0,0);
for my $hi (1..6) {
  for my $mid (1..$hi) {
    for my $low (1..$mid) {
      # The drop value can be anything from 1..$low
      $sum += ($hi + $mid + $low) * $low;
      $rolls += $low;
    }
  }
}
printf "The average value is %g\n", $sum/$rolls;

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I get a nice even 12, so obviously there's some hole in my methodology.
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Re: Re: Rolling DND Dice.
by tilly (Archbishop) on Feb 04, 2004 at 01:57 UTC
    The hole in your reasoning is that you count, for instance, 3-3-3-3 as one possibility, and also 1-3-4-6 as one possibility.

    In fact 3-3-3-3 can only happen one way, while 1-3-4-6 can happen 24 ways.

[reply]
      Right you are. Putting the possible combinations into play, I finally came up with a program that gets the right answer, and can be used for any number of sides (though I did not generalize it to be any number of dice, nor any number of discards -- it is possible to do so, but it's messy enough already). For 6 sided dice, it is probably working harder than the explicit method, but for 20 sided dice, it is much faster.

      In the code, L is the low value for the roll. The number of distinct permutations possible varies, depending on how many times L is repeated, and the average value for the permutations also changes based on that. 

      my $sides = 16;
my ($total, $rolls) = (0,0);
foreach my $L (1..$sides) {
  # Start with all others = $L
  my $combos = 1;
  my $sum = $L * 3;
  $rolls += $combos;
  if ($sides > $L) {
    # How many repeat $L twice, with one higher?
    $combos = 4 * ($sides-$L);
    my $avg = 2 * $L + ($L+1+$sides)/2;
    $rolls += $combos;
    $sum   += $avg * $combos;
    # How many repeat $L once, with two higher?
    $combos = 6 * ($sides-$L)**2;
    $avg = $L + ($L+1+$sides);
    $rolls += $combos;
    $sum   += $avg * $combos;
    # How many do not repeat $L?
    $combos = 4 * ($sides-$L)**3;
    $avg = 3*($L+1+$sides)/2;
    $rolls += $combos;
    $sum   += $avg * $combos;
  }
  printf "%d combos of three dice >= $L averaging %g\n", $combos, $sum
+/$combos;
  $total += $sum;
}
printf "Average of $rolls rolls is %g\n", $total/$rolls;

# Explicit method included for check
$sum = $rolls = 0;
for $i (1..$sides) {
  for $j (1..$sides) {
    for $k (1..$sides) {
      for $l (1..$sides) {
        my $min = $l;
        for ($i, $j, $k) { $min = $_ if $_ < $min }
        $sum += $i + $j + $k + $l - $min;
        ++$rolls;
      }
    }
  }
}
printf "Average of $rolls rolls is %g\n", $sum/$rolls;

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