Re: Rolling a biased die

[I0]

my %bias = (
   1 => 3.1,
   2 => 2.0234,
   3 => 1.7,
   4 => 1.542232,
   5 => 1.321249563,
   6 => 1.0142,
 );
my $sum = 0;
while( my($k,$v) = each %bias ){
    $rand = $k if rand($sum+=$v) < $v;
}

[tommyw]

The algorithm is the generalised version of How do I pick a random line from a file?, so see there for an explanation.

--
Tommy
Too stupid to live.
Too stubborn to die.

[ferrency]

Update:I'm totally wrong! Please ignore me! And take my advice: try the code before you claim it's broken :) Sorry, IO.

I don't think you want to put the rand() inside the loop. I think to get the correct results, you need to choose one random number outside the loop, and test it vs. the sum as you go along.

Try out a simple case by hand:

my %bias = (1 => 1, 2 => 1);

Assuming your first iteration picks up (1 => 1), you have:

$rand = $k if rand($sum += $v) <= $v; 
# This simplifies to: 
# $rand = 1 if rand(1) <= 1
# this is always true.  That's wrong.

What you really want is something like this:

my $sum = 0;
$sum += $_ foreach (values %bias);
my $target = rand($sum);
$sum = 0;
while (my ($k, $v) = each %bias) {
    if ($target <= ($sum += $v)) {
        $rand = $k;
        last;
    }
}

I'm sure there's a golfier way to do it, but this demonstrates the basic idea.

Update: Okay, after actually trying the code, I believe I'm totally wrong. I think that what I thought was a combination of two bugs may actually be a clever solution. Though I'm still not sure I believe it produces the correct result distribution. IO, would you care to describe how it works? Sorry about that.

Alan

[tomazos]

Cool algorithm. It's like a king of the hill match.

1 starts as king of the hill. ($rand = 1)

2 comes along and challanges it. Whoever wins stays on top (is assigned to $rand).

Just like 2, everyone else (3, 4, 5 and 6) gets a chance.

Whoever is left on top ($rand) is declared the winner. :)

---

To understand why the probabilities work you have to step through the algorithm backwards.

ie. What is the chance that 6 (the final iteration) is going to win it's match against the king of the hill? $bias{6} / sum(values %bias), which is obvious.

Now - consider the second last iteration (5). Given that 6 is going to have it's chance in a minute, and hence does not need to be included, what is the chance that 5 will win it's match? $bias{5} / (sum(values %bias) - $bias{6}). We remove 6 from the running by excluding it's weighting from the total.

Update: This explanation is awful. :)

[tomazos]

How about precaching a hash called %biasbase like this?

my $sum = 0;
$biasbase{$_} = ($sum += $bias{$_}) foreach (reverse keys %bias);

And then iterate like this:

while( my($k,$v) = each %bias and not defined $rand){
    $rand = $k if rand($biasbase{$k}) < $v;
}

Does that make sense? That way you can bail out early.