http://www.perlmonks.org?node_id=158490

in reply to Rolling a biased die

```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;
}

Replies are listed 'Best First'.
Re: Re: Rolling a biased die
by tommyw (Hermit) on Apr 12, 2002 at 11:32 UTC
Re: Re: Rolling a biased die
by ferrency (Deacon) on Apr 12, 2002 at 13:39 UTC
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

Re: Re: Rolling a biased die
by tomazos (Deacon) on Apr 13, 2002 at 13:56 UTC
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. :)

Re: Re: Rolling a biased die
by tomazos (Deacon) on Apr 13, 2002 at 14:38 UTC
How about precaching a hash called %biasbase like this?

```my \$sum = 0;
\$biasbase{\$_} = (\$sum += \$bias{\$_}) foreach (reverse keys %bias);
```while( my(\$k,\$v) = each %bias and not defined \$rand){