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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; }

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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);

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.