Scaling chance weights

AngleWyrm

Member
Introduction
Chance weights are integer values that represent outcomes to a random event. A flip of a coin has two outcomes, each with a chance weight of 1. A roll of a pair of dice has 36 chance weights partitioned across 11 possible outcomes like so: {1,2,3,4,5,6,5,4,3,2,1}.

Problem
I start with randomly chosen from this set of chance weights: {1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1}.
There are 216 chance weights, allotted to 16 possible outcomes.

But this is an engineering problem, and the author declares the 1/216 probability at the extremes isn't working well during testing, and they want it altered. For the sake of example, I'll just pick any old number and say projections show better system performance if the extremes were instead in the vicinity of 1/300 probability.

So that means there should be 300 chance weights instead of 216. I can multiply the set by 300/216 and get a new set but they are not integers:
{1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1} * 300/216 = {25/18, 25/6, 25/3, 125/9, 125/6, 175/6, 625/18, 75/2, 75/2, 625/18, 175/6, 125/6, 125/9, 25/3, 25/6, 25/18}
The greatest common divisor of this set is the extremes, 25/18.

Is the best solution just to multiply the new set by 18?

Dason

It's there a reason you're using "chance weight" as your vocabulary instead of working with the probabilities directly? It seems like that is really the source of your problem.

AngleWyrm

Member
It's there a reason you're using "chance weight" as your vocabulary instead of working with the probabilities directly? It seems like that is really the source of your problem.
Chance weights are the correct unit of measure, at the correct scale. Here is the math behind rolling random discrete outcomes

What I've discovered on the topic at hand is that the denominator of the outcome with the smallest chance count produces what I'm looking for, a set of integers describing the proportions of the outcome set.

18 x {25/18, 25/6, 25/3, 125/9, 125/6, 175/6, 625/18, 75/2, 75/2, 625/18, 175/6, 125/6, 125/9, 25/3, 25/6, 25/18} =
{25, 75, 150, 250, 375, 525, 625, 675, 675, 625, 525, 375, 250, 150, 75, 25}

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Dason

I'm just saying work with the probabilities directly. Heck the request you have is even framed in terms of probabilities. It sounds like all you really want to do is replace the extremes with 1/300. From there if you're working with the probabilities you could just scale the remaining probabilities so that the entire set sums to 1.

AngleWyrm

Member
I'm just saying work with the probabilities directly.
How many tosses of a six-sided die will it take to get a six? The question can be answered easily, but only if we accept that terms like "on average" and "Expected Value" aren't the best tools for the job. Because those terms would then describe a prediction designed to be wrong half the time.

When the problem is considered from the perspective of outcomes, the answer is much easier to achieve: Split the space of outcomes into two sets, those with sixes and those without. The answer is then seen more properly as the number of dice it takes to achieve any desired proportion.

outcomes without sixes = (5/6)^numDice
total outcomes = 6^numDice

So rolling 4 dice produces a set of outcomes where about half contain sixes, whereas about 95% of outcomes for rolling 17 dice contain sixes.

What is being exposed in the example is an aspect of games of chance, dependability of the prediction, which fails to surface in models of probability that mask prediction accuracy behind expected value.

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