"How to calculate distribution" Topic

How do you do calculate the number of ways to do this

I know 2d6
2- 1 Combo
3 – 2 combos
…
7 – 6 ways
8 – 5 ways

etc

How do you do it matematically on more dice. On poly dice
ie on xDy, how many will equal A?

anydice.com

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Martin Stephenson
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In a very general way, the calculation of probability for multiple events occurring is the probability of each of those individual events multiplied together.

So on a d20, the chance of rolling a "12" is 5% (.05). The chance of you rolling two d20 dice and getting "12, 12" is therefore .05 x .05 = .0025 or less than 1%.

The problem is the questions is rarely that simple when people start this discussion. They end up wanting to know the chance that if I roll XdY at least Z of them will be A or higher.

At that point, just use the web site listed.

Thanks Martin. Used it by multiplying the result by 216 then /100 to get number of combos (rounded)

Seems to work.

However does anyone know of a way to get NUMBER of combos easily

Basically I want to do some stats on IABSM fire tables- number of combos for a score x hits to give number of hits per y^x rolls. Divide by y^x to give ave number of hits- ie on a perfectly average set of rolls, how many hits will I on average give out.

Couple of mechanisms I want to check. (read "Tell Sunjester he's wrong")

Maybe this as well?

link

The number of combos on xDy that equal a equals the number of decompositions of a into x integers, multiplied by the number of permutations for x numbers. That requires some formulas from combinatorial mathematics.
Then for a %chance, divide by the number of all possible outcomes (y to the power x).

If you then want to know what the % is to beat a certain number, add up the %chance for all numbers that are valid. This is also called a cumulative probability distribution.

If you have a certain die rolling procedure (e.g. rolling xDy, scoring at least a) that gives you a specific chance for success (say s), and then want to know how many successes you will score with n independent rolls, you need a binomial distribution to calculate the results.

Any of these exercises are usually fairly standard in an undergraduate course on probability theory or combinatorial mathematics.

Or you can use a nice math program such as the ones mentioned above. These either use the closed formulas as mentioned above, or simulate the whole things by letting the computer conduct 1bazillion experiments using a random number generator.