Usually, I'm good at math. But for some reason, probability is my bane.

I've now gained an intuitive understanding of the odds of success (aka rolling 3 dice <> 100% success).

But the odds of a critical glitch are still beyond me.

A glitch is easy enough - it's simply the odds of 1/2 the pool equaling 1.

But a critical glitch? That's more complex.

How do you calculate that?

To me, you take the odds of getting no successes, and multiply that by the odds of getting 1/2 the pool as 1s. But I've learned that what my gut tells me is rarely correct when it comes to probability.

Probability is messy and a real PITA. Here is a link to a Wikipedia article on the math involved in calculating probabilities.

Okay, so I was bored at work and playing with Excel.

Here's what I figured out the chances of a critical glitch to be.

First I had to make a set of results where all the dice failed (2/3 chance per die).

http://www.serbitar.de/stuff/probabilities.xls

It's Serbitar. What do you expect?

For those who actually want to understand the math I'll explain. It requires a heck of a lot of summation at certain levels (which is just serial addition but can get problematic due the the sheer volume).

Okay, first of all you need to find the number of possibile combinations of dice. This is always 6^n where n is the number of dice in your pool.
So, if you had 6 dice in your pool the number of possible combinations is 6^6 or 6*6*6*6*6*6 which equals 46656.

6^n is an important number and will be required for calculating all probabilities.


Now, the probability of any single result is equal to the number of combinations that give the desired result divided by the number of possible results, with appropriate adjustments for weighting. Ideal, dice should not favor any particular result and these equations are for the ideal situations, so we can ignore weighting.

So, the probability of a Critical Glitch with a single die is 1/6 and the probability of a standard Glitch with a single die is 0. This is because are only six possible combinations, only one of which results in a critical glitch (1) and it is impossible to have a hit if one also has a 1 inthis case.

Now, calculating the probability for a glitch or a critical clitch with two dice is also easy.
6^2 = 36

There are 11 combinations that result in a glitch. (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (3,1) (4,1) (5,1) (6,1) Of those only 7 result in a critical glitch.

So with 2 dice the probability of a standard glitch is 4/36 and of a critical glitch is 7/36.

How 3 dice is where is becomes difficult. there are 6^3 or 216 possible combinations, so just counting is going to be tedious. Instead, we're going to have to use a formula.

The is 1 possible combination that has three 1s so well start with that.
Then there are 3*5 possible combinations that have two 1s; (since the die that isn't a 1 can have as many as five possible values)

So there are 1+3*5 = 16 possible combinations of 3 dice that result in a glitch. Of these, 2*3 are not critical. So, there is a 10/216 probability of a critical glitch and a 6/216 probability of a standard glitch.

6^4 = 1296
With 4 dice, you have 1 combination that gives 4 ones. You have 4 combinations that give 4 ones (111x) (x111) (1x11) (11x1). You have 6 combinations that give 2 ones (11xx) (1x1x) (1xx1) (x11x) (x1x1) (xx11).

Now, in the case of 2 ones, you have 5^2 possible combinations of the dice that are not 1s.

So your total number of possible glitches are 1 + 4*5 + 6*25 = 171. Of these 2*4 + 4*5*6 = 128 are standard glitches.

For a pool of 4 dice that is a 128/1296 chance of a standard glitch and a 43/1296 chance of a critical glitch.

It is pretty much the same as the pools increase but there is just more addition and more multiplication and you have to be careful to keep track everything.

Hey Serbitar, I got an error when trying to open your xls file. It pitched a circular reference warning.

Serb: You are my hero.
Hyz: Okay. That's what I needed - the full on breakdown of where the numbers come from. Perfect.
Eid: I suspect that's simply because you have columns with 0s- if you fill them in, all should be well.

There is a circular reference in it. I sent you a PM with the details.

Thanks for the hint. Its corrected now.