This was an original idea by Eroschilles, i believe. The purpose is to show people how to find out the statistics for making saves in the game and even pulls from a case or booster. To start out, I am only going to do the simple math involved in statistics. This will involve simple multiplication, division, addition, subtraction, and fractions. Later on, I will add the more difficult nCr probabilities. Those simply make it faster to figure out very large probabilities, such as the chances of getting your most wanted pieces in a case.

To Start: The d20 has, of course, 20 sides. This means that each number is essentially 5%, or 1/20. A save of 11 or higher basically means above 10, so the chance of success is 10/20, or 50%. That is the pretty much all you need to know about the d20.
Single Saves: Since you know how the d20 works, you also know how to figure out your chances of making saves.

Here they are:
save 6: 75% success, 25% failure
save 11: 50% success, 50% failure
save 16: 25% success, 50% failure

Multiple Saves: But what about multiple rolls for one save? Simple multiplication works good for this.

Here is an example of a roll where you must make two saves:
save 11: 50%
2nd save 11: 50%
.5*.5=.25
that means you have a 25% chance of succeeding both rolls, and a 75% chance of failing one or both of the rolls.

Here is an example in which you have two chances to make a save:
save 11: 50%
2nd save 11: 50%
This is trickier, since you must account for all outcomes. The simplest way to do this is to write them all out and then add up the different chances of success and failure.
1. succeed: .5=.5
2. fail, succeed: .5*.5=.25
3. fail, fail: .5*.5=.25
Success: 50%+25%=75%
Fail: 25%
That means you have a 75% chance of succeeding overall, and a 25% chance of failing.

The same logic can easily be applied to more than two saves.

Mettle Reroll: This essentially works the same as multiple saves, but you add 4/20 to the roll, or 20%.

Here is an example anyway:

Force Point reroll with Mettle
save 11: 50%
2nd save 11: 50%+20%=70%
1. succeed: .5=.5
2. fail, succeed: .5*.7=.35
3. fail, fail: .5*.3 =.15
Success: 50%+35%=85%
Fail: 15%
That means you have a 85% chance of succeeding overall, and a 15% chance of failing.

In Practice: Lets combine them all in a demonstration of ****'s infamous soresu style mastery rolls:

****'s Force Point rerolls with mettle
save 11: 50%
2nd save 11: 50%+20%=70%
3rd save 11: 50%+20%+20%=90%
1. succeed: .5=.5
2. fail, succeed: .5*.7=.35
3. fail, fail, succeed: .5*.3*.9=.135
4. fail, fail, fail: .5*.3*.1=.015
Success: 50%+.35%+13.5%=98.5%
Fail: 1.5%
That means you have a 98.5% chance of succeeding overall, and a 1.5% chance of failing.

I will do the probability on cases later. If you find any mistakes, please notify me.
P.S.: Please do not turn this into a **** argument, it was simply a demonstration since he had all of the different abilities necessary for me to show the scenario.

did you know that 86.356 percent of stats are made up on the fly?

I am thinking that you may have way too much time on your hands.

Were you listening to Styxx while compiling this info?

Those are interesting numbers, but I think those commanders, especially Leia, are easily dispatched. **** needs no commanders to get his numbers. I think I would give the edge to **** just because of that. Very interesting, nonetheless.

How about this:

HK-50 vs. Fully Boosted Shock Trooper
34 Defense
.05 chance to hit
Expected damage: 2

How about my method for cases?

With JA Leia, Wedge and Luke force spirt Kol almost has the soresu style and mettel.

Main problem there:

That's 96 points. The *ahem* Soresu Style Mastery figure was 55 points. By himself. No boosting. And wasn't owned by Shooters adjacent to him.

Thanks for doing this thread. It's good to get math practice.

I remember seeing over at the WOTC boards a few months back the probability of Boba, BH disintegrating some one being around 9%. Is that right? I thought each time he attacked it was treated as an independent occurrence, putting the probability of getting a disintegration at 5%. Or were they treating the twin as a single occurence?

(I know, random question, but at least it has to do with probabilities).

I think they got something like 9.75%, which I think they would have used the formal addition rule to do. But that doesn't make sense as each attack is a separate occurence or whatnot. I think they were confused, or I'm confused now. Maybe both...

Ok, I see how you came up with that for disintegration. Only problem is that doesn't account for if a bodyguard takes a disintegration from a first attack by Boba. Then he gets he second attack from the twin, even though the first was a disintegration. But, either way the math comes out the same with or without a body guard for his chance of getting a disintegration in a given turn. I always counted each roll independently at 5%.
If Boba was given extra attack, would that significantly increase his chances of getting the big D?