dice statistics help

There are much better people out there, who can help you with this. But I get the feeling you are stuck with me and my own kind of logic. I can't give you a formula. I am too stupid for that. I can only guide you into how I would calculate these things.

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Either way. You need to cut your problem in smaller pieces.

After all, you say that you roll 4 dice. And you roll them 3 times as long as you roll something bad.

Your main question would be, how much is the chance on every number of successes. And for every roll.

Ok, so, you want a list that has something like this:

G: 1st 2nd 3th
0: ##% ##% ##%
1: ##% ##% ##%
2: ##% ##% ##%
3: ##% ##% ##%
4: ##% ##% ##%

Right?

Or perhaps you want the average results of good rolls.
Which would simply become a list:
Rolls (of 4): Good
1: ##
2: ##
3: ##

The main problem here is the number of re-rolls on the individual dice. When a die is good, it will not be re-rolled. Thus reducing the chance by the number of dice.

If I where to use tables. I could calculate this. But I can't show it too you. If only we had a Discord for BGDF.

Either way. We start with...

0

You have no good rolls. So your first roll commences with 4 dice. We are going to consider the number of successes. And we can use anydice for this.

https://anydice.com/program/22ccb

The results:
0,48.225308642
1,38.5802469136
2,11.5740740741
3,1.54320987654
4,0.0771604938272

Now, let's look at the 0. We got a 48.2% chance that you have 0 good results. This means that the 2nd roll is going to have 4 dice.
If we look at for example the mentioned 4 good results. Then there is no 2nd roll.
We need to consider how much dice the second roll is going to have. And multiply the results of that with the chance that we even need to do a 2nd roll.

So, the next step is to consider the results with less than 4 dice. But we still include the 4 dice.

https://anydice.com/program/22ccc

4 Dice
0,48.225308642
1,38.5802469136
2,11.5740740741
3,1.54320987654
4,0.0771604938272

3 Dice
0,57.8703703704
1,34.7222222222
2,6.94444444444
3,0.462962962963

2 Dice
0,69.4444444444
1,27.7777777778
2,2.77777777778

1 Die
0,83.3333333333
1,16.6666666667

How to read this?
When we roll the dice, we add a number of good results on the previous roll.
So if the first roll yields 2 good results. The chance for this is 11.57%. The roll of 2 dice is also 11.57%.
We look at 2 Dice. And an example of rolling 2 more good results is 2.78%.
The true chance of this happening is 11.57% * 2.78% = 0.32%
So a chance of 0.32% that we roll 2 good results in the first roll and get to 4 in the 2nd roll. This chance however can be added up to the 0.077% of rolling 4 good dice right away. After all, we want to know how big the chance is in getting 4 good dice. But we need to calculate other results as well. The 4 good dice in the second round include:
4 in the first round +0 in the 2nd round
3 in the first round +1 in the 2nd round
2 in the first round +2 in the 2nd round
1 in the first round +3 in the 2nd round
0 in the first round +4 in the 2nd round

On a side note, +0 in the 2nd round with 0 dice is 100%. So the chance on the 4 good rolls in the first round is used again.

So you need to add up 5 percentages for a total chance on getting 4 in the 2nd round. Then you use this percentage and rinse and repeat. If you do this correctly. Every round of rolls will slowly shift the results from low to high.

In order to calculate the average per round. Simply have the result 0,1,2,3 or 4, multiplied by the chance it occurs. For example, the first round average is:

0 * 48.225308642%
1 * 38.5802469136%
2 * 11.5740740741%
3 * 1.54320987654%
4 * 0.0771604938272% +
________________________
0.6667

Yes, that's right, the average result in the first roll is 2/3th.

Now, perhaps you can make the table. Or show me a place where I can show it to you. Because then I would make the table in excel. And it would go...eventually simple and fast...
I don't do much formula's anymore in calculating chances. I have fallen in love with my table method.

Good luck!

I see you got it for 98% complete now. The little percentages differences are the key here for the missing parts.

I will see, if I got some time this weekend. And make the table like I intended. This table should be able to be expanded and stuff. Merely by copy pasting. Then you will see what I meant.

But yours is good too. It is just more specific for the described situation of 4d6 that have 3 rolls.

You had to sort it out, the ways of how a player can roll, didn't you? My original method helps in this.