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dice statistics help

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Juzek's picture
Joined: 06/19/2017

I'm looking for some good formulas for my dice.

I am currently having players roll 4 6-sided dice up to 3 times, saving or not saving any dice they roll until the last roll, where they must keep it. (fairly standard Yahtzee style)

Lets say I have one side that is very bad, and one side that is very good.

if the player is selecting for or against the dice, what is the likely average number of good/bad icons rolled?

I know the likely hood of the specific number you want never showing up is (5 misses/6 sides)^12 rolls total = 11.2%, thus the likelihood of seeing it at least once is 100%-11.2% = 88.8

and the likely hood of getting all the good specific side on the first roll is (1/6)^4 = 0.07%

but how do I calculate the in-between likelihoods? the average? preferably as a formula that I can alter the number of dice and rolls allowed until I can tune it.

Thanks for the help!

X3M's picture
Joined: 10/28/2013
My attempt

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.

X3M wrote:
I consider % as a factor of 1/100. Percentage means per one hundred. Cent is an old roman word for one hundred.
Ok, I hope my calculation at the end makes sense this way.


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: ##% ##% ##%


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...


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.

The results:

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.

4 Dice

3 Dice

2 Dice

1 Die

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% +

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!

Juzek's picture
Joined: 06/19/2017
I tried a table

I tried a table method

I broke down all the possible ways to get 4 good dice, 3 good, 2 good 1 good 0 good, and filled in with the chances from Anydice what the likelihood for each roll is.

But I must have a mistake somewhere! The total for all the chances for all the scenarios is only 55%, and it should be 100.

How does my table look?

@anyone else
There has got to be a formula out there for something like this.


X3M's picture
Joined: 10/28/2013
You modified later on, right?

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.

X3M's picture
Joined: 10/28/2013
I think you made a mistake

Which happens when you don't let the system correct you.

23,36% 0
37,38% 1
26,17% 2
10,47% 3
2,62% 4

I copied this from my sheet. I prepared the 4d6 from a sheet that has ALL number of dice.

I expanded the sheet for 4d6 from 1 to 12 rolls.

I have 1,32 while you have 1,75? something ain't right there. I know that I made no mistake. Because the 2nd roll again gives 0,67.

With infinite rolls, you get at an average good result of 4. My table goes towards the 4, with 3,43 at the 12th roll.

Now, making the tables for less or more dice will be easy. It is simply copy pasting the stuff. And then change the size. Perhaps.... I could make one up to 12 dice. Which would simply be used for <12 dice as well. Because if we build up. Some calculations have to be re applied. And a field of 12*12 is a bit much to re-apply per sheet.

So, if you want to see it. I want to know 2 things:
- How to share it anyway? PM me on this one.
- Would you like me to prepare for other number of dice as well?

Juzek's picture
Joined: 06/19/2017
I have updated the

I have updated the permissions in the above link, if you want to show me the kind of thing you were talking about.

Feel free to edit it. Or make a new tab etc

X3M's picture
Joined: 10/28/2013
Hard to share if copy paste fails...

I corrected some mistakes I made. Just to give you an idea on how difficult keeping an overview might be.
My claim about 3 rolls above wasn't right. But after correction, I am more sure. I got to compare with your results. And we have almost the same now. I could even spot the differences. I coloured these red for you and put in my list right next to it.

I copied both sheets that I have into your file.
But the copy paste makes no sense...
Can you PM me your email? Then I simply send the file.
That should help more.
And things like, colours, formula's etc would be in there.

The file is rather complex for new eye's. (Especially when only the text is visible) But I might update it another time to answer questions.
But you can ask me in PM, through email or here.

Maybe I should make some sort of tutorial on how I do things in a spreadsheet. But Excel is copyright. So no video's on that.

And you are free to ask me to calculate different things.

Cheers, X3M

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