ForeverTangent
Topic Author
Posts: 1
Joined: Tue 13 Jul 2021, 23:21

### Year Zero Success Roll Table question.

Hello All

In all the Year Zero system books, there is printed the 'Chance to Succeed' table. The first column is the number of Dice, the second column is the success percentage, and the third column the the pushed roll success percentage.

From my limited memories of my college statistics class, I understand how the second column is calculated for base success.

However, for the life of me I cannot figure out how the third column is being calculated.

Can someone enlighten me on the formula being used? I just cannot figure out how Free League are getting those numbers.

Staque

(Example of table below.)
Tales from the Loop Success Table
Screen Shot 2021-07-13 at 5.27.00 PM.png (17.31 KiB) Viewed 496 times

cheeplives
Posts: 104
Joined: Sat 28 Nov 2020, 09:39

### Re: Year Zero Success Roll Table question.

If you use the second one, remember the total dice rolled are 2xpool (so to get the 3 row, you'd use y=6, not y=3).

https://math.stackexchange.com/question ... iple-rolls

https://math.stackexchange.com/question ... ls/1894724

Spat
Posts: 64
Joined: Sat 04 Jan 2020, 13:32

### Re: Year Zero Success Roll Table question.

Can someone enlighten me on the formula being used? I just cannot figure out how Free League are getting those numbers.
As said in the articles, it is easier to calculate the probability of failure, and subtract from 1 to know the one for success.
It is the case here as a "success" is validated if you have 1 or several 6 rolled, and this probability is hard to calculate.
If we make a calculation for 2 dice for example (as it gets complicated real quick) :

- your probability to fail 2 dice roll is (5/6)^2
you need to roll 1 to 5 on all dice.
That's 0.694, thus your probability of success is 0.31

- your probability to fail twice (you can push) a 2 dice roll is :
(probability to fail 1st roll) x (probability to fail 2nd)
As an approximation you can say the 2 probabilities are the same and you find 0.694^2 = 0.48
Thus your chance of success is approximately 52%.

But that's not entirely true as you don't re roll 1 on the Attribute Dice (AD).
then you have to count all the combinations, hopefully with a computer :

if you rolled 1 Attribute Dice + 1 Skill Dice (you are wounded already)
If you rolled a 1 on the AD, then you re roll 1 SD and you have 5 possible failures out of 6 occurrences.
If not, then you re roll 2 Dice, and you have 5x5 possible failures out of 6x6 occurrences.

And if you rolled 2 AD : the same than before
+ the case where you rolled twice 1, and then you have 1 failure out of 1 occurence (you don't re roll)

Thus the count of failures is : (5 + 5x5) + (5 + 5x5 + 1) = 61
and the count of occurrences is : (6 + 6x6) + (6 + 6x6 + 1) = 85
And a probability of failure of 61/85 = 0,718 (failing the pushed roll only)

And then the probability of failing both is : 0,694 * 0,718 = 0,498

Hence the 50% to succeed a 2 dice roll with the possibility to push.

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