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.