Love the video, and recognize it isn't a math explainer, but there is some algebra or algorithms that make the formula easier: Edit: (21 + A + .5(H-M+1)(H-M+1)(H-M)/H) - D)/20 when H=help die size & M = minimum success on help die (min 1) ((21+A-D)(8-M) + .5(H)(H-1) - .5(M)(M-1))/(H)20 when H=help die size M = greater of 0 or D-A-19 Edit: There is an easier way to get to the same answer. Looking at the average, adding a help die is identical to adding the average of the die if you could have passed the threshold without the help die. However, when you need the help die, for example needing at least a 2 from a d8, then, contenuing the example, you can find the average of a d7 +1, then multiply by 7/8, the chance you'l roll within the success range. That means for any singular help die, you can take the normal (21 + A - D)/ 20 and slot in a + .5(H-M+1)(H-M+1)(H-M)/H (M=Minimum roll to succeed, min 1) in the numerator making the formula (21 + A + .5(H-M+1)(H-M+1)(H-M)/H) - D)/20. To do multiple dice, you still need to find the average if they roll high enough and the chance they roll that high. First, when you have 1/8( (21 + A + 8 - D)/20 + (21 + A + 7 - D)/20 + ...), you can pull the numerators together and simplify it into 1/8 * (21(8) + A(8) - D(8) + 8 + 7 ...)/20 which can be rewritten as (21(8) + A(8) - D(8) + .5(8)(8-1)) / ((8)20) using the fact sum from 1 to N = .5N(N-1) and multiplying the denominators. This however falls into the Nieve issue of forgetting about minimum roll requirements. To fix this in the series you simply leave off the terms that are less than 1 in the numerator, but when you compact them you can do the same by subtracting .5(M)(M-1) (1 to M) and subtracting M from the 8 multiples, where M is the minimum roll required to hit on a 19, AKA D-A-19 with a minimum of 0. Additionally, the 8 can be substituted for any other help die size and the 21 for any other heavy/brutal test, with the caveat that you ignore negative. So that's ((21+A-D)(8-M) + .5(8)(8-1) - .5(M)(M-1))/(8)20 when M = greater of 0 or D-A-19
Ayyooogreat video Claudio!
I'm used to see a Help Die just as +4.5 and completely forgot about anything beyond 4. Thanks!
Glad you could appreciate the main point of the video
yoooooo help dice time! They're one of my fav mechanics
Yup, it's a tricky one
This is why I read basic math for dummies! 😊
@@Rick-bardAggro that's enough for the gameplay side of things
You rolled an 8 on your Help Die on helping us understand the math of Help Die 😁
@@Knucklebone.of.Fickle.Fortune i space out my videos so I don't suffer the multiple help penalty
You are the Treantmonk of DC20!
@@LuizCesarFariaLC that's a great honor, he was an inspiration for starting my channel
Love the video, and recognize it isn't a math explainer, but there is some algebra or algorithms that make the formula easier:
Edit: (21 + A + .5(H-M+1)(H-M+1)(H-M)/H) - D)/20 when H=help die size & M = minimum success on help die (min 1)
((21+A-D)(8-M) + .5(H)(H-1) - .5(M)(M-1))/(H)20 when H=help die size M = greater of 0 or D-A-19
Edit: There is an easier way to get to the same answer. Looking at the average, adding a help die is identical to adding the average of the die if you could have passed the threshold without the help die. However, when you need the help die, for example needing at least a 2 from a d8, then, contenuing the example, you can find the average of a d7 +1, then multiply by 7/8, the chance you'l roll within the success range. That means for any singular help die, you can take the normal (21 + A - D)/ 20 and slot in a + .5(H-M+1)(H-M+1)(H-M)/H (M=Minimum roll to succeed, min 1) in the numerator making the formula (21 + A + .5(H-M+1)(H-M+1)(H-M)/H) - D)/20.
To do multiple dice, you still need to find the average if they roll high enough and the chance they roll that high.
First, when you have 1/8( (21 + A + 8 - D)/20 + (21 + A + 7 - D)/20 + ...), you can pull the numerators together and simplify it into 1/8 * (21(8) + A(8) - D(8) + 8 + 7 ...)/20 which can be rewritten as (21(8) + A(8) - D(8) + .5(8)(8-1)) / ((8)20) using the fact sum from 1 to N = .5N(N-1) and multiplying the denominators. This however falls into the Nieve issue of forgetting about minimum roll requirements. To fix this in the series you simply leave off the terms that are less than 1 in the numerator, but when you compact them you can do the same by subtracting .5(M)(M-1) (1 to M) and subtracting M from the 8 multiples, where M is the minimum roll required to hit on a 19, AKA D-A-19 with a minimum of 0. Additionally, the 8 can be substituted for any other help die size and the 21 for any other heavy/brutal test, with the caveat that you ignore negative.
So that's ((21+A-D)(8-M) + .5(8)(8-1) - .5(M)(M-1))/(8)20 when M = greater of 0 or D-A-19
@@IlstrawberrySeed excellent breakdown to simplify the equations
Insano video
@@lautaroalvarenga9537 hopefully in a good way lol
This one is totally going to hurt poor Wolves' brain :)
@@Fred-drk it's not as bad as the resistance video on my opinion
second
@King_Hodop not bad
@@claudio_wild eh it was an hour late.
How the sausage is made.
@Bardicaggravation it's a tasty sausage tho