Ability contest math help please

[Galithar]

If I am initiating an ability contest where I know my bonus is significantly higher then my opponents is there a statistical difference between giving myself advantage or my opponent disadvantage?

For example if I have a +12 and my opponent has a +4 I can lose if I roll 8 lower then them. (9 if I win a tie)

Can someone show me the math to determine which method (adv/disadv) would make my chance of failure the smallest, or if they are mathematically equivalent?

[Diego]

If I am initiating an ability contest where I know my bonus is significantly higher then my opponents is there a statistical difference between giving myself advantage or my opponent disadvantage?

For example if I have a +12 and my opponent has a +4 I can lose if I roll 8 lower then them. (9 if I win a tie)

Can someone show me the math to determine which method (adv/disadv) would make my chance of failure the smallest, or if they are mathematically equivalent?

Assuming that ability contest is "roll, see how much you pass by, they roll, see how much they pass by, see who passed by most" then the answer is they are equivalent, basically by symmetry.

You can think of the contest as being YourD20 - TheirD20 + [difference in stats]. When looking at an average d20 roll, advantage improves a roll exactly as much as disadvantage hurts it because there is nothing special about numbers 1-10 compared to 11-20. So [YourD20 + advmod] - TheirD20 will give the same result as YourD20 - [TheirD20 - advmod]

[MoiMagnus]

Yes, it is symmetrical.

To prove it mathematically, you just need to remark that "d20" and "21 - d20" are equivalent.

Small proof:

From the remark we deduce "max of two d20" = "max of two (21-d20)" = "21 - min of two d20"
So "max of two d20 + my bonus VS d20 + his bonus" is the same as "21 - min of two d20 + my bonus VS 21 - d20 + his bonus"
We can get rid of the 21 on both side, and we obtain "my bonus - min of two d20 VS his bonus - d20"
Subtracting something on one side is the same as adding it on the other side, so we obtain "d20 + my bonus VS min of two d20 + his bonus"

[NNescio]

If I am initiating an ability contest where I know my bonus is significantly higher then my opponents is there a statistical difference between giving myself advantage or my opponent disadvantage?

For example if I have a +12 and my opponent has a +4 I can lose if I roll 8 lower then them. (9 if I win a tie)

Can someone show me the math to determine which method (adv/disadv) would make my chance of failure the smallest, or if they are mathematically equivalent?


Let us ignore modifiers and percentages for a moment, and focus only on the random results of three dice.

Let us define the result of an arbitrary fair die as the random, uniform, discrete variable X. It doesn't matter how many sides the die has, but for the purposes of the following equations we are going to assume all the dice share the same number of faces.

Next we define E(X) as the expectation (AKA the expected value) of rolling such a die. In common layman terms, this is what most people would call the average roll of a die.

We then define Max(X, X) as the greater of two dice rolls when rolling two identical fair dice (or the same die rolled twice). Likewise we define Min(X, X) as the lower of two dice rolls for the same procedure. Analogously, we define E(Max(X, X)) as the expectation of such a maximum (i.e 'average' result when rolling a die with advantage) , and likewise for E(Min(X, X)) (i.e 'average' result when rolling a die with disadvantage).

The overall 'average result' of rolling with advantage against an opponent with neither advantage nor disadvantage can be represented by the following equation:

E(Advantage vs. Neutral) = E(Max(X,X) - X)

As for rolling with neither advantage nor disadvantage against an opponent with disadvantage:

E(Neutral vs. Disadvantage) = E(X - Min(X, X))

By Linearity of Expectation (http://www.cse.iitd.ac.in/~mohanty/col106/Resources/linearity_expectation.pdf) (which basically says that the 'average' of a sum of random numbers is equal to the sum of their individual averages), we can expand the inner terms:

E(Advantage vs. Neutral)
= E(Max(X,X) - X)
= E(Max(X, X)) - E(X)

Note that the maximum of two die rolls is the sum of both rolls minus the minimum:

= E(2X - Min(X, X)) - E(X)
= 2E(X) - E(Min(X, X)) - E(X) { Expanding again, as per Linearity of Expectation
= E(X) - E(Min(X, X))
= E(X - Min(X, X))
= E(Neutral vs. Disadvantage)
Q.E.D.

Thus both are equivalent.

Now, actually converting this into success percentages is somewhat complicated, but we can just treat it as a function that maps from E to to p:

Let f:E → p
Given that E(Max(X,X) - X)) = E(X - Min(X, X))
∴ f(E(Max(X,X) - X)) = f(E(X - Min(X, X))

...that is, we get the same % chance on both sides of the equations, because we are feeding the same 'input' into the same function.

With modifiers, we are still adding the same constants on both sides (let's call them c1 and c2), so the equality again still holds, because Linearity of Expectation still applies:

That is, E(Max(X,X) - X + c1 - c2 )
= E(Max(X,X) - X) + c1 - c2
= E(X - Min(X, X)) + c1 - c2 { We already established earlier that E(Max(X,X) - X) = E(X - Min(X, X))
= E(X - Min(X, X) + c1 - c2)

Again, to get actual success percentages, we can just feed this into the function earlier, and since both inputs are the same, both would still yield the same % success.

As seen above, this proof applies for dice of any arbitrary sides, and for any arbitrary combination of integer modifiers c1 and c2, so long as all the die rolls are made with identical fair dice with the same number of sides.

Edit: Made a mistake with f. Fixed.

[Galithar]

Thanks guys. All makes sense I just could not for the life of me wrap my head around that last night lol

For whatever reason my brain was trying to convince me that giving them disadvantage would make them more likely to not be able to match my roll. But that's not the case when both sides are variable like it is against a static DC.

I knew that the difference between disadvantage and straight roll on a DC 19 goes from 10 percent success to 1 percent success. Which makes it 10 times harder to pass the check.

And against a DC 11 it goes from 50% to 25% a bigger absolute change but smaller relative change.

It makes perfect sense why this is a completely different problem then when both sides are variable I just couldn't figure out why until y'all explained it to me haha

[Chronos]

MoiMagnus, that's the exact same argument that I use, but you managed to put it much more succinctly than I.

The only catches in the calculation come when there's a possibility of something else imposing advantage and/or disadvantage to one or the other of the contestants. That might mean that your choice instead turns into a choice between both at advantage and both normal, or between both normal and both at disadvantage. In that case, what it comes down to is, if you have the better modifier, then you want the option with more rolls, while if you have the worse modifier, then you want the option with fewer rolls.

And of course, if you can manage it, you at advantage AND your opponent at disadvantage is best of all.

[Galithar]

MoiMagnus, that's the exact same argument that I use, but you managed to put it much more succinctly than I.

The only catches in the calculation come when there's a possibility of something else imposing advantage and/or disadvantage to one or the other of the contestants. That might mean that your choice instead turns into a choice between both at advantage and both normal, or between both normal and both at disadvantage. In that case, what it comes down to is, if you have the better modifier, then you want the option with more rolls, while if you have the worse modifier, then you want the option with fewer rolls.

And of course, if you can manage it, you at advantage AND your opponent at disadvantage is best of all.

The initial question for me was "Is it better to enlarge myself, or reduce my opponent if attempting to grapple"
I now definitively know that enlarge is better. The math on a successful grapple is the same, and Enlarge can't fail whereas reduce can.