Stacking advantage: doing the math

Morrus

Well, that was fun
Staff member
One thing I've always liked the idea of in D&D, but which has never been strongly enough rewarded, is the idea of characters maneuvering for advantageous positions. In the past we've had +2 flanking bonuses, in some editions a measly +1 higher ground bonus (which was too small and fiddly for anyone to care about), and bonus/penalties for standing/prone.

For example, I kinda imagine a difficult shot which, say, a guy with a crossbow normally would find it very hard to make. But by stacking the odds in his own favour - findind a decent spot on higher ground, being hidden, various other things - he can make that shot easier and attempt a shot that he normally wouldn't.

Bonuses of +2 were never really enough for that. Advantage sounds like it will be. However, advantage doesn't stack, so he is not rewarded for taking multiple measures to improve his chances. He just has to pick one.

So my question is this:

We've got a thread on the bonus equivalent of advantage (+3/+4, IIRC). I assume that if advantage were to stack (roll three dice, roll four dice) the statistical benefit would suffer from diminishing returns, which sounds good. But what would the statistics be? Would allowing advantage to stack create virtual certainties most of the time?

What if the target was taking his own actions to give you disadvantage, thus reducing the number of dic you roll - that could lead to some interesting tactical play with movement and positioning being important. I get higher ground, but he is behind a waist-high rock (at the simplest level - that's how it would work without stacking); but what if each of us were trying to take a couple of other measures, too? I'm also hiding, and he crouches. And so on.
 

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I assume that if advantage were to stack (roll three dice, roll four dice) the statistical benefit would suffer from diminishing returns, which sounds good. But what would the statistics be? Would allowing advantage to stack create virtual certainties most of the time?

What if the target was taking his own actions to give you disadvantage, thus reducing the number of dic you roll - that could lead to some interesting tactical play with movement and positioning being important. I get higher ground, but he is behind a waist-high rock (at the simplest level - that's how it would work without stacking); but what if each of us were trying to take a couple of other measures, too? I'm also hiding, and he crouches. And so on.

This effectively turns D&D into a dice pool system, with a bunch of other weird maths layered in as well. Seems like a very complicated mix.
 

This effectively turns D&D into a dice pool system, with a bunch of other weird maths layered in as well. Seems like a very complicated mix.

It might do the former, but I'm not seeing the "complicated" part.

But what I'm really interested in is what mathematical effect it would have - whether it would break things or not. I'm not enough of a statistician to tell.
 

I was thinking, too, of letting advantages and disadvantages stack!!

After all, advantages never let you achieve a result better than the maximum you could roll anyway. They only make you closer to that (the opposite for disadvantages of course).

So, you total up all advantages granted by favourable conditions and subtract disadvantages from negative ones. The net result is the number of rerolls you can make (good or bad).

Similar to the Alternity system but much simpler. I don't think it can overwhelm the game. After all a 1 is still a failure and a 20 a success.
 

But what I'm really interested in is what mathematical effect it would have - whether it would break things or not. I'm not enough of a statistician to tell.

Time to be more helpful.

Let's assume that the base success chance is 60%.

The chance of one success from 1 dice is: 60%
The chance of one success from 2 dice is: 84%
The chance of one success from 3 dice is: 93%

Something like that for each additional dice.
 

Time to be more helpful.

Let's assume that the base success chance is 60%.

The chance of one success from 1 dice is: 60%
The chance of one success from 2 dice is: 84%
The chance of one success from 3 dice is: 93%

Something like that for each additional dice.

So from three dice onwards, we're looking at near certainty? Yeah, that does sound too much. I was hoping the diminishing returns would be more pronounced.
 

So from three dice onwards, we're looking at near certainty? Yeah, that does sound too much. I was hoping the diminishing returns would be more pronounced.

Nah, it's non-linear, because you only need one success from any of the dice and each has individually a 60% chance.

My maths might be a bit wonky, but it seems right.
 

So from three dice onwards, we're looking at near certainty? Yeah, that does sound too much. I was hoping the diminishing returns would be more pronounced.

In the "middle ground" of percentages, i.e. when the single die success percentage is between 40% - 60% yes, you get a big bump, because the probability curve becomes bell-shaped.

On the extremes the deviation is less pronounced.
 

Well, it's still a bell curve -- so, those numbers are right if the chance of success is 60%, but if the chance of success is 20% (you need to roll a 17 or better to hit) it looks like this:

1 Die: 20%
2 dice: 36%
3 dice: 49%

And if it's 10 %

1 die: 10%
2 dice: 19%
3 dice: 27%

So, it drops off sharply the harder it is to hit.

IMO, stacking adv/dis doesn't really make enough difference in the game to really be worth the effort.

It's important to note in the RAW that I think we're meant to read Advantage and Disadvantage as conditions, not currencies -- you don't build up more than one. And advantage and disadvantage cancel each other out.


So, if you have advantage from three sources, you have advantage. If you have advantage from three sources and disadvantage from one, they cancel each other out (it doesn't require three sources of disadvantage to counter three sources of advantage).

I'm not sure how I feel about that, but in the long run it's probably simpler, and that's probably better.

-rg
 

Stacking Advantage Math

The die rolls are all independent events, so the formula to figure out if at least one of them is successful is this:

P(A or B) = P(A) + P(B) - P(A and B)

So if we use a success chance of 60% as Justinhalliday did then the formula looks like this:

.6 + .6 - .36 = .84

[P(A and B) is the product of P(A) and P(B). .6 * .6 = .36]

It gets a bit more complicated once you're looking at more then 2 events. So...

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

.6 + .6 + .6 - .36 - .36 - .36 + .216 = .936

Math!
 

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