What is a decent dice rolling strategy for partial/half advantage?

Question

I understand that 5e is all about advantage and cancelling, etc. I understand advantage well, and the statistics. I was hoping there would be a commonly known dice rolling strategy with similar statistical distribution as regular advantage, just less so. Something perfectly between a regular 1d20 roll and an advantage roll (2d20 take highest) https://anydice.com/program/24913.

What dice rolling strategies do you know of that can create something halfway between a regular d20 roll and an advantage roll (especially when it comes to criticals)?

Extra dice are fine (up to 6 I won't even blink at), having to use maths is fine (let's limit it to 6 operations, and nothing that needs a calculator).

I'd like something that isn't curved like 2@4d20 (roll 4d20 and take second highest) or 3d6 etc. Though I'm always up to learn about experiences playing with any other rolling strategies. (To illustrate: https://anydice.com/program/24917)

What I'm hoping it does is halve the difference in the crit probabilities (something more like 2.5% for 1 and 7.5% for 20, and then generally bias towards larger numbers. Anything that can do that I think will be potentially usable as something between advantage and regular roll. But that perfect linear result would make it useful in more applications than just DC rolls (roll higher than X). For instance, loot table rolling can be more interesting (obviously the table would have to designed for it).

I've spent tens of hours searching and read all the discussions I could find, and haven't yet come across anything compelling. My assumption is that I'm missing something obvious. I've also spent some time tinkering in anydice hoping to stumble on an option, to no avail. The closest I got before posting is this abomination: https://anydice.com/program/2491d

Dislaimer I understand that this is strictly core changing house-rules/homebrewing, maybe highly discouraged, and that many will scoff... so let me clarify that I'm not looking for answers that direct me to better understand 5e RAW. I'm aware of how Lucky interacts with the dice as well as Elven Accuracy, etc. I am not yet worried about edge cases, I'm just looking for options. I know what I am asking, and am wondering if anyone has found something like I'm describing.

Answer 1

The most straightforward way to add "half Advantage" is to only make it work half of the time. You can do this by using different colored D20s and throw in another d6 to decide whether you're allowed to choose or not.

For example, a red D20, a green D20, and a green d6. If you roll 4-6 on the d6 you pick the highest die, if you roll 1-3 on the d6 you are stuck with whatever the red D20 rolled. Same for Disadvantage except you pick the lowest die on a 4-6 on the d6. (Or on a 1-3, if you think that's more thematic, as long as you make it clear as a group how it'll work)

Answer 2

Advantage when the dice Match

I like Eric's answer, and agree that "The most straightforward way to add "half Advantage" is to only make it work half of the time." However, that particular solution involves the addition of a third die. Granted, a pretty low increase in complexity, but strictly speaking unnecessary.

An even simpler 'half Advantage' mechanic would be to roll the 2d20¹ as usual, but permit Advantage only when the two dice were 'matched', that is, both were odd or both were even, since this would happen at random half the time.

You could use the same principle for 'half Disadvantage', where you have Disadvantage only when the two dice match².

When throwing physical dice, simply adding in an extra d6 as Eric suggests is easy enough, but the matching/not matching method may have some extra utility if playing on a VTT where you don't want to take the time to program in (or click on) an extra die roll.

¹_{Note that strictly speaking, this is not just a 2d20. It is an ordered pair where you know which die is the default, and which is the extra that may or may not count. Eric expresses this as "red die, green die", and in a VTT the default die is typically the one displayed first. Ilmari Karonen's answer, which says that there is no simple way to generate half advantage with just 2d20 is correct, because they are considering a "pure" 2d20 without any information distinguishing between the two dice.}

²_{My instinct was to call for Disadvantage on a mismatch, but as DqwertyC points out in comments, this results in a different expectation.}

Answer 3

Erik's suggestion of effectively flipping a coin to choose between a normal roll and a roll with advantage is pretty much the ideal "half advantage" roll, as it produces a distribution exactly halfway between normal and advantage. Its only down side is that you end up having to roll an extra die (or flip an actual coin, I guess).

If you'd instead prefer an approximate solution that works with just two d20 rolls, Kirt's solution — roll two distinguishable d20, pick highest if both are are even or both are odd, else pick an arbitrary predesignated one — is almost as good. Its only (minor) disadvantage is that you do need to have the two dice be distinguishable (e.g. of different colors) or rolled at different times so that you're able to pick one of them to be your "normal" d20 roll in the case where one of them rolls an even number and the other an odd one.

Honestly, I could almost just stop here and just tell you to upvote those two answers. But before Kirt posted their answer, I did come up with my own solution which, while being in other ways worse, does have one minor advantage over Kirt's: it doesn't require the two dice to be distinguishable in any way. Which is why I'll leave it here for completeness, at least for now.

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Here's how the method I came up with works: roll 2d20 and take the higher one, just like with normal advantage, but count even numbers as higher than odd numbers. (You can also do half disadvantage the same way, but taking the lower roll instead.)

The resulting distribution looks kind of weird and jagged, as you might expect, since even numbers will be a lot more likely than odd ones:

But if you look at the cumulative probability of rolling at least a given target number, you can see that it still lies quite close to half-way between normal and (dis)advantage, at least for most of the range:

This approximation does break down a bit at the extremes of the 1–20 range: a 20 with half advantage rolled this way is just as likely as with full advantage, and a 1 is just as unlikely.

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If you don't like that, you could always tweak the rule to count odd numbers as higher than even instead. However, this swings the pendulum all the way to the other side — with this tweak to the rule, rolling with half (dis)advantage doesn't change the probability of rolling 1 or 20 at all:

Honestly, though, I'd still suggest going with either Erik's or Kirt's solution instead. They're just better than what I came up with.

Answer 4

I would suggest rolling one or two d12 and a d20 for partial advantage. Here are the odds for "at least" results using Anydice.

• a straight d20 (black)
• highest of a d12 and a d20 (orange)
• highest of two d12 and a d20 (blue)
• normal advantage, highest of two d20 (green)

Analysis:

• Likelyhood of getting 13 or more is exactly same as with straight roll.
• The blue line makes result less than 8 slightly less likely than with normal advantage.
• Makes easy things (high bonus or low DC/AC) easier, does not help with succeeding on hard things (raw roll of 13 or more needed).
• Not exactly what you asked for (a line exactly between black and green lines in this graph, if I understood correctly).
• Does not increase chances of critical hits, when used for attack rolls.
• Does not change the mechanic of roll the dice and pick highest, so it is very simple to use.
• If you take "half advantage" to mean that you are less likely to not fail badly, but not more likely to do really well, then this is exactly that.
• Somewhat similar to rogue class feature "Reliable talent" (treat a roll less than 10 as 10).
• Without doing extra math, this only works with advantage... If used directly for disadvantage, it makes 12 the maximum roll, and makes natural 1 more likely than with 2d20 disadvantage.

The biggest reason to choose this for your purpose would be the simplicity, I think. The whole advantage/disadvantage with no stacking aims for that, so this tries to keep with that spirit.

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Credits: "highest" Anydice function adapted from this answer.

Answer 5

TLDR: Roll regular advantage, then also roll a d4, if d4 rolls 1 then flip the value (20->1, 1->20, 19->2, 11->10, etc)

Inpired by the answers I saw...

So after I posted the question I continued experimenting and did manage to come to the realization that to get the distribution I was looking for, the simplest way was probably just to modify the advantage roll some of the time.

After seeing the simplicity of Eric's answer I was flabbergasted how I could miss the obvious, and I put away what I was working on. However, after all the creative answers and seeing that they all have benefits in specific situations (eg: what if you don't have an easy way to distinguish the dice, or if using an online roller, or to avoid rolling extra dice), I was inspired to continue and see if I was close to something, and I managed to create another strategy...

It's not as elegant as all those shared here, it requires rolling another dice, and it even involves some simple math... but at least it has the right distribution, and the procedure might even provide a fun suspense/drama to the game.

Strategy

What I came up with is a minimally complicated way to dampen the distribution around the middle. The way I'll demonstrate works like Erik's answer in terms of probability. It's easy enough to perform, only involves 1 extra dice, the math is minimal (but that is more to complicate it). And what's great about this is you can easily adjust the flattening effect. The mechanic is to roll a "flip" die.

Let me explain what I mean by "flip"...

To flatten the probabilities, we want to balance around the middle (10.5). If some low percentage of the high values became the corresponding low value (reflected across the middle) it would flatten a bit. And we can apply this flip to both high and low values because there are less of the low values to flip. So when we flip: 20 becomes 1, 1 becomes 20, 19 becomes 2, 2 becomes 19, 18 becomes 3, 3 becomes 18, and so on until 11 becomes 10, and 10 becomes 11. The calculation I'm using is { M=10.5, X+(M-X)*2 }. We are "flipping" the number over the mid point. We do this at a set probability, say 1 in 4 (that's rolling a 4 sided dice and avoiding a 1, or hoping for a 4, depending on how you want to run it).

As for the math, it's not hard if you think about the number bonds for 10 (first bit of number sense people learn: 1+9, 2+8, 3+7, etc), because the flip value is just that plus 1. I don't think that's hard (then again I like math), but it will definitely put off some people.

Here are a couple examples

Conceptualizing the flip dice as "keeping" a good roll (just don't roll 1):

• Roll with advantage ( 18 & 4 becomes 18), we like this roll...
• Now roll the flip die to avoid the flip, a d4 (2)
• Our flip die did not fail us and we keep our 18!

Conceptualizing the flip dice as "saving" a bad roll (get the crit):

• Roll with advantage ( 1 & 6 becomes 6), we still have a chance...
• Now roll the flip die to get that flip, a d4 (4)
• Our flip die came through! That 4 becomes 17!

More Thoughts

The procedure could add good drama (d4 is dramatic either way), but also has the opportunity to be anticlimactic (rolling a 10 or 11, and generally close to the middle), but the statistics work out (AFAIK, and this is the only way to accomplish this so easily that I've found yet). Using a d4 as the flip die is perfect for "half" advantage, but you could use a d6 to be much closer to regular advantage with just a bit of flattening but still a rather significant addition of risk that shifts the chance of a crit fail from 1-in-400 to near 1-in-50. And of course you can play with it further, for instance a flip on 1 OR 6 of a d6 withh result in a 1/3 chance of a flip and a much flatter advantage closer to a regular d20.

Here's an anydice implementation of this method: https://anydice.com/program/2492e

Flexibility

What's interesting about this technique is it seems to work over almost any other roll. Anything you want to dampen, just use a flip dice to flatten how much you want. Only want a little bit of dampening: use a d12. Want to flatten it by half: d4. I'm still playing with it, but I like that it can be customized for any roll.

Here's a version that illustrates it's flexibility: https://anydice.com/program/24936

Here's an image of dampening 2@4d20 (roll 4d20, drop high and low, use remaining 2 for advantage... aka take the second highest number)

Though keep in mind this isn't actually dampening towards equal probability, it's bringing the reflections closer. So for a curve that is symmetrical around the middle it will do nothing. So the "sometimes apply a normal roll" is better at a true "dampening" effect. This method remove bias towards one end, but not bias towards the ends/middle.

Answer 6

I thought I would have a go at trying to make something as this seems like a fairly interesting problem.

I looked into it and if you only want to roll 2d20 it is impossible to get a distribution perfectly between no advantage and advantage but you can at least in theory make something really close, even if the dice are indistinguishable.

I originally made something like that but it was really messy and I thought it would be better to make something a bit cleaner although it is still a bit odd.

The basic idea is to roll 2d20 and to choose the lower of the two as long as the result is not too much lower. Specifically you take the lower dice unless it would reduce the result by more than 6, then you take the higher dice.

Examples:

1. Roll a 4 and a 7 and the result is 4
2. Roll a 5 and a 17 and the result is 17
3. Roll a 14 and a 20 and the result is 14

Admittedly it isn't the nicest solution but it gets fairly good results. The error in terms of cumulative probability is at most 2%.

Mostly I just did this for fun and honestly most of the other methods seem better and easier to use but I enjoyed the challenge.

Thanks for the interesting question.