simple #s question: what is ADV worth?

[da newt]

In term of a simple rule of thumb, how much is ADV worth? If an average D20 roll is 10.5, what is the average result of a D20 roll with ADV? I assume something like 13, but I'm not sure how to calculate it.

[Dork_Forge]

In term of a simple rule of thumb, how much is ADV worth? If an average D20 roll is 10.5, what is the average result of a D20 roll with ADV? I assume something like 13, but I'm not sure how to calculate it.

I think the rough rule is advantage is like +5, but a more mathy type will give a better breakdown.

[BloodSnake'sCha]

The average is 13.825, a 3.325 improvement from the 10.5 of a regular roll.

You can calculate it like this:
There is one option for a 1 (1,1)
3 for 2 (1,2)(2,2)(2,1)
5 for 3 (1,3)(2,3)(3,3)(3,1)(3,2)
...
39 for 20 (I am not going to write it).

We will some up the number times the number of options for this number.
1*1+2*3+3*5+4*7....20*39 and will devaide by the total number of options (400 for a two 20 sides dice).

We will get 5530/400=13.825

[Chronos]

"What's the average roll?" is the wrong question to ask, because d20 rolls are success or failure, and it doesn't matter by how much you succeed or fail. The important question to ask is, by how much does advantage increase your chance of success? And that depends on how difficult the roll was to begin with.

If you ordinarily have a 50% chance of success, then advantage will turn that into a 75% chance. That's equivalent to getting +5 on the roll. That's probably about the most typical situation to come up in the game, but it's also the situation where advantage has the biggest impact. On the other hand, if you ordinarily have a 95% chance of success, then advantage will turn that into a 99.75% chance, making it worth slightly less than a +1 on the roll. Or at the other end of the scale, if you ordinarily have a 5% chance, then advantage will turn that into a 9.75% chance, again worth slightly less than +1. And if you have no chance at all of success, or are guaranteed success, then advantage won't matter at all.

Another complication that can come up, though, is that with attack rolls, no matter your plusses or the enemy's AC, a 1 always misses, and a 20 always hits. If the enemy's AC outclasses your attack bonuses by so much that the natural 20 rule is the only way that you can hit, or is so much lower than your bonus that the natural 1 rule is the only way that you can miss, then a +1 or -1 won't matter at all, but advantage or disadvantage will still make a slight difference.

Yet another complication comes from opposed rolls, which don't have a fixed DC. There's a lot more to keep track of there in counting the probabilities, but I can say that if you start from normal rolls on both sides, then giving yourself advantage is exactly equivalent to giving your opponent disadvantage, or vice-versa (though obviously, it's better yet to do both). And if the base roll is (approximately) 50-50, then advantage will turn it into (approximately) a 67% chance (depending on how you handle ties).

[Contrast]

Not as simple a question as you might think as the relative benefit depends on what number you're trying to roll.

This page (http://zerohitpoints.com/Articles/Advantage-in-DnD-5) helps break it down a bit.

[da newt]

Bloodsnake - thanks for the specific answer I was looking for.

Chronos and Contrast - thanks for all the other stuff that comes along with it. I figured there had to be a point of max and min return, and had sort of figured some of it out, but the info provided is way better than my rough estimations.

[LudicSavant]

Advantage changes your hit rate to 1-(1-P)^2, where P is the probability of hitting without Advantage. For example, if your hit rate was 60%, your Advantage hit rate would be 1-(1-.6)^2 = 84%.

It also changes your crit rate in the same fashion. For example, if your previous crit rate was 5%, your new crit rate would be 1-(1-0.05)^2, or 9.75%.

If you'd instead like the formula for Elven Accuracy, it's just 1-(1-P)^3 instead of ^2.

[Contrast]

If you haven't use it before you can also mess around with AnyDice which allows you to simulate roll outcomes. Here it is for advantage (http://anydice.com/program/b38).

[Keravath]

This thread shows how it works but shows the actual % change in to hit (for both advantage and disadvantage). A 5% change in to hit is the same as a +1 modifier.

https://www.enworld.org/threads/the-math-of-advantage-and-disadvantage.654979/

From the table, if you need to roll an 11 to hit, advantage increases your chances to hit by 25% which is equivalent to +5. However, if you needed to roll a 15 then advantage is more like +4, and if you needed to roll a 19 to hit then advantage increases your to hit chance by 9% (10%->19%) or about a +2 (i.e. about the same as needing a 17 to hit instead of 19).

A good rule of thumb is that Advantage is +4 to +5 over the middle range of target numbers you need to roll 8-14 and a bit less outside that.

[bid]

We will some up the number times the number of options for this number.
1*1+2*3+3*5+4*7....20*39 and will devaide by the total number of options (400 for a two 20 sides dice).
Disadvantage is easier to calculate:
(1^2 + 2^2 + 3^2 + ... + 20^2) / 20^2 = 2870/400 ~ 7.175
{python: sum(x**2 for x in range(21)) / 20.0**2}

Still, you should expect a better bonus on average use since you almost never try the far ends of the odds. If you need 20 to succeed, you'll need more than advantage to risk it. If you consider most rolls will be in the 4-17 range, you will get a better weighted average.


Note:
Same thing applies to 4d6 drop 1, where the dropped value is (1^4 + ... + 6^4) / 6^4 ~ 1.1755
{python: sum(x**4 for x in range(7)) / 6.0**4}

[CheddarChampion]

Advantage is +3.325 on 'average.' It is not equivalent to that number though, as the range of possible results stays the same.

Crit chance boosts by 4.75% and crit fail chance decreases by 4.75% if applicable.
(just less than +1 or -1 on a d20 each, +1 is worth 5%)

[LudicSavant]

Disadvantage is easier to calculate

You can simplify those methods to the formulas I posted, in which case you should be able to do it in your head on the fly.

[NNescio]

In term of a simple rule of thumb, how much is ADV worth? If an average D20 roll is 10.5, what is the average result of a D20 roll with ADV? I assume something like 13, but I'm not sure how to calculate it.

The expected value (i.e. average) is 13.825. But this is not the right question to ask, because the value of advantage changes depending on what target number you are trying to roll.

Most of the time, for most rolls that matter, advantage is worth a +4 or a +5.

Here's a more in-depth breakdown:




Roll Required
d20 probability
2d20b1 probability
Difference
Effective† Bonus


≥1*
100%
100%
0
0


≥2
95%
99.75%
4.75%
+0.95


≥3
90%
99.00%
9.00%
+1.8


≥4
85%
97.75%
12.75%
+2.55


≥5
80%
96.00%
16.00%
+3.2


≥6
75%
93.75%
18.75%
+3.75


≥7
70%
91.00%
21.00%
+4.2


≥8
65%
87.75%
22.75%
+4.55


≥9
60%
84.00%
24.00%
+4.8


≥10
55%
79.75%
24.75%
+4.95


≥11
50%
75.00%
25.00%
+5


≥12
45%
69.75%
24.75%
+4.95


≥13
40%
64.00%
24.00%
+4.8


≥14
35%
57.75%
22.75%
+4.55


≥15
30%
51.00%
21.00%
+4.2


≥16
25%
43.75%
18.75%
+3.75


≥17
20%
36.00%
16.00%
+3.2


≥18
15%
27.75%
12.75%
+2.55


≥19
10%
19.00%
9.00%
+1.8


20
5%
9.75%
4.75%
+0.95


*Assuming Nat 1 ≠ autofail (e.g. Saves and Ability Checks)

†Strictly speaking, actual decimal bonuses would be useless [rounded down, basically]. The numbers here aren't actual bonuses though; they are just chosen to convey the 'effective' improvement in success rate.

Put in another way, when you roll d20 with advantage, 51% of your rolls will be at least 15. That number (15) is also the smallest median which falls within the range of possible values, which is a more useful 'metric' of average for this particular purpose (binary success/fail outcomes), instead of the expected value.

(I suppose some people would just simplify and call this the 'median' of a 2d20b1, but technically it's an abuse of terminology.)

[bid]

You can simplify those methods to the formulas I posted, in which case you should be able to do it in your head on the fly.
This is the average of your formulas, not the same thing.

Although the average is not that useful.

[NNescio]

Disadvantage is easier to calculate:
(1^2 + 2^2 + 3^2 + ... + 20^2) / 20^2 = 2870/400 ~ 7.175
{python: sum(x**2 for x in range(21)) / 20.0**2}

It's exactly 7.175, not approximately.

For advantage, just do 20 - (1^2 + 2^2 + 3^2 + ... + 19^2) / 20^2 = 13.825

print(20 - sum(x**2 for x in range(20)) / 20.0**2)
...which is equivalent to 21 - 7.175. Generally, Max(dX, dX) = X+1 - Min(dX, dX), because of the Linearity of Expectation.

Proof: E(Max(dX, dX)) = 2E(dX) - Min(dX, dX)
= 2*(X+1)/2 - Min(dX, dX) = X+1 - Min(dX, dX)

Alternatively, one can solve E(2dXb1) ('average' value of 2dX pick highest) analytically with the following formula:

E(2dXb1) = (4X - 1) (X + 1)/6X

And for E(2dXw1) ('average' value of 2dX pick lowest):

E(2dXw1) = (X + 1)(2X + 1)/6X





Note:
Same thing applies to 4d6 drop 1, where the dropped value is (1^4 + ... + 6^4) / 6^4 ~ 1.1755
{python: sum(x**4 for x in range(7)) / 6.0**4}

It's ~1.755, assuming you're talking about E(Min(4d6)).

[bid]

Alternatively, one can solve E(2dXb1) ('average' value of 2dX pick highest) analytically with the following formula:

E(2dXb1) = (4X - 1) (X + 1)/6X

And for E(2dXw1) ('average' value of 2dX pick lowest):

E(2dXw1) = (X + 1)(2X + 1)/6X
Is there a general formula for 3d6, 4d6 or more?

[NNescio]

Is there a general formula for 3d6, 4d6 or more?

Yes but you need Faulhaber's formula for the term involving the sum of powers. It's one of those... "you need LaTeX to write it out clearly" things. (There's a 'shorthand' 'cheaty' way of writing it, but that's hardly useful.)

You can implement this in Python using scipy.stats.binom and Fraction, which might be faster than a simple lambda iteration (especially if you def Faulhaber's as a recursive function), but it's much more complicated to write out.

Probably someone has already implemented it in a math library somewhere though.

--

Edit: There's a quick and dirty approximation you can use. The min for nDx can be approximated by min(nDx) ≈ 1/(n+1) * x + 0.5. Basically what we are doing here is fully expanding the power sum terms but then taking only the first two while dropping the rest. This gives a fast approximation that should be 'semi'-accurate to one decimal place for small values of n (which is quite bad, but eh, good enough for a quick ballpark figure done mentally).

Likewise the 'average' of nDx drop lowest can be approximated by:
E(nDxb[n-1]) ≈ n(x + 1)/2 - x/(n+1) - 0.5

Basically, heh, you know how the average (i.e. expected value) of a dX die is given by X/2 + 0.5, right?

Well, due to a quirk in the math:
The 'average' minimum of 2 dX dice is roughly equivalent to X/3 + 0.5.
And for 3 dX dice, it's X/4 + 0.5.
For 4 dX dice, it's X/5 + 0.5.
...And so on.

To approximate the average of "n dX dice drop lowest", we take the average of ndX and subtract the average minimum of n dX dice, which is how we arrive at E(nDxb[n-1]) ≈ n(x + 1)/2 - x/(n+1) - 0.5

Now, as for the approximate 'average' maximums:
For 2 dX dice, it's 2/3*X + 0.5.
For 3 dX, it's 3/4*X + 0.5
For 4 dX, 4/5*X + 0.5
...And so on.

The average of "n dX dice drop highest" can be likewise approximated in a similar manner.

(Just be aware that these are very rough approximations, and they get relatively worse as you roll more and more dice.)

[KorvinStarmast]

Depending on your target number, from +2 to +5.

Details here. (https://rpg.stackexchange.com/q/14690/22566)

From the Answer by RS Conley at that link:

Advantage 16+ to hit, goes from 25% chance of success to roughly 43% chance of success. (.75 time .75)
Disadvantage 16+ to hit, goes from 25% chance of success to roughly a 6% chance of success (.25 times .25)
Why target number matters (If you are trying to roll a 9, 14, or 19 influences how much advantage will help the roll)

The general rule of thumb that in the mid range of the d20 (from success on a 9+ to 12+) advantage grant roughly a equivalent to a +5 bonus and disadvantage a -5 penalty. The increase and decrease in odds tappers off when your odds of success approach 1 or 20. For example a advantage on a 19+ your chance of failure goes from 90% to 81% not quite a +2 bonus on a d20.

A useful application of knowing the odds of rolling two dice is that you can just convert it to a straight bonus when rolling for a large number of NPCs.

A bunch of goblins with an advantage from surprise that need 13+ to hit the players you can just apply a +4 (or +5 if you round up) bonus instead of rolling the second dice.
This is because they have a 60% chance of failure on 13+.
Taking .6 times .6 yields .36 a drop of 24%.
Not quite a +5 bonus on a d20 dice.

[bid]

Now, as for the approximate 'average' maximums:
For 2 dX dice, it's 2/3*X + 0.5.
For 3 dX, it's 3/4*X + 0.5
For 4 dX, 4/5*X + 0.5
...And so on.
That's very nice napkin maths, thank you!

[Dirclaw]

Your question has already been answered, but if you are interested in the math of it, and not just the results, then advantage for an x sided die will always be equal to ((x^2)-1)/6x, or about x/6. However, this is the absolute difference, not the difference depending on the number needed to hit

[strangebloke]

If your chances of success without advantage are 50% it's a +5, unless you're making an attack roll in which case it's functionally +6 because it roughly doubles the 1/20 chance that you'll deal double damage.

At the far end (you need a natural 20) it's a +1 for a save or a skill check or a +2 for an attack roll.

Elven accuracy is a +8 at best and a +2 at worst.