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How do I compare the result of “1d20+x, with advantage” to “1d20+y, without advantage”, assuming x
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$begingroup$
I roll a d20.
If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?
Assume x < y.
dnd-5e statistics advantage-and-disadvantage
$endgroup$
add a comment |
$begingroup$
I roll a d20.
If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?
Assume x < y.
dnd-5e statistics advantage-and-disadvantage
$endgroup$
add a comment |
$begingroup$
I roll a d20.
If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?
Assume x < y.
dnd-5e statistics advantage-and-disadvantage
$endgroup$
I roll a d20.
If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?
Assume x < y.
dnd-5e statistics advantage-and-disadvantage
dnd-5e statistics advantage-and-disadvantage
edited 5 hours ago
V2Blast
29k5105177
29k5105177
asked 14 hours ago
EradashEradash
32213
32213
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add a comment |
4 Answers
4
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oldest
votes
$begingroup$
Theory
The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).
begin{array}{r|llll}
text{Natural DC} & text{Advantage} & text{No Advantage} & text{Difference} & text{Eq. Flat Modifier} \ hline
text{DC 1-} & text{100.000%} & text{100.000%} & text{0.000%} & 0 \
text{DC 2} & text{99.750%} & text{95.000%} & text{4.750%} & 0.95 (1)\
text{DC 3} & text{99.000%} & text{90.000%} & text{9.000%} & 1.8 (2) \
text{DC 4} & text{97.750%} & text{85.000%} & text{12.750%} & 2.55 (3) \
text{DC 5} & text{96.000%} & text{80.000%} & text{16.000%} & 3.2 (4) \
text{DC 6} & text{93.750%} & text{75.000%} & text{18.750%} & 3.75 (4) \
text{DC 7} & text{91.000%} & text{70.000%} & text{21.000%} & 4.2 (5) \
text{DC 8} & text{87.750%} & text{65.000%} & text{22.750%} & 4.55 (5) \
text{DC 9} & text{84.000%} & text{60.000%} & text{24.000%} & 4.8 (5) \
text{DC 10} & text{79.750%} & text{55.000%} & text{24.750%} & 4.95 (5) \
text{DC 11} & text{75.000%} & text{50.000%} & text{25.000%} & 5 \
text{DC 12} & text{69.750%} & text{45.000%} & text{24.750%} & 4.95 (5) \
text{DC 13} & text{64.000%} & text{40.000%} & text{24.000%} & 4.8 (5) \
text{DC 14} & text{57.750%} & text{35.000%} & text{22.750%} & 4.55 (5) \
text{DC 15} & text{51.000%} & text{30.000%} & text{21.000%} & 4.2 (5) \
text{DC 16} & text{43.750%} & text{25.000%} & text{18.750%} & 3.75 (4) \
text{DC 17} & text{36.000%} & text{20.000%} & text{16.000%} & 3.2 (4) \
text{DC 18} & text{27.750%} & text{15.000%} & text{12.750%} & 2.55 (3) \
text{DC 19} & text{19.000%} & text{10.000%} & text{9.000%} & 1.8 (2) \
text{DC 20} & text{9.750%} & text{5.000%} & text{4.750%} & 0.95 (1) \
text{DC 21+} & text{0.000%} & text{0.000%} & text{0.000%} & 0 \
end{array}
A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.
There are a few casual observations we can make:
- It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.
- At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.
- It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.
- But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).
- In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.
Practice
So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:
- Calculate the difference between y and x
- Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"
- Look at the Eq. Flat Modifier for that row in the table
- If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.
Examples
- DC19, 1d20+5/Adv vs 1d20+7/NoAdv
- Difference is 7 - 5 == 2
- DC19 - 5 is DC14
- DC14 has a Eq. Flat Modifier of 4.55
- Therefore, the Advantage roll is better than the non-Advantage roll.
- DC3, 1d20+1/Adv vs 1d20+2/NoAdv
- Difference is 2 - 1 == 1
- DC3 - 1 is DC2
- DC2 has a Eq. Flat Modifier of 0.95
- Therefore, the non-advantage roll is better than the Advantage roll
- DC17, 1d20+9/Adv vs 1d20+14/NoAdv
- We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...
- Difference is 14 - 9 == 5
- DC17 - 9 is DC8
- DC8 has a Eq. Flat Modifier of 4.55
- Therefore, the non-advantage roll is better than the Advantage roll
Attack Rolls
Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.
Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.
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If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
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– Yakk
10 hours ago
add a comment |
$begingroup$
It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.
Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.
(source: Zero Hit Points)
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In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
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– illustro
13 hours ago
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@illustro thanks, my experience is limited to 3(.5)E ...
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– Glorfindel
13 hours ago
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(and even there you didn't need to beat a DC, but it's been a while ...)
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– Glorfindel
13 hours ago
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I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
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– Carcer
12 hours ago
1
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@Carcer: Have some more AnyDice. :)
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– Ilmari Karonen
10 hours ago
add a comment |
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If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.
Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.
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1
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This answer is great! And shows me we didn't need actual numbers. +1
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– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
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– Carcer
13 hours ago
1
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The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
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– Yakk
11 hours ago
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
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– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
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– Carcer
11 hours ago
|
show 1 more comment
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First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:
In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.
(Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)
Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...
$y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;
$y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;
$y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;
$y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or
$y ge x + 5$.
(As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)
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Here is what I think to be a simpler graph OP can plug numbers into. Merely changeX:1
to what X you want and the same forY:2
. E.g., you want 5 and 10,X:5
andY:10
.
$endgroup$
– Captain Man
9 hours ago
add a comment |
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4 Answers
4
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4 Answers
4
active
oldest
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active
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active
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$begingroup$
Theory
The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).
begin{array}{r|llll}
text{Natural DC} & text{Advantage} & text{No Advantage} & text{Difference} & text{Eq. Flat Modifier} \ hline
text{DC 1-} & text{100.000%} & text{100.000%} & text{0.000%} & 0 \
text{DC 2} & text{99.750%} & text{95.000%} & text{4.750%} & 0.95 (1)\
text{DC 3} & text{99.000%} & text{90.000%} & text{9.000%} & 1.8 (2) \
text{DC 4} & text{97.750%} & text{85.000%} & text{12.750%} & 2.55 (3) \
text{DC 5} & text{96.000%} & text{80.000%} & text{16.000%} & 3.2 (4) \
text{DC 6} & text{93.750%} & text{75.000%} & text{18.750%} & 3.75 (4) \
text{DC 7} & text{91.000%} & text{70.000%} & text{21.000%} & 4.2 (5) \
text{DC 8} & text{87.750%} & text{65.000%} & text{22.750%} & 4.55 (5) \
text{DC 9} & text{84.000%} & text{60.000%} & text{24.000%} & 4.8 (5) \
text{DC 10} & text{79.750%} & text{55.000%} & text{24.750%} & 4.95 (5) \
text{DC 11} & text{75.000%} & text{50.000%} & text{25.000%} & 5 \
text{DC 12} & text{69.750%} & text{45.000%} & text{24.750%} & 4.95 (5) \
text{DC 13} & text{64.000%} & text{40.000%} & text{24.000%} & 4.8 (5) \
text{DC 14} & text{57.750%} & text{35.000%} & text{22.750%} & 4.55 (5) \
text{DC 15} & text{51.000%} & text{30.000%} & text{21.000%} & 4.2 (5) \
text{DC 16} & text{43.750%} & text{25.000%} & text{18.750%} & 3.75 (4) \
text{DC 17} & text{36.000%} & text{20.000%} & text{16.000%} & 3.2 (4) \
text{DC 18} & text{27.750%} & text{15.000%} & text{12.750%} & 2.55 (3) \
text{DC 19} & text{19.000%} & text{10.000%} & text{9.000%} & 1.8 (2) \
text{DC 20} & text{9.750%} & text{5.000%} & text{4.750%} & 0.95 (1) \
text{DC 21+} & text{0.000%} & text{0.000%} & text{0.000%} & 0 \
end{array}
A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.
There are a few casual observations we can make:
- It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.
- At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.
- It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.
- But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).
- In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.
Practice
So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:
- Calculate the difference between y and x
- Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"
- Look at the Eq. Flat Modifier for that row in the table
- If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.
Examples
- DC19, 1d20+5/Adv vs 1d20+7/NoAdv
- Difference is 7 - 5 == 2
- DC19 - 5 is DC14
- DC14 has a Eq. Flat Modifier of 4.55
- Therefore, the Advantage roll is better than the non-Advantage roll.
- DC3, 1d20+1/Adv vs 1d20+2/NoAdv
- Difference is 2 - 1 == 1
- DC3 - 1 is DC2
- DC2 has a Eq. Flat Modifier of 0.95
- Therefore, the non-advantage roll is better than the Advantage roll
- DC17, 1d20+9/Adv vs 1d20+14/NoAdv
- We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...
- Difference is 14 - 9 == 5
- DC17 - 9 is DC8
- DC8 has a Eq. Flat Modifier of 4.55
- Therefore, the non-advantage roll is better than the Advantage roll
Attack Rolls
Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.
Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.
$endgroup$
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
add a comment |
$begingroup$
Theory
The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).
begin{array}{r|llll}
text{Natural DC} & text{Advantage} & text{No Advantage} & text{Difference} & text{Eq. Flat Modifier} \ hline
text{DC 1-} & text{100.000%} & text{100.000%} & text{0.000%} & 0 \
text{DC 2} & text{99.750%} & text{95.000%} & text{4.750%} & 0.95 (1)\
text{DC 3} & text{99.000%} & text{90.000%} & text{9.000%} & 1.8 (2) \
text{DC 4} & text{97.750%} & text{85.000%} & text{12.750%} & 2.55 (3) \
text{DC 5} & text{96.000%} & text{80.000%} & text{16.000%} & 3.2 (4) \
text{DC 6} & text{93.750%} & text{75.000%} & text{18.750%} & 3.75 (4) \
text{DC 7} & text{91.000%} & text{70.000%} & text{21.000%} & 4.2 (5) \
text{DC 8} & text{87.750%} & text{65.000%} & text{22.750%} & 4.55 (5) \
text{DC 9} & text{84.000%} & text{60.000%} & text{24.000%} & 4.8 (5) \
text{DC 10} & text{79.750%} & text{55.000%} & text{24.750%} & 4.95 (5) \
text{DC 11} & text{75.000%} & text{50.000%} & text{25.000%} & 5 \
text{DC 12} & text{69.750%} & text{45.000%} & text{24.750%} & 4.95 (5) \
text{DC 13} & text{64.000%} & text{40.000%} & text{24.000%} & 4.8 (5) \
text{DC 14} & text{57.750%} & text{35.000%} & text{22.750%} & 4.55 (5) \
text{DC 15} & text{51.000%} & text{30.000%} & text{21.000%} & 4.2 (5) \
text{DC 16} & text{43.750%} & text{25.000%} & text{18.750%} & 3.75 (4) \
text{DC 17} & text{36.000%} & text{20.000%} & text{16.000%} & 3.2 (4) \
text{DC 18} & text{27.750%} & text{15.000%} & text{12.750%} & 2.55 (3) \
text{DC 19} & text{19.000%} & text{10.000%} & text{9.000%} & 1.8 (2) \
text{DC 20} & text{9.750%} & text{5.000%} & text{4.750%} & 0.95 (1) \
text{DC 21+} & text{0.000%} & text{0.000%} & text{0.000%} & 0 \
end{array}
A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.
There are a few casual observations we can make:
- It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.
- At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.
- It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.
- But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).
- In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.
Practice
So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:
- Calculate the difference between y and x
- Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"
- Look at the Eq. Flat Modifier for that row in the table
- If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.
Examples
- DC19, 1d20+5/Adv vs 1d20+7/NoAdv
- Difference is 7 - 5 == 2
- DC19 - 5 is DC14
- DC14 has a Eq. Flat Modifier of 4.55
- Therefore, the Advantage roll is better than the non-Advantage roll.
- DC3, 1d20+1/Adv vs 1d20+2/NoAdv
- Difference is 2 - 1 == 1
- DC3 - 1 is DC2
- DC2 has a Eq. Flat Modifier of 0.95
- Therefore, the non-advantage roll is better than the Advantage roll
- DC17, 1d20+9/Adv vs 1d20+14/NoAdv
- We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...
- Difference is 14 - 9 == 5
- DC17 - 9 is DC8
- DC8 has a Eq. Flat Modifier of 4.55
- Therefore, the non-advantage roll is better than the Advantage roll
Attack Rolls
Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.
Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.
$endgroup$
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
add a comment |
$begingroup$
Theory
The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).
begin{array}{r|llll}
text{Natural DC} & text{Advantage} & text{No Advantage} & text{Difference} & text{Eq. Flat Modifier} \ hline
text{DC 1-} & text{100.000%} & text{100.000%} & text{0.000%} & 0 \
text{DC 2} & text{99.750%} & text{95.000%} & text{4.750%} & 0.95 (1)\
text{DC 3} & text{99.000%} & text{90.000%} & text{9.000%} & 1.8 (2) \
text{DC 4} & text{97.750%} & text{85.000%} & text{12.750%} & 2.55 (3) \
text{DC 5} & text{96.000%} & text{80.000%} & text{16.000%} & 3.2 (4) \
text{DC 6} & text{93.750%} & text{75.000%} & text{18.750%} & 3.75 (4) \
text{DC 7} & text{91.000%} & text{70.000%} & text{21.000%} & 4.2 (5) \
text{DC 8} & text{87.750%} & text{65.000%} & text{22.750%} & 4.55 (5) \
text{DC 9} & text{84.000%} & text{60.000%} & text{24.000%} & 4.8 (5) \
text{DC 10} & text{79.750%} & text{55.000%} & text{24.750%} & 4.95 (5) \
text{DC 11} & text{75.000%} & text{50.000%} & text{25.000%} & 5 \
text{DC 12} & text{69.750%} & text{45.000%} & text{24.750%} & 4.95 (5) \
text{DC 13} & text{64.000%} & text{40.000%} & text{24.000%} & 4.8 (5) \
text{DC 14} & text{57.750%} & text{35.000%} & text{22.750%} & 4.55 (5) \
text{DC 15} & text{51.000%} & text{30.000%} & text{21.000%} & 4.2 (5) \
text{DC 16} & text{43.750%} & text{25.000%} & text{18.750%} & 3.75 (4) \
text{DC 17} & text{36.000%} & text{20.000%} & text{16.000%} & 3.2 (4) \
text{DC 18} & text{27.750%} & text{15.000%} & text{12.750%} & 2.55 (3) \
text{DC 19} & text{19.000%} & text{10.000%} & text{9.000%} & 1.8 (2) \
text{DC 20} & text{9.750%} & text{5.000%} & text{4.750%} & 0.95 (1) \
text{DC 21+} & text{0.000%} & text{0.000%} & text{0.000%} & 0 \
end{array}
A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.
There are a few casual observations we can make:
- It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.
- At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.
- It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.
- But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).
- In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.
Practice
So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:
- Calculate the difference between y and x
- Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"
- Look at the Eq. Flat Modifier for that row in the table
- If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.
Examples
- DC19, 1d20+5/Adv vs 1d20+7/NoAdv
- Difference is 7 - 5 == 2
- DC19 - 5 is DC14
- DC14 has a Eq. Flat Modifier of 4.55
- Therefore, the Advantage roll is better than the non-Advantage roll.
- DC3, 1d20+1/Adv vs 1d20+2/NoAdv
- Difference is 2 - 1 == 1
- DC3 - 1 is DC2
- DC2 has a Eq. Flat Modifier of 0.95
- Therefore, the non-advantage roll is better than the Advantage roll
- DC17, 1d20+9/Adv vs 1d20+14/NoAdv
- We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...
- Difference is 14 - 9 == 5
- DC17 - 9 is DC8
- DC8 has a Eq. Flat Modifier of 4.55
- Therefore, the non-advantage roll is better than the Advantage roll
Attack Rolls
Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.
Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.
$endgroup$
Theory
The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).
begin{array}{r|llll}
text{Natural DC} & text{Advantage} & text{No Advantage} & text{Difference} & text{Eq. Flat Modifier} \ hline
text{DC 1-} & text{100.000%} & text{100.000%} & text{0.000%} & 0 \
text{DC 2} & text{99.750%} & text{95.000%} & text{4.750%} & 0.95 (1)\
text{DC 3} & text{99.000%} & text{90.000%} & text{9.000%} & 1.8 (2) \
text{DC 4} & text{97.750%} & text{85.000%} & text{12.750%} & 2.55 (3) \
text{DC 5} & text{96.000%} & text{80.000%} & text{16.000%} & 3.2 (4) \
text{DC 6} & text{93.750%} & text{75.000%} & text{18.750%} & 3.75 (4) \
text{DC 7} & text{91.000%} & text{70.000%} & text{21.000%} & 4.2 (5) \
text{DC 8} & text{87.750%} & text{65.000%} & text{22.750%} & 4.55 (5) \
text{DC 9} & text{84.000%} & text{60.000%} & text{24.000%} & 4.8 (5) \
text{DC 10} & text{79.750%} & text{55.000%} & text{24.750%} & 4.95 (5) \
text{DC 11} & text{75.000%} & text{50.000%} & text{25.000%} & 5 \
text{DC 12} & text{69.750%} & text{45.000%} & text{24.750%} & 4.95 (5) \
text{DC 13} & text{64.000%} & text{40.000%} & text{24.000%} & 4.8 (5) \
text{DC 14} & text{57.750%} & text{35.000%} & text{22.750%} & 4.55 (5) \
text{DC 15} & text{51.000%} & text{30.000%} & text{21.000%} & 4.2 (5) \
text{DC 16} & text{43.750%} & text{25.000%} & text{18.750%} & 3.75 (4) \
text{DC 17} & text{36.000%} & text{20.000%} & text{16.000%} & 3.2 (4) \
text{DC 18} & text{27.750%} & text{15.000%} & text{12.750%} & 2.55 (3) \
text{DC 19} & text{19.000%} & text{10.000%} & text{9.000%} & 1.8 (2) \
text{DC 20} & text{9.750%} & text{5.000%} & text{4.750%} & 0.95 (1) \
text{DC 21+} & text{0.000%} & text{0.000%} & text{0.000%} & 0 \
end{array}
A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.
There are a few casual observations we can make:
- It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.
- At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.
- It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.
- But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).
- In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.
Practice
So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:
- Calculate the difference between y and x
- Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"
- Look at the Eq. Flat Modifier for that row in the table
- If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.
Examples
- DC19, 1d20+5/Adv vs 1d20+7/NoAdv
- Difference is 7 - 5 == 2
- DC19 - 5 is DC14
- DC14 has a Eq. Flat Modifier of 4.55
- Therefore, the Advantage roll is better than the non-Advantage roll.
- DC3, 1d20+1/Adv vs 1d20+2/NoAdv
- Difference is 2 - 1 == 1
- DC3 - 1 is DC2
- DC2 has a Eq. Flat Modifier of 0.95
- Therefore, the non-advantage roll is better than the Advantage roll
- DC17, 1d20+9/Adv vs 1d20+14/NoAdv
- We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...
- Difference is 14 - 9 == 5
- DC17 - 9 is DC8
- DC8 has a Eq. Flat Modifier of 4.55
- Therefore, the non-advantage roll is better than the Advantage roll
Attack Rolls
Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.
Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.
edited 8 hours ago
answered 12 hours ago
XiremaXirema
26.1k373152
26.1k373152
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
add a comment |
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
$begingroup$
If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
$endgroup$
– Yakk
10 hours ago
add a comment |
$begingroup$
It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.
Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.
(source: Zero Hit Points)
$endgroup$
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
1
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
add a comment |
$begingroup$
It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.
Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.
(source: Zero Hit Points)
$endgroup$
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
1
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
add a comment |
$begingroup$
It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.
Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.
(source: Zero Hit Points)
$endgroup$
It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.
Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.
(source: Zero Hit Points)
edited 13 hours ago
answered 13 hours ago
GlorfindelGlorfindel
6821614
6821614
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
1
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
add a comment |
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
1
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
$endgroup$
– illustro
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
@illustro thanks, my experience is limited to 3(.5)E ...
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
(and even there you didn't need to beat a DC, but it's been a while ...)
$endgroup$
– Glorfindel
13 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
$begingroup$
I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
$endgroup$
– Carcer
12 hours ago
1
1
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
$begingroup$
@Carcer: Have some more AnyDice. :)
$endgroup$
– Ilmari Karonen
10 hours ago
add a comment |
$begingroup$
If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.
Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.
$endgroup$
1
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
1
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
|
show 1 more comment
$begingroup$
If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.
Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.
$endgroup$
1
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
1
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
|
show 1 more comment
$begingroup$
If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.
Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.
$endgroup$
If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.
Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.
edited 11 hours ago
answered 13 hours ago
CarcerCarcer
28.1k583148
28.1k583148
1
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
1
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
|
show 1 more comment
1
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
1
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
1
1
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
This answer is great! And shows me we didn't need actual numbers. +1
$endgroup$
– NautArch
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
$begingroup$
@NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
$endgroup$
– Carcer
13 hours ago
1
1
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
$begingroup$
The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
$endgroup$
– Yakk
11 hours ago
1
1
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
$begingroup$
@Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
$endgroup$
– Carcer
11 hours ago
|
show 1 more comment
$begingroup$
First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:
In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.
(Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)
Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...
$y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;
$y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;
$y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;
$y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or
$y ge x + 5$.
(As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)
$endgroup$
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely changeX:1
to what X you want and the same forY:2
. E.g., you want 5 and 10,X:5
andY:10
.
$endgroup$
– Captain Man
9 hours ago
add a comment |
$begingroup$
First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:
In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.
(Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)
Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...
$y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;
$y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;
$y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;
$y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or
$y ge x + 5$.
(As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)
$endgroup$
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely changeX:1
to what X you want and the same forY:2
. E.g., you want 5 and 10,X:5
andY:10
.
$endgroup$
– Captain Man
9 hours ago
add a comment |
$begingroup$
First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:
In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.
(Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)
Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...
$y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;
$y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;
$y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;
$y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or
$y ge x + 5$.
(As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)
$endgroup$
First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:
In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.
(Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)
Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...
$y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;
$y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;
$y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;
$y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or
$y ge x + 5$.
(As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)
answered 9 hours ago
Ilmari KaronenIlmari Karonen
11.5k33448
11.5k33448
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely changeX:1
to what X you want and the same forY:2
. E.g., you want 5 and 10,X:5
andY:10
.
$endgroup$
– Captain Man
9 hours ago
add a comment |
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely changeX:1
to what X you want and the same forY:2
. E.g., you want 5 and 10,X:5
andY:10
.
$endgroup$
– Captain Man
9 hours ago
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely change
X:1
to what X you want and the same for Y:2
. E.g., you want 5 and 10, X:5
and Y:10
.$endgroup$
– Captain Man
9 hours ago
$begingroup$
Here is what I think to be a simpler graph OP can plug numbers into. Merely change
X:1
to what X you want and the same for Y:2
. E.g., you want 5 and 10, X:5
and Y:10
.$endgroup$
– Captain Man
9 hours ago
add a comment |
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