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2014-07-08, 07:24 AM
I'd like to know how to calculate the odds of success behind rolls with advantage or disadvantage.
Anyone know the formula or can provide some examples?

Is it true that advantage is much more "advantageous" when you need to roll a 5 to hit, versus when you need to roll a 15?

Doug Lampert
2014-07-08, 08:06 AM
I'd like to know how to calculate the odds of success behind rolls with advantage or disadvantage.
Anyone know the formula or can provide some examples?

Is it true that advantage is much more "advantageous" when you need to roll a 5 to hit, versus when you need to roll a 15?

On 1d20 your chance of rolling X or higher is (21-X)/20 or if you prefer percent: (21-X)*5 %.

With disadvantage your chance of hitting with both dice, and thus your chance of hitting at all, is that value squared: (21-X)^2/400 or (21-X)^2/4 %.

On 1d20 your chance of FAILING to roll X or higher is (X-1)/20 or if you prefer percent: (X-1)*5 %.

With advantage your chance that both dice fail is that value squared: (X-1)^2/400
But you asked about success, so subtract from one to get chance of success with advantage and needing an X or better: 1-[(X-1)^2/400]
Converted to %: 100-[(X-1)^2/4] %

If you need an 11 to succeed then your chances are 25%, 50%, or 75% based on advantage and disadvantage.
If you need a 5 to succeed then your chances are 64%, 80%, or 96% based on advantage and disadvantage.
If you need a 15 to succeed then your chances are 9%, 30%, or 51% based on advantage and disadvantage.

Lycoris
2014-07-08, 08:14 AM
Work is slow and I like numbers, so go first post! :smalltongue:

This is roughly the same work as Doug has posted above, but may be a little easier to parse for some people, and doesn't provide ways to calculate failure (as calculating success naturally gives you the failure rate by nature of P(Success) + P(Failure) = 100%).

Disadvantage is a little easier to model than Advantage, so we'll start with that. As well, I'm going to ignore Criticals and Fumbles to look at general cases, if only because it's again easier, and because I think that the odds of one should roughly cancel out the odds of another in the general case.

To determine the effects of Disadvantage, remember that you can look at it as needing to succeed on two dice rolls. Therefore, the probability of succeeding with Disadvantage should be equal to the odds of succeeding twice at the same task.

Assuming you need a 5 or higher, this would give you:
P(Success Normal) = 16/20 = 80%
P(Success w/Disadvantage) = 16/20 * 16/20 = 256/400 = 64%

Assuming you need a 15 or higher, this would give you:
P(Success Normal) = 6/20 = 30%
P(Success w/Disadvantage) = 6/20 * 6/20 = 36/400 = 9%

In the case of Disadvantage, succeeding at something twice is harder when said task is harder to begin with.

To determine the effects of Advantage, you look at it as needing to succeed on one of two dice rolls. For this, only two probabilities are required: The probability of succeeding on the first roll (as the 2nd roll doesn't matter if you pass), and the probability of succeeding on the second roll, assuming you've failed the first. The sum of these two probabilities is your result.

Assuming you need a 5 or higher, this would give you:
P(Success Normal) = 16/20 = 80%
P(Success w/Advantage) = 16/20 + (4/20 * 16/20) = 384/400 = 96%

Assuming you need a 15 or higher, this would give you:
P(Success Normal) = 6/20 = 30%
P(Success w/Advantage) = 6/20 + (14/20 * 6/20) = 204/400 = 51%

In the case of Advantage, your odds of success will increase more when the odds of success are lower, especially in terms of proportion. Going from 3/10 to 1/2 will generally be more impactful than going from 8/10 to 9.5/10. This also applies to Disadvantage, so a good rule of thumb is "The harder the roll is to begin with, the more valuable Advantage is, and the more crippling Disadvantage will be". Hopefully this helps. :smallsmile:

obryn
2014-07-08, 08:16 AM
I'd like to know how to calculate the odds of success behind rolls with advantage or disadvantage.
Anyone know the formula or can provide some examples?

Is it true that advantage is much more "advantageous" when you need to roll a 5 to hit, versus when you need to roll a 15?
The others covered the math, so I'll bring pretty pictures and charts.

http://i.imgur.com/GHsiZir.png

The way you read this is, you figure out your target number first and find it along the bottom. Move upwards to see the probability of rolling at least that number with Advantage, Disadvantage or on a normal d20.

If you need a high number, Disadvantage dramatically reduces your chances. If you need at least a low number, Advantage dramatically increases your chances.

# %
1 100
2 99.75
3 99
4 97.75
5 96
6 93.75
7 91
8 87.75
9 84
10 79.75
11 75
12 69.75
13 64
14 57.75
15 51
16 43.75
17 36
18 27.75
19 19
20 9.75

# %
1 100
2 90.25
3 81
4 72.25
5 64
6 56.25
7 49
8 42.25
9 36
10 30.25
11 25
12 20.25
13 16
14 12.25
15 9
16 6.25
17 4
18 2.25
19 1
20 0.25

2014-07-08, 08:36 AM
I think you guys covered everything perfectly! This i suppose will be a useful reference post for the future

Particle_Man
2014-07-08, 01:48 PM
So turning that into how much of a bonus or penalty, in terms of "plusses/minuses", it would be equal to we get something like:

0.95
1.8
2.55
3.2
3.75
4.2
4.55
4.8
4.95
5
4.95
4.8
4.55
4.2
3.75
3.2
2.55
1.8
0.95

For numbers between 2 and 20 (no bonus or penalty if you need a 1 to succeed, since that is already at 100% regardless of advantage, disadvantage or neither)?

Yorrin
2014-07-08, 01:52 PM
So turning that into how much of a bonus or penalty, in terms of "plusses/minuses", it would be equal to we get something like:

0.95
1.8
2.55
3.2
3.75
4.2
4.55
4.8
4.95
5
4.95
4.8
4.55
4.2
3.75
3.2
2.55
1.8
0.95

For numbers between 2 and 20 (no bonus or penalty if you need a 1 to succeed, since that is already at 100% regardless of advantage, disadvantage or neither)?

Yeah, I've found this the easiest way to think about it. It's basically a +/- 1 at the extremes and a +/- 5 when you're starting at 50/50.

Job
2014-07-08, 04:23 PM
Obryn beat me to the punch but had it laying around for ages.