Average increase/decrease from rolling twice?

[deuxhero]

Beyond the crit range in/20 chance to turn an attack into a critical and 19/20 chance to avoid a critical miss (or a 20-crit range/20 chance to negate a critical hit/extra 1/20 of a critical miss for roll twice take worse result debuffs) on to hit rolls, has someone calculated who much the average increase (or decrease) for abilities that let you roll twice on a d20 roll is?

I noticed the PF spell Honeyed Tongue which gives this for Diplomacy rolls. Bards get a LOT of diplomacy increasing spells, and I have no idea how it stacks up to them. Skill rolls don't fail on a 1/succeed on 20 so the math should be straight forward.

[GilesTheCleric]

In the 5e playtest/discussion surrounding it, I believe it was equated to a +5. I don't have a source for that, though.

[chaos_redefined]

It averages to a +3ish. But, that is somewhat deceptive... it depends on what you need to pass. If there is a 5% chance of needing any given number to pass, it's 3ish. If you need to roll a natural 10 or higher on the dice to succeed (which 5e tries to make the normal), then it is equivalent to a +5. If, on the other hand, you only fail if you roll a 1, it's a tad less than a +1. Similarly, if you can only succeed by rolling a natural 20, it's a tad less than a +1.

So... needs more information?

[deuxhero]

Straight +X bpmises are "wasted" if you go over too.

What you are saying applies if you know what you need to hit and rerolls are finite (which is not an unusual situtation), but it's not exactly what I'm looking for.

[HunterOfJello]

Thinking about it as a static bonus is completely wrong and shouldn't be attempted.

Think about it as, if i get to roll 2d20 and take the better one I:
reduce rolling a 1 from 5% to 0.2%
AND
increase the chance of rolling a 20 from 5% to 9.8%
at the same time.

This is a massive difference and change in your odds in general.

[Temotei]

Snowbluff linked this (http://onlinedungeonmaster.com/2012/05/24/advantage-and-disadvantage-in-dd-next-the-math/) in a similar discussion a while back.

Long story short: Depends on the die size, how many times you reroll, and your goal number, with the above posts explaining why pretty well.

[grarrrg]

Since no one has given the exact answer yet...

1d20 averages to 10.5
2d20 'pick best' averages to 13.825, for an increase of 3.325
So if you can get a +3 or a 'roll twice', the roll twice is probably* better.
If you can get a +4 or a 'roll twice', the +4 is probably* better.

*Depends on target number(s), things get iffy near the high/lows due to roll twice ranging from 1-to-20 and "+3" ranging from 4-to-23.

And just for funsies
3d20 'pick best' averages to 15.4875 which is 4.9875 better than 1d20 and 1.6625 better than 2d20
4d20 'pick best' averaged to 16.4833, which is ~1 better than 3d20, and ~6 better than 1d20

[deuxhero]

Thanks grarrrg!

[Chronos]

You simply can't say which is better without knowing the target number. This does not just apply near the ends where "things get iffy", but everywhere. If you're starting off with a 50% chance of success, then rolling twice and taking the better is exactly equivalent to a +5, since it turns a 50% chance into a 75% chance. If you're starting with a DC where you literally can't fail or literally can't succeed, then rolling twice is equivalent to +0. In practice, the effect is somewhere in between these two extremes, but where, you can't say, without knowing what sorts of challenges you typically face.

[deuxhero]

Straight bonuses are also worthless if the DC is too high.

There's also stuff like opposed rolls where there isn't a fixed DC

[Chronos]

Straight bonuses eventually become worthless for DCs too high, but not until after the point where rolling twice becomes worthless. If you're one short of the DC, then rerolling is useless, but a +1 bonus at least gives you a 5% chance.

[grarrrg]

You simply can't say which is better without knowing the target number.

That's kind of the whole point.
We DON'T KNOW the target number. And if it's something like AC, then it can swing around wildly from one roll to the next.

Roll Twice winds up being slightly better than a +3 overall, yes sometimes it will be better than a straight bonus, other times it will be worse, but without knowing what we're trying to hit all we can say is that Roll twice is worth more than +3, but less than +4.

[chaos_redefined]

The OP stated he is looking at Diplomacy, which has a DC based on charisma in PF (which I believe he is using). While charisma isn't a constant value, there is small variations in a lot of cases. Additionally, it has an upper limit, that I assume he wants to aim for. So, let's say you are a level 10 character. Most creatures will have a fairly low charisma (+0 bonus), as will most NPCs. The exception is things like sorcerers, oracles and bards. It seems reasonable to give them a +6 bonus.

Against a normal target, assuming they are unfriendly to begin with, the DC is gonna be 20. As a level 10 bard, he will most likely have a +13 bonus from ranks/trained, and a +4 bonus, minimum, from charisma. So, that's a +17 bonus with nothing else. So, the target number is 3. Rolling twice is better than a +2 bonus, but worse than a +3 bonus, according to the link above.

Against a high charisma target, the same situation will have a DC of 26. The target number becomes 9, and rolling twice is somewhere between a +4 and +5.

So, a +5 bonus or higher is strictly better than roll twice. A +3 or +4 is better against low charisma creatures, and worse against high charisma creatures. A +2 bonus is inferior.

Obviously, you'll need to recalculate this with whatever values you have, I'm just providing an example.

[Spore]

Why make it this complicated? grarrrg has the math for AVERAGES, your odds increase significantly more if you need a very high roll, and they increase significantly less if you need a low roll.


Since no one has given the exact answer yet...

1d20 averages to 10.5
2d20 'pick best' averages to 13.825, for an increase of 3.325

[deuxhero]

While there was a problem that prompted it, I was more curious about the general math

The specific example that prompted it was level 2 bard spells which is answered by it, on average, being bit better than +3 meaning Honeyed Tounge is better than Seducer’s Eyes till level 12 (which is more versatile but requires the target find the user attractive)

[Chronos]

Quoth grarrrg:

...but without knowing what we're trying to hit all we can say is that Roll twice is worth more than +3, but less than +4.
No, without knowing, we can't say even that. If we don't know, then all we can say is that it's worth no more than +5, and no less than +0.

[grarrrg]

No, without knowing, we can't say even that. If we don't know, then all we can say is that it's worth no more than +5, and no less than +0.

No.
Using your logic, it actually provides no more than a +19, and no less than a +0 bonus.

The average bonus, which is just that an AVERAGE bonus, not a "typical" bonus, not a "likely" bonus, but an AVERAGE, is +3.325.
+3.325 is the AVERAGE of your "supposed" +0 to +5

[Chronos]

It's never worth more than +5 (the value it takes when you're starting with a 50% chance). And we don't know enough to determine the average. If most of your rolls are ones where you'd ordinarily have a 50% chance to succeed, then the average is close to +5. If most of your rolls are ones where you need a natural 20, then the average is close to +1.

[grarrrg]

It's never worth more than +5 (the value it takes when you're starting with a 50% chance). And we don't know enough to determine the average. If most of your rolls are ones where you'd ordinarily have a 50% chance to succeed, then the average is close to +5. If most of your rolls are ones where you need a natural 20, then the average is close to +1.

You really don't get what "average" means, do you?

[Chronos]

Yes, in fact, I do. Do you?

Quick example: What's the average of 5, 5, 5, 1? How about 5, 1, 1, 1? Here's a hint: Neither one is 3.

If you take all of the situations in your game where you roll, find the benefit of rolling twice for each of those situations, add up all of the benefits, and then divide that by the number of rolls you made, you'll get the average benefit. What that average is will depend on what rolls you make.

[Coventry]

Sadly, now I have to contradict myself from over in the Roleplaying Forum. But I have had a few days to sleep on my earlier response, and have decided my answer was incomplete/wrong.


There are three kinds of averages - Mean, Median and Mode.

The mean of "2d20, take the best" is 13.825, as stated above.

However, the numbers are skewed towards the high end. There are 39 combinations out of 400 that result in a natural 20 ... almost but not quite 10%. Thus, the Mode (the most common result) is 20. The least common result only occurs one time in 400.

The Median (where half of the possible rolls fall below the value, and half of the possible rolls fall above) is actually 15. There are {14x14=} 196 combinations out of the 400 possible die rolls that are 14 or less. The other 204 possible combinations are 15 or higher.

Here's a chart:



Best 2 of Die
Avg - Mean
Avg - Median
Avg - Mode
Mode % of total


d4
3.1250
3
4
43.75%


d6
4.4722
5
6
30.56%


d8
5.8125
6
8
23.44%


d10
7.1500
8
10
19.00%


d12
8.4861
9
12
15.97%


d20
13.8250
15
20
9.75%

[grarrrg]

Yes, in fact, I do. Do you?

Quick example: What's the average of 5, 5, 5, 1? How about 5, 1, 1, 1? Here's a hint: Neither one is 3.

:smallannoyed:
Really? Can you at least temporarily act like an adult?


If you take all of the situations in your game where you roll, find the benefit of rolling twice for each of those situations, add up all of the benefits, and then divide that by the number of rolls you made, you'll get the average benefit.

Since you're being dense, have a table.

Left column is "normal roll", which I hope we can agree is going to average to 10.5.
Top row is the second die
The grid itself is how much of a bonus the extra die gets you over the base die.




1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20


1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19


2


1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18


3



1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17


4




1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16


5





1
2
3
4
5
6
7
8
9
10
11
12
13
14
15


6






1
2
3
4
5
6
7
8
9
10
11
12
13
14


7







1
2
3
4
5
6
7
8
9
10
11
12
13


8








1
2
3
4
5
6
7
8
9
10
11
12


9









1
2
3
4
5
6
7
8
9
10
11


10










1
2
3
4
5
6
7
8
9
10


11











1
2
3
4
5
6
7
8
9


12












1
2
3
4
5
6
7
8


13













1
2
3
4
5
6
7


14














1
2
3
4
5
6


15















1
2
3
4
5


16
















1
2
3
4


17

















1
2
3


18


















1
2


19



















1


20























20 * 20 = 400
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+1+ 2+3.........+1 = 1330
1330 / 400 = 3.325

The average of 2d20 take best is 10.5 + 3.325 + 13.825

How about that, doing exactly what you said, considering all potential rolls, and adding up the bonuses, and dividing by the number of rolls, I got the exact same answer that I've been saying since the beginning.



The mean of "2d20, take the best" is 13.825, as stated above.

Thank you.

[chaos_redefined]

OK... A couple of points...

First off, the effective average benefit is not +3.325 in most games. Emphasis on effective. To get an indication of the effective average benefit, ask your DM to keep a record of the target number of each time you rolled in a game. At the end of the game, figure out the probability you would succeed rolling once, and the probability that you would succeed rolling twice. Then, find the difference, and divide by 0.05. You now have the effective average benefit. The average benefit of +3.325 is the effective average benefit if the chance of needing any number on the dice is 5%.

Second... the average is actually misleading. Let's say that I have 3 judges. They were all going to give me a result of 5 out of 10, but I bribed one of them into giving me 10 out of 10. Because of this, my average is 6.67. In comparison, the median is still 5. As an even more extreme scenario, what is the average of 1, 2, 3, 10000. The outlier can severely impact the average. More reliable is the median: This value indicates the 50% cutoff. Where possible, it is used in computing and engineering, however, it's difficult because the median is difficult to optimize and difficult to calculate in multiple dimensions.

Put simply, when rolling a d20 twice and taking the better(1d20d1), you have a chance slightly higher than 50% of getting a 15 or higher. So, it isn't unreasonable to treat 1d20d1 as an "average" of 15.

[squiggit]

1d20d1 is a weird looking notation and I'm not quite sure what it's trying to convey.

What's wrong with 2d20k1?

[Chronos]

Quoth grarrrg:

How about that, doing exactly what you said, considering all potential rolls, and adding up the bonuses, and dividing by the number of rolls, I got the exact same answer that I've been saying since the beginning.
I didn't say "all potential rolls", because nobody cares about all potential rolls. I said "all rolls that you actually make at the gaming table".

It might help to realize that the effect of a d20 roll is not "1" or "10" or "20". It's either "success" or "failure". Rolling twice and keeping the better one won't increase your results by 1, or by 3.25, or by 5. It'll either increase your result from "failure" to "success", or it won't increase your result at all. Similarly, a numerical bonus will also either increase your result from "failure" to "success", or it won't increase your result at all. So if we want to know what bonus re-rolling is equivalent to, then we need to find the amount of bonus that increases your chance of success by the same amount.

[grarrrg]

Second... the average is actually misleading....More reliable is the median

Any of mean/median/mode can be useful or misleading, depending on the data.
For the given results of "roll 2 take best" the Mode is quite misleading (20 out of a possible 20? YAY!), but the both the Mean and Median are fairly representative.


I didn't say "all potential rolls", because nobody cares about all potential rolls.
Well, for starters there's the OP:

While there was a problem that prompted it, I was more curious about the general math

I said "all rolls that you actually make at the gaming table".

As for rolls you make at the gaming table, that greatly varies, now doesn't it?
If it's something like AC it could be anywhere, so we have to take ALL possibilities into consideration (of course, if it IS AC, there is the added issue of Nat 20's, Nat 1's, and Crit-threat range to worry about...)
If it's a Skill Check those are usually multiples of 5, but your other modifiers are still fairly fluid, so all possibilities remain in play (unless your modifiers get you close enough to the maximum useful range of that particular Skill, but we have to assume that isn't the case).
And then there are opposed checks, which can pretty much be anything, as you have no way of knowing what your opponent has, so all possibilities are still in play.



.It's either "success" or "failure"....So if we want to know what bonus re-rolling is equivalent to, then we need to find the amount of bonus that increases your chance of success by the same amount.

If you have a SPECIFIC roll in mind, then we can do the math for that.
The whole point of this was that we DON'T have a specific roll or situation.
Since we do not have a specific roll in mind, then the equivalent "amount of bonus" winds up being between +3 and +4 (+3.325 to be exact, but there are no fractions in game). If you have a choice between choosing a +3 (or +4) bonus or 'rolling twice', then your options are fairly close in the long run.


And you want to know what I find entertaining?
Stealing borrowing the table from what was linked (http://onlinedungeonmaster.com/2012/05/24/advantage-and-disadvantage-in-dd-next-the-math/) earlier (which is where I suspect you got your "it's between +0 and +5" that you keep blabbing about from).


Target
1d20
Roll 2
Difference
+ Equivalent


1
100%
100%
0
+0


2
95
99.75
4.75
+.95


3
90
99
9
+1.8


4
85
97.75
12.75
+2.55


5
80
96
16
+3.2


6
75
93.75
18.75
+3.75


7
70
91
21
+4.2


8
65
87.75
22.75
+4.55


9
60
84
24
+4.8


10
55
79.75
24.75
+4.95


11
50
75
25
+5


12
45
69.75
24.75
+4.95


13
40
64
24
+4.8


14
35
57.75
22.75
+4.55


15
30
51
21
+4.2


16
25
43.75
18.75
+3.75


17
20
36
16
+3.2


18
15
27.75
12.75
+2.55


19
10
19
9
+1.8


20
5
9.75
4.75
+.95



Now take the average of all those numbers on the right side.
I'll give you 2 guesses what the average of your "+0 to +5 bonus" is.
Here's a hint: It's not 3.

[Anlashok]

Surprised that this thread became so heated. Figured the answer would be as simple as "A d20 rolls X on average, 2d20b1 rolls Y on average" and... that's it.

[chaos_redefined]

Any of mean/median/mode can be useful or misleading, depending on the data.
For the given results of "roll 2 take best" the Mode is quite misleading (20 out of a possible 20? YAY!), but the both the Mean and Median are fairly representative.

The mean is more strongly representative of outliers than the median is. This is a Bad Thing (TM) in pretty much all cases. For example, it doesn't matter if you roll a 20 or a 19 in pretty much all cases. The difference is negligible in most games, and it comes up less than .1% of situations in my experience. Similarly, the difference between a 1 and a 2 is negligible. If, however, I was to replace the 1 and 20 on the d20 with a 2 and a 19, and do the same maths we said earlier, we would end up with a value higher than 3.325. This is a Bad Thing (TM), as it means that two dice that give the same results in practice give different values in theory.

More representative is: If you roll 2d20 and keep the best, there is a 50% chance of getting a 15 or higher. There is a 25% chance of getting an 18 or higher, and there is a 75% chance of getting an 11 or higher.

[bekeleven]

The mean is more strongly representative of outliers than the median is. This is a Bad Thing (TM) in pretty much all cases. For example, it doesn't matter if you roll a 20 or a 19 in pretty much all cases. The difference is negligible in most games, and it comes up less than .1% of situations in my experience. Similarly, the difference between a 1 and a 2 is negligible. If, however, I was to replace the 1 and 20 on the d20 with a 2 and a 19, and do the same maths we said earlier, we would end up with a value higher than 3.325. This is a Bad Thing (TM), as it means that two dice that give the same results in practice give different values in theory.

More representative is: If you roll 2d20 and keep the best, there is a 50% chance of getting a 15 or higher. There is a 25% chance of getting an 18 or higher, and there is a 75% chance of getting an 11 or higher.

In other words:
The five number summary of 1D20 is 1, 5, 10, 15, 20.
The five number summary of 2D20b1 is 1, 11, 15, 18, 20.

You could show these on a box plot (http://en.wikipedia.org/wiki/Box_plot) to make the point that it's not equivalent to a simple numeric increase.

If I have +0 to a roll with a DC of 21, I'd rather take a +1 than the ability to roll 50D20 and keep the best one.

If I have a +0 on a roll with a DC of 10, 2D20B1 is exactly equivalent to a +5.

So, as others stated, the answer is "Somewhere between 5 and 0."

[Spore]

Surprised that this thread became so heated. Figured the answer would be as simple as "A d20 rolls X on average, 2d20b1 rolls Y on average" and... that's it.

Human perception of probability is a trained skill as is the math behind it. It takes a smart guy to do the math behind it. But it takes time to understand how different variants of generating random numbers influence the result.

[chaos_redefined]

In other words:
The five number summary of 1D20 is 1, 5, 10, 15, 20.
The five number summary of 2D20b1 is 1, 11, 15, 18, 20.

You could show these on a box plot (http://en.wikipedia.org/wiki/Box_plot) to make the point that it's not equivalent to a simple numeric increase.

If I have +0 to a roll with a DC of 21, I'd rather take a +1 than the ability to roll 50D20 and keep the best one.

If I have a +0 on a roll with a DC of 10, 2D20B1 is exactly equivalent to a +5.

So, as others stated, the answer is "Somewhere between 5 and 0."

Box plots work, but I closed matlab before posting this and didn't want to open it up again. I've done my work for the day. I do like the 5-number summary, but I don't think many people would know about it. (I swear I learnt about them in uni, and I generally don't assume most people have done first year stats). The other thing that this shows, however, is the spread. The Inter-Quartile Range (IQR) is smaller when using 2d20b1 than 1d20, which means your result is more predictable. While you might not get 15 every time, you are more likely to get near 15 with 2d20b1 than you are to get near 10 when rolling 1d20.