Quote from Rezilia >>

Again, where are you getting these averages? You can clearly find that the chance of getting the full 12 on DBS is lower than getting the full 12 on GS. I already showed you the percentages.

And no, DBS and 2LS do not have the same average. 2LS deals 2-16 + STR*2 damage. DBS, without RB, deals only 3-12 + STR*2 damage. If you miss one LS, you're guaranteed 1-8 + STR. If you miss once on DBS, you have a chance of 2-8+STR OR 1-4+STR.

So again, where are you getting these averages from?

And DBS still needs the RB feat to contend, so yes it does need a feat, Houligan.

We're getting these averages by, you know, calculating the averages. Expected value of a function is the mean of all possible results weighted by the likelihood of that result. For normal die rolls, we can take advantage of the symmetry of the results and just take the average of the highest and lowest values. For a d4, that's (1 + 5) / 2 = 2.5. For 3d4, it's 3 * 2.5 = 7.5. For 1d6 it's (1 + 6) / 2 = 3.5. For 2d6 it's 2 * 3.5 = 7.

For two longswords, it's 2d8, which is 2 * 4.5 = 9. With Great Weapon Fighting, the expected value of 1d4 is 3: 1/4 * 4 + 1/4 * 3 + 1/2 * 2.5. So the average damage of 3d4 is 3 * 3 = 9. So without the Revenant Blade feat, DBS matches two longswords for damage.

Hopefully now you have a better understanding of statistics and how averages work.

Quote from Rezilia >>

Again, where are you getting these averages? You can clearly find that the chance of getting the full 12 on DBS is lower than getting the full 12 on GS. I already showed you the percentages.

And no, DBS and 2LS do not have the same average. 2LS deals 2-16 + STR*2 damage. DBS, without RB, deals only 3-12 + STR*2 damage. If you miss one LS, you're guaranteed 1-8 + STR. If you miss once on DBS, you have a chance of 2-8+STR OR 1-4+STR.

So again, where are you getting these averages from?

And DBS still needs the RB feat to contend, so yes it does need a feat, Houligan.

 you can get the average value of a dice roll by adding up all the possibilities and diving by the number of possibilities 

so for a d4 (1+2+3+4)/4=2.5 for a d6 (1+2+3+4+5+6)/6=3.5 If you notice there is a pattern, half the maximum roll, plus.5 is the average roll for a 1-X dice

so the average of rolling 2 d6s is the average roll 3.5 * 2 = 7 the average of rolling 3d4 is 2.5*3=7.5 the average of rolling 1d12 is 6.5

you are thinking about things in terms of chance to roll max damage when you should be thinking in terms of average damage rolling 1d12 gives you a 1/12 chance of hitting a 12 and 3d4 gives you only a 1/64 chance so 1d12 is better right?  but the d12 also has a 1/6 chance of rolling 1 or 2, whereas it is impossible to roll below a 3 using the 3d4 so 3d4 is better right?

just look at the average damage it doesn’t lie

edit: guy above me beat me to it and explained it better haha

I think we will just have to agree to disagree here.

I think that DBS does not need the RB feat to contend (finesse, 1/2 ASI, AC bonus). DBS without the feat gives you a viable TWF option, especially in cases where you do not always want to tie up your bonus action to attack.

And there is more to damage than just the chance of getting max. If that were the case, the Greataxe would be the best weapon in the game. Yet most people pick the Greatsword, despite the "worse" odds.

That's honestly not how averages work. You can't combine all 3 d4s as if they are in one attack roll. They're two separate attack rolls. 1d12 and 2d6 with one attack roll do not have the same average as 3d4 when 2d is one roll and 1d is another roll. 

An important note that I think was missed. You can add your ability modifier to the damage roll of the bonus action attack granted by the DBS without needing the TWF Fighting Style. Because you are not using the TWF bonus action to attack but the special property of the DBS.

Quote from Rezilia >>

That's honestly not how averages work. You can't combine all 3 d4s as if they are in one attack roll. They're two separate attack rolls. 1d12 and 2d6 with one attack roll do not have the same average as 3d4 when 2d is one roll and 1d is another roll. 

 You can for averages

Making it easy, say we attack an AC of 14 with a to hit bonus of plus 2 (lvl , 10str)

we hit with rolls of 12-20 and miss with rolls of 1-11 We miss 11/20 times and hit 9/20 times (45%)

using a great axe we deal an average damage of 6.5 damage on a hit, and hit 45% of the time, so we deal an average of 6.5*0.45 = 2.925 average damage per round

for the dbs you deal 5*.45(first attack)+2.5*.45(bonus action attack) = 3.375 average damage per round

and if you remember from algebra you can rewrite 5*.45+2.5*.45 as .45*(5+2.5) which equals .45*7.5

so what ever our chance to hit is we just multiply it by our damage on a hit

the numbers get better for the dbs as str goes up since you add your str modifier to both attacks, it gets worse as you add more attacks per round since you only get one bonus action. So for fighters dbs isn’t that great, but for rogues (need to be elf with the feat so it’s a finesse weapon for sneak attack) it is two chances to apply sneak attack (so if you miss the first attack you can still get the sneak attack die on the second attack) and just an extra chance to apply your Dex modifier on a weapon attack since you usually don’t get extra attacks. 

Edit: when I get home from work I can go into the math of this in more detail. 

Quote from Rezilia >>

That's honestly not how averages work. You can't combine all 3 d4s as if they are in one attack roll. They're two separate attack rolls. 1d12 and 2d6 with one attack roll do not have the same average as 3d4 when 2d is one roll and 1d is another roll. 

That is literally how averages work. You’re right about 1d12 and 2d6 not having the same average as 3d4 (1d12 is 6.5 and 2d6 is 7, while 3d4 is, as established, 7.5), but your reasoning is wholly incorrect.

First of all, we’re calculating average damage of a hit, so we’re already assuming that we hit; number of attack rolls is irrelevant. But even if we say there’s only a 50% chance of hitting, the math doesn’t change. Averages are weighted by the likelihood of each event, so we just multiply the expected damage by 50%. Multiplication is distributive. 0.5 * 3d4 is exactly the same as 0.5 * 1d4 + 0.5 * 2d4.

Except that my whole argument is that number of attack rolls DOES matter.

The only one that has a legit argument is Houligan because he points out that the second attack roll gets bonus damage.

But once feats and styles are added in, DBS can't compete unless you're a Valenar Elf. It doesn't take long to get feats and styles, you get them in early levels. What weapons are at for level 1 doesn't matter.

Quote from Rezilia >>

Except that my whole argument is that number of attack rolls DOES matter.

And that argument is incorrect. All that matters is the dice. I’ve already demonstrated this with concrete numbers.

Let’s try this. I’ve shown you my numbers. Show me yours. What do you think the average damage for these options is, and how do you arrive at those results? Be as complete and rigorous as you can so we can better help you to understand where you’re making mistakes.

You're just talking about damage though, so the assumption is that you're hitting. If you want factor in attack rolls (hit or miss) then you need to either do algebra or provide modifiers and AC.

Let's go the algebra route and call the chance to hit H, which is based on AC, to hit bonuses, advantage etc, we have to assume H is consistent between attacks. We can use this as an expectation value, which means we can multiply our averages by it.

Let's also say you have damage bonus B, which for this situation is identical for strength or dexterity, removing finesse from the equation.

We have three scenarios as you've described; Double-Bladed Scimitar, 2 Scimitars and 2 Longswords using the Dual Wielder feat. We should also apply this feat to the two scimitars, and the Revenant Blade feat for Double-Bladed Scimitar

Double-Bladed Scimitar (Revenant Blade)

1. Damage; 2d4+B on standard attack, 1d4+B on second attack
2. Average on d4 is 2.5 [(min+max)/2 = (1+4)/2 = 5/2 = 2.5]
3. Average damage on base hit = (2*2.5+B)H = 5H + BH
4. Average damage on bonus hit = 2.5H + BH
5. Total Average Damage D(D) = 7.5H + 2BH

Notes; +1 AC, Double-Blade Scimitar counts as finesse

Two Scimitars (Dual Wielder)

1. Damage; 1d6+B on standard attack, 1d6 on second attack
2. Average damage on d6 is 3.5
3. Average damage on base hit = (3.5 + B)H = 3.5H + BH
4. Average damage on base hit = 3.5H

Total Average Damage D(S) = 7H + BH

Notes; +1 AC, draw or stow two weapons

Two Longswords (Dual Wielder)

1. Damage; 1d8+B on standard attack, 1d8 on second attack
2. Average damage on d8 is 4.5
3. Average damage on base hit = (4.5 + B)H = 4.5B +BH
4. Average damage on base hit = 4.5H

Total Average Damage D(L) = 9H + BH

Notes; +1 AC, draw or stow two weapons

Comparison

• Double-Bladed Scimitar: D(D) = 7.5H + 2BH
• Two Scimitars: D(S) = 7H + BH
• Two longswords: D(L) = 9H + BH

Now, controlling for a constant hit probability of H, we can see that the double-bladed scimitar beats out the two scimitars by 0.5H + BH;

d(D-S) = (7.5H + 2BH) - (7H +BH)
            = 0.5H + BH

Where things get interesting is the difference between the Double-Bladed Scimitar and two Longswords. Now, it should be noted that the Dual Wielder feat is mandatory to use two longswords, but no feat is needed for the double-bladed scimitar. As such, there's a reasonable chance that an ASI could be taken instead, giving the double-bladed scimitar a higher modifier B. But for the sake of easier math we've given the double-bladed scimitar a handicap by taking a feat.

d(L-D) = (9H + BH) - (7.5H + 2BH)
            = 1.5H - BH

We can thus determine a break even value;

          0 = 1.5H - BH
       BH = 1.5H
          B = 1.5

So for a damage bonus less that 1.5, the Longsword wins out, but for a damage bonus greater than 1.5 the Double-Blade Scimitar wins out. This is evident by the fact you're able to add the bonus to the second attack

Disclaimer, it is late and I haven't done algebraic probability in a long while, so if someone wants to check my math, that'd be great.

Quote from SagaTympana >>

 Let’s try this. I’ve shown you my numbers. Show me yours. What do you think the average damage for these options is, and how do you arrive at those results? Be as complete and rigorous as you can so we can better help you to understand where you’re making mistakes.

 
I'm not a calculator, but when determining the likelihood of result, you determine the probability of roll results in combination with each other. The chance of getting a 12, for instance, is much lower than getting a 7 because there are many, many combinations that can result in 7. You don't just add possible rolls together and divide by the number of results.

If you want to go calculate the results with actual probability including each combination of roll results, that's great, and I'd be inclined to believe your argument. But just adding possible results together isn't real probability.

When talking about average damage, you're talking about expectation values. The expectation value of a die with N sides is (N+1)/2. So this means that the average result on a d4 is 2.5, or on a d20 it's 10.5.

When you're totally dice, you can add the expectation values, so for example the expectation on 2d6 is 3.5 + 3.5 = 7. This also works with different dice, the average outcome on d12 + d6 is 10 (6.5 + 3.5)

You can also multiply expectation values by probabilities of success (if we doing binary outcomes). Say an attack has a 75% to hit, that means we can multiple the damage by 0.75 to determine it's expectation value. So if an attack has a 75% chance to hit and does d12 + d6 damage, we can work out the average damage to be 0.75 * (6.5 + 3.5) = 7.5

What this means is that over a suitably large sample of rolls, the average outcome will be 7.5. Obviously no single roll will be that, that's impossible, but the average will be.

Again, that doesn't work, because that's not how roll probability works.

You need to determine:

- Chance of each possible result of each roll. So for instance, on 2d6 you have a far greater chance of 7 than 12. You can't just add all the numbers together and divide.
- Number of rolls required to get the damage you want, factors into the above.
- Number of attack rolls required to get the damage you want.

The second thing boosts the first, it gives you a better minimum.
The third thing massively nerfs your chances overall.

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4, and less damage even than someone with a straight 2d6. However, 2d4+1d4 gives a better chance than 1d6+1d6.

With dual longsword, you have 1d8+1d8. With DBS (+RB), you have 2d4+1 + 1d4+1. You do get better odds on the first roll, slightly better, but the second roll still sucks.

I don't know why this is so hard for you all to understand.

Quote from Rezilia >>

Again, that doesn't work, because that's not how roll probability works.

You need to determine:

- Chance of each possible result of each roll. So for instance, on 1d12 you have a far greater chance of 7 than 12. You can't just add all the numbers together and divide.
- Number of rolls required to get the damage you want, factors into the above.
- Number of attack rolls required to get the damage you want.

The second thing boosts the first, it gives you a better minimum.
The third thing massively nerfs your chances overall.

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4, and less damage even than someone with a straight 2d6. However, 2d4+1d4 gives a better chance than 1d6+1d6.

With dual longsword, you have 1d8+1d8. With DBS (+RB), you have 2d4+1 + 1d4+1. You do get better odds on the first roll, slightly better, but the second roll still sucks.

I don't know why this is so hard for you all to understand.

Try doing this the trial and error way. go to a dice rolling app and do a roll of 3000d4. You will get somewhere around 7500. Then roll 2000d4+1000d4 which will also give you around 7500. Then roll 2000d6 you will get around 7000 then roll 1000d12 and you will get around 6500. This experiment shows a sample of 1000 attack rounds with all these different weapon types and gives the Total damage a character would have done with these weapons over the course of a 1000attack career assuming a 100% hit chance and no bonus damage on hit.

dave has already shown you the math for why a varying hit chance doesn’t matter as it averages out over time. But try to think about it like this

if you use your unarmed strike with 20 strength, you deal 6 damage per hit, assuming an armor class of 16 you will hit 50% of the time. So if you attack 1000 time you will hit about 500 times and you will deal about 500*6=3000 damage.

Now say they changed the rules for unarmed strikes and you rolled twice and did half damage each hit, so one attack round you would do 2 attacks and deal 3 damage per hit. so 1000 attack rounds later you would have rolled to hit 2000 times. Your hit chance for each roll was 50% so you would have hit around 1000 times doing 3 damage each hit, for about 3000 damage same as the first example.

for the first example each round you had a 50% chance to deal 6 damage and a 50% chance to deal 0 damage. In the second example you had a 25% chance to deal 6 damage, a 50% chance to deal 3 damage, and a 25% chance to deal 0 damage

i think what is confusing you is that on any given round where you happen to hit, you will deal less average damage (6 in scenario 1, 4 in scenario 2) but in scenario 2 you will be dealing “some” damage more often and it averages out to the same total damage over the life of the character

now in the situation where the bad guy has 5hp left, you have a 50% chance to end the combat on your round in scenario 1 and only 25% chance in scenario 2. But if you had a bad guy at 1 hp you have a 75% chance to ko him your round with scenario 2 and only a 50% chance with scenario 1. If you had 2 bad guys with 2hp each you’d have a 50% chance of killing one of them in scenario 1 and a 0% chance of killing both of them but you would have a 50% of killing exactly 1 of them in a scenario 2 and a 25% chance where you kill both of them, so you would kill at least 1 bad guy 75% of the time.

they each have some situational advantages, but they deal the same average damage in the long run. 

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4

That's not true, you can simplify the situation to see why;

An attack that does 12 damage and hits 50% of the time will do, on average, 6 damage.

An attack that does 8 damage and hits 50% of the time does 4 damage average, and an attack that does 4 damage 50% of the time does an average of 2. The total for this is 6.

They're the same because P(2N) + P(N) = P(2N + N) = P(3N).

You seem to have a misunderstand of how averages, expectation values and probability works.

Lets just do the math out for Longsword vs DBS. 1d8 vs 2d4 and 1d4. And use 400 attacks as the baseline (since it will divide evenly)

Longsword (Main): 50 attacks to 1 damage, 50 do 2 damage, 50 do 3 damage, etc. Total damage dealt is:
Total Damage Dealt = (50*1)+(50*2)+(50*3)...+(50*8) = 1800
Divide by the number of attacks made (400) = 4.5 damage dealt on average, assuming all attacks hit
If you wanted to factor in hit chance (assume X chance to hit): (X * 50 * 1) + (X * 50 * 2) + ... (X * 50 * 8), you can factor the X out onto its own:
X*[(50*1)+(50*2)+(50*3)...+(50*8)], so you can just take the expected damage if all attacks hit, and then apply the hit chance

And then the same math for the bonus attack with the longsword

DBS Main: 25 attacks do 2 damage, 50 attacks do 3 damage, 75 attacks do 4 damage, 100 attacks do 5 damage, 75 do 6, 50 do 7, 25 do 8
Total Damage Dealt = (25*2)+(50*3)+(75*4)+(100*5)+(75*6)+(50*7)+(25*8) = 2000
5 damage on average, reduced by hit chance

DBS Bonus: 100 do 1 damage, 100 do 2 damage, 100 do 3 damage, 100 do 4 damage.
Total Damage Dealt = 1000, 2.5 damage on average, reduced by hit chance.

Now, since both weapons are making the same number of attacks (one main hand, one bonus), something very helpful happens when comparing the equations:
If you leave the hit chance as a variable (X) and then also include ability mod (A)
Longsword: ([4.5 average main damage + A] * X hit chance) + (4.5 average bonus damage * X hit chance)
DBS:              ([5 average main damage + A]* X hit chance)   + ([2.5 average bonus damage + A] * X hit chance)
You can again factor out the hit chance
Longsword: X * (4.5 + A + 4.5)
DBS:             X * (5 + A + 2.5 + A)
And since both of these equations are scaled by the same hit chance (X), we can just drop it when comparing the results.

Hence why when comparing two damages where the player is making the same number of attacks in each case, hit chance can be ignored.

And so at what point is (9+A) (the longsword) better or worse than (7.5+2A) (the DBS)
A = 1: 10 vs 9.5
A = 2: 11 vs 11.5
A = 3: 12 vs 13.5

So without the two-weapon fighting STYLE, the DBS out damages the longsword for all ability modifiers greater than 1, regardless of your hit chance, since it would be the same hit chance whether you are making your two attacks with longswords or the DBS.

The DBS allows the wielder efficient TWF without needing a feat or fighting style.

Quote from Davedamon >>

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4

That's not true, you can simplify the situation to see why;

An attack that does 12 damage and hits 50% of the time will do, on average, 6 damage.

An attack that does 8 damage and hits 50% of the time does 4 damage average, and an attack that does 4 damage 50% of the time does an average of 2. The total for this is 6.

 

 
Actually, no, because on the first you have a 50% chance of doing 6 damage on average. On the second you have a 25% chance of doing 6 on average, 25% of 2, and 25% of 4.

Quote from Rezilia >>

Quote from SagaTympana >>

 Let’s try this. I’ve shown you my numbers. Show me yours. What do you think the average damage for these options is, and how do you arrive at those results? Be as complete and rigorous as you can so we can better help you to understand where you’re making mistakes.

 
I'm not a calculator, but when determining the likelihood of result, you determine the probability of roll results in combination with each other. The chance of getting a 12, for instance, is much lower than getting a 7 because there are many, many combinations that can result in 7. You don't just add possible rolls together and divide by the number of results.

You absolutely do just add the possible rolls together and divide by the number of results. That is the very definition of average. On 2d6, a lot MORE possible rolls result in 7 than in 12, so there are more 7s in that sum than there are 12s. If you won't even present an actual average for anything, I don't know what you expect, but let's keep trying.

Quote from Rezilia >>

 
If you want to go calculate the results with actual probability including each combination of roll results, that's great, and I'd be inclined to believe your argument. But just adding possible results together isn't real probability.

I already calculated the results with actual probability. This isn't an argument. It's a teachable moment, and it's up to you whether or not you want to learn.

Quote from Rezilia >>

Again, that doesn't work, because that's not how roll probability works.

You need to determine:

- Chance of each possible result of each roll. So for instance, on 2d6 you have a far greater chance of 7 than 12. You can't just add all the numbers together and divide.
- Number of rolls required to get the damage you want, factors into the above.
- Number of attack rolls required to get the damage you want.

Each die is completely independent of the other. What you get on one die has no bearing on any other. 2d6 is just 2 * 1d6.

Quote from Rezilia >>

The second thing boosts the first, it gives you a better minimum.
The third thing massively nerfs your chances overall.

You're half right. The higher minimum is really important to bumping up the average. But the number of attack rolls is 100% irrelevant if the to-hit chance is constant; again, multiplication is distributive. 0.5 * (1 + 1) is the same thing as 0.5 * 1 + 0.5 * 1. If you can't wrap your head around that, I don't know what to tell you.

Quote from Rezilia >>

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4, and less damage even than someone with a straight 2d6. However, 2d4+1d4 gives a better chance than 1d6+1d6.

This is completely false. Let's use d2s to demonstrate. I really hope we can agree that the average of 1d2 is 1.5. You either get a 1 or 2, (1 + 2) / 2 = 3 / 2 = 1.5. So if we do that twice, we sum them and get 3. Here's what happens if we roll 2d2:

1 1 = 2

1 2 = 3

2 1 = 3

2 2 = 4

(2 + 3 + 3 + 4) / 4 = 12 / 4 = 3

Miraculously, the result is the same. This is because, as I said above, the two dice are independent of each other. They do not affect one another. Average(1d6 + 1d6) is the same thing as Average(1d6) + Average(1d6).

Quote from Rezilia >>

Quote from Davedamon >>

Someone who rolls 2d4+1d4 will, on average, deal less damage than someone with a straight 3d4

That's not true, you can simplify the situation to see why;

An attack that does 12 damage and hits 50% of the time will do, on average, 6 damage.

An attack that does 8 damage and hits 50% of the time does 4 damage average, and an attack that does 4 damage 50% of the time does an average of 2. The total for this is 6.

 

 
Actually, no, because on the first you have a 50% chance of doing 6 damage on average. On the second you have a 25% chance of doing 6 on average, 25% of 2, and 25% of 4.

Dave is correct. Why on earth are you halving the hit chance from 50 to 25%?

I have no Feats in my game. I suspect that would make the DBS better.

I personally would take two chances to crit every round over a single chance to crit.