D&D 5E What if weapon damage had advantage: Common damage dice

[Yunru]

I guess it already can with Savage Attacker (or is that keep the second even if lower?).
Anyway, most of these were easy 1dX with advantage deals an average damage of (2(i^2)-i)/(X^2), where i is 1 through X.
2d6 is far, far trickier (to do in excel, real easy in java or the likes), due to the varying probabilities of a given number of damage occurring.

Anyways, here's the die size and the average damage with advantage. A number surrounded by brackets means that number repeats infinitely.
1d4: 3.125
1d6: 4.47(2)
1d8: 5.8125
1d10: 7.15
1d12: 8.486(1)
2d6: 8.3719135802

That last one may be automatically rounded by excel but... It's probably precise enough.

 

[thethain]

I might be tempted to let 2d6 with advantage roll 4d6 and keep best 2, as its whole stick is being ever so slightly better than 1d12 and if you used this system it would not be in those situations.

 

[Yunru]

You still get more consistent damage with 2d6, just not as high an average.

EDIT: Here's an any dice link, you're looking for the means and deviations on the summary tab: http://anydice.com/program/b9c4

 

[thethain]

Actually, looking at it again, why would 2d6 not be 8.94. It would be the addition of two 1d6 adv (mean) the rolls wouldn't affect each other.

Here is an online tool that lets you share it a little easier.
http://anydice.com/program/b9c2

 

[Yunru]

Because it's not the addition of two 1d6 with advantage. It's the higher of 2(2d6).

Any dice can't map that (I think).

2 has a 1 in 36 chance of being rolled.
3 has a 2 in 36 chance of being rolled.
etc. up to:
7 has a 6 in 36 chance of being rolled.
and then back down to:
12 has a 1 in 36 chance of being rolled.

That's why it's not just 2(highest of 2d6), because they have uniform chances of each number being rolled.
Highest 2 of 4d6 is also different again.

 

[TwoSix]

Actually, looking at it again, why would 2d6 not be 8.94. It would be the addition of two 1d6 adv (mean) the rolls wouldn't affect each other.

Here is an online tool that lets you share it a little easier.
http://anydice.com/program/b9c2

I think the OP was doing 2d6 advantage as 3d6 keep highest 2.

 

[Yunru]

I think the OP was doing 2d6 advantage as 3d6 keep highest 2.

Nope.

 

[TwoSix]

Because it's not the addition of two 1d6 with advantage. It's the higher of 2(2d6).

Any dice can't map that (I think).

2 has a 1 in 36 chance of being rolled.
3 has a 2 in 36 chance of being rolled.
etc. up to:
7 has a 6 in 36 chance of being rolled.
and then back down to:
12 has a 1 in 36 chance of being rolled.

That's why it's not just 2(highest of 2d6), because they have uniform chances of each number being rolled.
Highest 2 of 4d6 is also different again.

You can, actually, in anydice, the command is "output [highest of 2d6 and 2d6]". Your calculation of 8.372 was correct.

 

[Yunru]

Yeah I did it the hard way:
https://1drv.ms/x/s!At-zPv0cZTn6hio5oe3k1vRzMfsN

Would of been so much easier as a program:
X[11]=[2,3,4,5,6,7,8,9,10,11,12]
Y[11]=[1,2,3,4,5,6,5,4,3,2,1]

for (i=0, i>11, i=i+1)
for (j=0, j>11, j=j+1)
if (X>X[j]) {Z=X}
else {Z=X[j]}
Total = Total + (Y*Y[j]*Z)
Count = Count + (Y*Y[j])
end
end
Total = Total/Count