4
\$\begingroup\$

I'm trying to figure out how to calculate the above and keep getting stuck on the greatsword (mace too I suppose) average damage specifically.

I'm trying to estimate how much each weapon would benefit if a high level homebrewed feature allowed you to replace the roll of a weapon's die with your proficiency bonus.

For example, normally if your attack hits with a greatsword it does STR + 2d6 damage. With this feature you could replace the value of the die roll with your proficiency bonus. Giving a minimum damage of STR + 6 at level 17.

In particular, how would this interact with the great weapon fighting style for greatswords with a proficiency bonus of 6?

With the math I've done, I have the following numbers.

Replacing weapon damage die roll with proficiency bonus the new average of the weapon die becomes:

  • D4 = 6
  • D6 = 6
  • D8 = 6.375
  • D10 = 7
  • D12 = 7.75
  • 2d6 = 7.9090~

If Great Weapon Fighting were also used with the proposed feature, the new average die values increase to:

  • D8 = 6.46875
  • D10 = 7.2
  • D12 = 8.0416
  • 2D6 = ?????

I know I'm missing some part of the logic to do the calculation for the 2d6 weapons, but I can't figure it out.

Improve this question
\$\endgroup\$
12
  • 2
    \$\begingroup\$ Oh, are you saying "you have a choice of the roll or your proficiency bonus after seeing the roll?" \$\endgroup\$
    – Thomas Markov
    May 26 at 16:55
  • 1
    \$\begingroup\$ Welcome to RPG.SE! Have a look at the tour and at the help center for some guidance about posting in this site! I still do not understand your calculation: if you are replacing the damage roll with the PB, why are you getting different numbers in the first list? \$\endgroup\$
    – Eddymage
    May 26 at 16:57
  • 1
    \$\begingroup\$ @Eddymage I think what we're doing here is "roll damage, choose either the dice result or your proficiency bonus, then add your STR mod". Then, the second list is trying to figure out what the expected value is after the Great Weapon Fighting style is applied to the roll. \$\endgroup\$
    – Thomas Markov
    May 26 at 17:01
  • 1
    \$\begingroup\$ @ThomasMarkov Ok, that's clearer now: I think that the question should rephrased more clearly.. \$\endgroup\$
    – Eddymage
    May 26 at 17:05
  • 2
    \$\begingroup\$ Additionally, while I appreciate the check mark, it is a common courtesy on this site to wait about 24 hours before accepting an answer. That way, people who haven't yet seen the question aren't discouraged from leaving an answer that may be even better than mine. \$\endgroup\$
    – David Coffron
    May 26 at 19:41

3 Answers 3

Reset to default
8
\$\begingroup\$

Without Great Weapon Fighting

This is a simple problem, so you could very quickly solve it discretely in a tool of you choice by finding the distribution of results and replacing any that is less than 6 with 6.

However, for more complex problems, I like to use anydice to get a baseline and then create a spreadsheet that shows the rest. I know you could do it all in anydice1, or do it all in a spreadsheet2, but a split job works best for me. To start with, let's generate the 2d6 results table with anydice:

enter image description here

Now let's put that data in spreadsheet with the following function: Max(#,6). Then we simply multiply these results by the % from anydice and sum these weighted results and we get this spreadsheet showing a result of 7.556:

enter image description here

With Great Weapon Fighting

With this more complex version, we can see the benefit I gain from splitting the work between anydice and spreadsheet math.

Let's do the same thing for your Great Weapon Fighting case. Here is our new anydice set (taken from this Q&A):

enter image description here

Doing the same task in our spreadsheet, we get the result 8.469:

enter image description here

1 This anydice function shows exactly that with the added flexibility of customizing the proficiency bonus, from the user minnmass if you have more anydice expertise than myself

2 If you wanted to do it all in a spreadsheet, it becomes quite complicated when adding Great Weapon Fighting. I did it anyway for demonstration purposes. Here are the formulas used; I only included the top-left most formula in each section for readability.

Improve this answer
\$\endgroup\$
4
  • \$\begingroup\$ Thank you very much, that's what I was looking for. I also really appreciate those tool recommendations. I actually did the calculations manually (ugh) for all 36 die combinations while I was away and got very close to that result (8.4688~). Though I'm sure the above is more accurate. Thanks again! \$\endgroup\$
    – Dominic Murphy
    May 26 at 19:43
  • 1
    \$\begingroup\$ And, here's an AnyDice program that will calculate the damage with and without GWF; just change the bits in the middle to change the die size, die count, proficiency bonus, and other damage modifier (eg., STR, +1 from a weapon, etc.): anydice.com/program/2fa82 - ed, rpg.stackexchange.com/a/104788/20971 helped a lot \$\endgroup\$
    – minnmass
    May 26 at 19:53
  • 1
    \$\begingroup\$ @minnmass I'm happy to include this in my answer since you left it as a comment on mine, but I think it has the making of a great answer if you wanted to spend the time explaining how to create and use such a function. If you do, I'll just link to your answer instead of to the footnote. \$\endgroup\$
    – David Coffron
    May 26 at 20:00
  • 2
    \$\begingroup\$ @DavidCoffron: the answer I linked had most of the hard bits already done, especially for GWF; I just duct-taped them together with the "min" function. \$\endgroup\$
    – minnmass
    May 26 at 20:11
7
\$\begingroup\$

Your result is the average of all the individual possible results, weighted by their probability. To get it, multiply the result by the probability and sum. With a single die, that's easy. Pick a d4, rolled normally:

1: 25% 
2: 25% 
3: 25% 
4: 25%

(1 x 0.25) + (2 x 0.25) + (3 x 0.25) + (4 x 0.25) = (.25) + (.5) + (.75) + (1) = 2.5

It's exactly the same for multiple dice, but the probabilities are a little less obvious. For 2d6:

2: 1/36
3: 2/36
4: 3/36
5: 4/36
6: 5/36
7: 6/36
8: 5/36
9: 4/36
10: 3/36
11: 2/36
12: 1/36

When you sum (2 x 1/36) + (3 x 2/36) + ... + (12 x 1/36), you get 7.

You want to replace everything 6 or below with 6. That's easy enough:

6: 1/36
6: 2/36
6: 3/36
6: 4/36
6: 5/36
7: 6/36
8: 5/36
9: 4/36
10: 3/36
11: 2/36
12: 1/36

And summing that gives you (6 x 15/36) + (7 x 6/36) + ... + (12 x 1/36) = 272/36 = 7.5555...

7 and five ninths.

Improve this answer
\$\endgroup\$
1
  • \$\begingroup\$ +1 for a great method I used until I found anydice. However, this answer could be improved by offering an explanation of how to solve the problem with Great Weapon Master. I personally would use your approach if an anydice answer wasn't present for that case (as my anydice expertise is lacking), so I think explaining how to generate the discrete probabilities would go along way. \$\endgroup\$
    – David Coffron
    May 26 at 19:43
4
\$\begingroup\$

Warning!

This answer is fuled by dark excel magic which is provided to you as a spreadsheet to look at here. I will not grant access to that sheet, but if you want to work with it, you can grab a copy for your own work.

Without Rerolls

To do this, you need to make a spreadsheet first: One of the dice goes 1 to 6 to the right, the other goes 1 to 6 down. This will fill out 36 squares with their sums. Or rather, an Array. For simplicity, we call the field in the top left [1,1] and the top right will be [1,6], bottom left is [6,1] and bottom right is [6,6].

Each of the fields has a number of 2 to 12, and each field stands for 1/36th chance. You can sum up all the fields (252) and divide by the number of fields (36) and get the average: perfectly 7!

enter image description here

From here, we can adjust to the other methods, and using a spreadsheet program, things get easy (if you know your black excel magic)

Replacing any outcome of <6 with 6

If the sum of 2d6 is replaced for any outcome below 6 with a 6, then all outcomes in the triangle [1,1] to [1,5] to [5,1] (or where the sum of the array indices is lower than 6) are replaced with 6. This leads to the top left corner being all 6s and an average damage of 7,55. The distribution of such a modified roll is this

$$6 = \frac {15} {36}\ ;\ 7 = \frac 6 {36}\ ;\ 8=\frac 5 {36}\ ;\ 9= \frac 4 {36}\ ;\ 10= \frac 3 {36}\ ;\ 11= \frac 2 {36}\ ;\ 12= \frac 1 {36}$$

enter image description here

If one of the 2d6 gets replaced

If only the lower of the two dice will be replaced by 6, the effect is much more powerful, the average damage gets boosted to 10.47. The distribution is now $$7 = \frac 1 {36}\ ;\ 8=\frac 3 {36}\ ;\ 9= \frac 5 {36}\ ;\ 10= \frac 7 {36}\ ;\ 11= \frac 8 {36}\ ;\ 12= \frac {11} {36}$$ Or in math terms: 7 has a chance of 2.78 %, 8 has a chance of 8.33 %, 9 has a chance of 13.89 %, 10 has a chance of 19.44 %, 11 has a chance of 25.00 % and 12 has a chance of 30.56 %.

enter image description here

Great Weapon Fighting...

Grand Weapon Fighting increases the average to 8.33 and thus by 1.33.

Excel Magic

Replacing the Combined Dice

To make the table compatible with Great Weapon Fighting, you need to employ sub-tables.

Fields [1,1], [1,2], [2,1] and [2,2] are filled with the average of the whole table we already had, in the graphics in orange - so the value there is \$\frac{272}{36}\$

Fields [1,3] to [2,6] and its swapped counterparts need simulated rerolls: the 1 could be 1 to 6 after all. We can simulate that as the average outcome of the static die plus a roll of 1 to 6 (average 3.5), so the fields are as follows: $$[1,3],[3,1],[2,3],[3,2]=\frac{42}{6}\ ;\ [1,4],[4,1],[2,4],[4,2]=\frac{46}{6}$$ $$[1,5],[5,1],[2,5],[5,2]=\frac{51}{6}\ ;\ [1,6],[6,1],[2,6],[6,2]=\frac{57}{6}$$

The average damage output boosts to \$\frac{304.89}{36}=8.4691\$, an increase of 0.9136 compared to the no-reroll variant. enter image description here

Replacing one of the 2d6

The same is true for the other method, we in effect need to have subcells for each rerolled chunk. Just the sum chunks are slichtly different. The average damage is \$\frac{394.56}{36}=10.96\$ - an increase of only about 0.488 compared to the no-reroll variant.

enter image description here

Improve this answer
\$\endgroup\$
6
  • \$\begingroup\$ This is a fantastic spreadsheet only answer (+1), but using spreadsheets (at least with my level of expertise) becomes more cumbersome with rerolls. I've experienced this a lot when trying to evaluate the utility of feats like Lucky and Portent in optimization calculations. Do you have a better way to handle rerolls than nesting results in these dice tables (such as are present by the Great Weapon Master fighting style posed in the question)? If so, it would add a lot to your answer! \$\endgroup\$
    – David Coffron
    May 26 at 19:52
  • \$\begingroup\$ @DavidCoffron Rerolls would require multiple spreadsheets that will be combined, this method can't incorporate such feats easily... especially since OP does not tell us how the existing ones interacts with his feat. \$\endgroup\$
    – Trish
    May 26 at 19:54
  • \$\begingroup\$ @DavidCoffron There! Figured it out, but... 7.56 to 8,.46 is a meaningful increase, while 10.47 to 10.96 isn't really noticeable anymore. \$\endgroup\$
    – Trish
    May 26 at 20:32
  • \$\begingroup\$ That's funny. I finished mine at almost exactly the same time. (see my answer for how I did it). Basically I did the chance calculation at the end, rather than getting expected results within the 2d6. \$\endgroup\$
    – David Coffron
    May 26 at 20:33
  • 1
    \$\begingroup\$ @DavidCoffron pulling the results in from sub-tables means I didn't need to modify the main grid evaluation - and could just swap the tables of the parts that make up the main grid^^ \$\endgroup\$
    – Trish
    May 26 at 20:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .