# Can you help me with the math on Great Weapon Fighting style on Greatswords?

I know there is already a question regarding the average damage impact on this style and I have been successful to calculate this damage increase myself.

Yet I was more interested in knowing the cumulative chance of rolling <= a certain amount of damage.

I did some spreadsheet calculation and I only got to a total chance of 99,77% to roll <=12. The average only shows 8,31, so I there is a small error which I am overlooking.

Can you explain me how to calculate the odds for each damage result?

How I calculated this: per damage outcome I calculated the chance of rolling it. 'Natural roll' + from re-rolling other results which have at least 1x '1' or '2' in the result.

For example the chance for 2 result with GWF:

P2gwf = P2* (P2+P3+P4/3)

P3gwf = P3 (P2+P3+P4/3)

with PX = chance to roll X on 2d6

I assume that all 1's and 2's that show up in a roll are re-rolled

So far the number crunching did confirm my intuition that 2d6 we roll 1's and 2's far more often than a d12, so will have more use for the Great Weapon Style.

• Welcome to RPG.SE! Take the tour if you haven't already. Can you clarify what exactly your question is? It seems like you're asking a few different questions in your post.
– V2Blast
Oct 25 '18 at 3:19
• Hello and welcome to the stack, please could you clarify exactly what your issue is though i did notice that your chance to role less than or equal to 12 on a d12 was not 100% which indicates some error there but without the actual spreadsheet itself it is very difficult to see what the issue is. If you are interested in percentage chances for dice then you may find www.anydice.com a useful tool. other than that happy gaming and good luck with your miscalculations. Oct 25 '18 at 3:50
• I'm not sure how far goes your math knowledge, but I've given my answer using some basic concepts of probability theory - first, conditional probabilities and then sum of independent random variables. There should be other - probably easier - ways to do that, but this is the most straightforward way for me. You can ping me in chat if I can help with something here :) Oct 25 '18 at 4:20

One way to calculate it: First, you calculate the probabilities of one 1d6. The probability vector can be calculated as a sum of probabilities $$\P_1 + P_2 + P_3\$$, where $$\P1 = 1/6 * [1/6, 1/6, 1/6, 1/6, 1/6, 1/6]\$$, $$\P_2 = 1/6 * [1/6, 1/6, 1/6, 1/6, 1/6, 1/6]\$$ and $$\P_3 = [0, 0, 1/6, 1/6, 1/6, 1/6]\$$. Essentially, the first two probabilities are the probabilities of the reroll (i.e., first rolling a 1 or 2 and then rolling something). The third probability is the probability of the first dice, which doesn't provide you a reroll.

That gives us the following probability values:

0.055555555555556

0.055555555555556

0.222222222222222

0.222222222222222

0.222222222222222

0.222222222222222

You can check they sum up to 1. Note that this is not a cumulative function, it's a probability function, i.e. the probability of a roll equal to some value. It's ordered by that value (i.e. the first value is the probability of a 1, second is the probability of a two, etc).

Now you can calculate the probability of the sum of these two discrete random variables. That probability is given by $$\ P_{\textrm{sum}}[j] = \sum_{i = 1}^{6} P[i] P[j - i] \$$1 where $$\j\$$ goes from 2 to 12, representing the probabilities of a sum from 2 to 12. This will yield the following probability vector:

0.003086419753086

0.006172839506173

0.027777777777778

0.049382716049383

0.098765432098765

0.148148148148148

0.172839506172840

0.197530864197531

0.148148148148148

0.098765432098765

0.049382716049383

If you want the cumulative function instead of the mass function, you can just sum up to the value you want. The final result is

0.003086419753086

0.009259259259259

0.037037037037037

0.086419753086420

0.185185185185185

0.333333333333333

0.506172839506173

0.703703703703704

0.851851851851852

0.950617283950617

1.000000000000000

1 The values of $$\P[i]\$$ for $$\i < 1\$$ are zero. Same goes for $$\i > 6\$$.

• ps: I will probably clear this later if the querent is fine with this approach. Oct 25 '18 at 4:21
• Many thanks HellSaint. The method of probability vectors is way more elegant than mine approach for calculating all the different scenario's. This has enabled me to pinpoint my calculation error. So both methods work, which is what I like about math: there are many ways to come up with an answer. Yet as a personal note: after 14 years of no notable math usage my knowledge has become rusty and making these calculations sleep-deprived, isn't the best either. I finally ironed the last error out: rolling an '8' has 5 possible combinations, not 4 :P. Oct 25 '18 at 5:07
• @JorisVanderCammen Forgot 4+4? :P Either way, if the answer has solved your problem, you can accept it by clicking the green "check" under the down-arrow. I recommend waiting 24h though, as other people might show up with a better answer. Oct 25 '18 at 5:14
• I just was getting lost with all the different parts of the sum. Like I said, your solution is a lot cleaner and easier to calculate. Would you mind explaining me (or pointing to) how this probability vector works. So it describes to the chance of a 1,2, etc to show up on the actual roll. That's why there are three parts: your either re-roll in two cases and in the third you don't re-roll. The sum is just the combinations that make up how a number is rolled by this process of 2d6 re-roll all 1's and 2's. It's a lot clearer after some sleep and yeah, I'm waiting before I check you answer. Oct 25 '18 at 14:03
• @JorisVanderCammen The vector actually represents a discrete function which is the PMF of the random variable which is the result of the rolls and rerolls. You can think of the rolls and rerolls as always being two dice, and you are interested in the results of the second dice. If the first dice rolls 3+, the second dice uses the same value as the first dice. If the first dice rolls 1 or 2, the second dice rerolls. That gives you a conditional PMF $P(d_2 | d_1)$. Then you are interested in $P(d_2)$, which can be obtained by averaging $P(d_2|d_1)P(d_1)$ for all $d_1$. Oct 25 '18 at 20:51