So our DM cajoled one paladin to go two handed although he wanted to go sword and board to be a little tanky.  We have two paladins and the other one already decided to go sword and board, except it is hammer and board, but who's counting?

Well I didn't know this but he gave the two-handed paladin a d12 weapon but told him he'd make it 2d6 instead.  Then he took a trait that gave him the ability to reroll 1s and 2s.  So I wanted to convince the two-hander he was getting a pretty good deal.  To do this I wanted to compute the damage output for a regular 1d12 roll, a 1d12 reroll 1s and 2s, a 2d6 roll and a 2d6 reroll 1s and 2s.  Well the first calculation is a piece of cake.  The second calculation is a piece of cake.  The third calculation is a piece of cake.  But the fourth calculation, where you may have to reroll one or two dice and then accept the second result is quite involved because there are a tremendous number of permutations that could happen before you get to an answer.  It is easier to work it from the top back to the lowest result.

To roll a 6-6, you have to roll a 6-6 outright, or roll a 1-6 or 2-6 and then get a 6 on the reroll, or roll a 6-1 or 6-2 and get a 6 on the reroll, or roll any of the four 1s & 2s combined and then roll a 6-6 on the reroll.  Chance of rolling a 6-6 then becomes 1/36+2/36*1/6+2/36*1/6+4/36*1/36 = 4.9383%.  And trust me, that is one of the easiest cases to compute. 

The case for 1-1 is also easy, because it requires you to roll any of the four 1s & 2s combined and then roll snake eyes.  so it is a very simple 4/36*1/36 = 0.3086%.  The case for a 1-2 or a 2-1 is similarly easy with 4/36*2/36 = 0.6173%. 

The case for a 3-1 result or a 2-2 result illustrates how the calculation starts to get crazy. You can get a 2-2 result the same way you get a 1-1 result.  Start with all 1s & 2s and then you reroll the unlucky result of 2-2.  But getting a 3-1 or a 1-3 gets involved.  Now there is a 4/36 chance of rolling either a 1-3, 2-3, 3-2, or 3-1.  From this you reroll only one die and you have to roll a 1, so this is a 1/6th chance.  So to achieve 4 damage you compute 4/36*3/36 + 4/36*1/6 = 0.9259% + 1.8519% = 2.7778%.

The case for a "5" result [1-4, 2-3, 3-2, 4-1] is a similar set of hurdles.  For the 1-4 or 4-1 it is the same as a 1-3 or 3-1, so that is 1.8519% and the 2-3 or 3-2 is the same as that so you only need double the result.  So the chance of getting 5 damage in this set of circumstances is 3.7037%.

Now things start to take a turn for the worse.  To get a "6" result, you can do this by rolling a 3-3 outright.  But you can get this result by rolling a 1-3, 2-3, 1-4, 2-4, 1-5, 2-5 or the reverse.  So there are 12 combinations that could happen on the first roll that result in you rolling one die over.  But with the first die fixed, you have a specific number you must roll on the second roll to end with a "6" result.  So the 3-3 result outright is 1/36 = 2.7778%.  The final result of a "6" when rerolling one die is 12/36*1/6 = 5.5556%.  BUT, we forgot one.  What if you roll a combination of 1s and 2s on the first roll?  Then you roll both dice over, and you have a 5/36 chance of rolling a "6" so we have to add the probability of 4/36*5/36 = 1.5432% to get all the different permutations in the mix.  The final result ... 9.8765%.

Now if you are a sharp student you figured out already that I made the same mistake on the "5" result.  I forgot to add in the chance of rolling the 1s and 2s and rerolling to get a "5" on the second roll.  So let's go back and add that in.  2 * 1.8519 + 4/36*4/36 = 3.7037% + 1.2346% = 4.9383%.

After all this calculation I have the chances for these cases:

       2d6 rr1&2     2d6       1d12 rr1&2   1d12

"1"      0%               0%        1.3889%     8.3333%

"2"  0.3086%    2.7778%   1.3889%     8.3333%

"3"  0.6173%    5.5556%   9.7222%     8.3333%

"4"  2.7778%    8.3333%   9.7222%     8.3333%

"5"  4.9383%   11.1111%   9.7222%     8.3333%

"6"  9.8765%   13.8889%   9.7222%     8.3333%

...

"12"  4.9383%  2.7778%   9.7222%     8.3333%

If there is a website that computes results like this when re-rolling 1s and 2s can be factored in, I'd like to visit it.  I couldn't find it.

Try Anydice dot com.

If you just want expected damage output, you don’t need the chance of rolling each number. The averages are as follows:

1d12: 6.5

2d6: 7

1d12 re-roll 1/2: 7.3

2d6 re-roll 1/2: 8.3

Thanks. Of course he first two are known and the third is not a hard number to compute. One reason I need each chance is to also compute the standard deviation, which my two-handed Palo friend likes to see. 8.3 is a pretty good average. I was surprised to see how many permutations there are to consider and I enjoy doing the math myself once in a while.

Quote from MusicScout >>

Thanks. Of course he first two are known and the third is not a hard number to compute. One reason I need each chance is to also compute the standard deviation, which my two-handed Palo friend likes to see. 8.3 is a pretty good average. I was surprised to see how many permutations there are to consider and I enjoy doing the math myself once in a while.

 Considering that 1d6 reroll 1s and 2s gives the following options: (123456)(123456)3456 with the sets in parentheses representing 1 and 2 respectively, that gives you 16 possible results. Taken to 2d6, that would bump to 256 possible options. Factor in a crit? That's 256x256 which is north of 62500+3000+36=65536 possible results. That's a fair amount of math. 

Well, you are right that it is a fair bit of math, but not quite as you describe.  You have to come at this as a problem with two (possibly modified) inputs that will result in 11 possible outcomes.

The lowest outcome is a result of two, which can only be attained in one method.  The player makes their first roll and the result is all 1s & 2s and then the player rerolls those dice and rolls two ones.  The probability of this outcome is 4/36 x 1/36 = 4/1296 = 0.3086%.

To attain a result of three, the calculation is almost the same.  4/36 x 2/36 = 8/1296 = 0.6173%

Then you come to the first twist.  To get a result of 4, you can roll two 2s, or roll a 1 & 3 or a 3 & 1.  Two get a result of two 2s the calculation is again simple, because it is the same calculation as getting two 1s.  So we need to add the possibility of having a result of 1 & 3 (or the reverse).  To get this result, you have to roll a 3 and either a 1 or a 2.  Just to do this, the chances are 4/36.  With one 3 fixed, the second die has to roll a 1, so this is a 1/6 chance.  We combine these and get 4/36 x 1/6 = 4/216 = 1.8519%.  We combine these two cases and the total probability is 2.1605%.

The calculation continues to balloon from there until you get close to the top results.  But even the 6-6 result can be obtained by three paths.  You get a natural 6-6, 1/36 times.  You roll 1s and 2s first and then roll a 6-6, 4/36 x 1/36 times.  And then there is the possibility you roll one 6 together with a 1 or 2, and then reroll that as a 6, 4/36 x 1/6 times.  I think the cases I still can't get right are the results of 6, 7 or 8, where there is "overlap" in the table.  But I'm going to a little more searching online and see if I can find the answer.  Right now, I've got most of it figured out, but the sum of all my probabilities are 101.2%, so I have a mistake somewhere.  My math is certainly tight enough to say the average result is 8.4, but I'm also looking for the standard deviation.