Brutal

[Eric Finley]

Heh. Keterys, if you're going to make a die from scratch, make it specific; it's not like you're going to use it for anything except your d12b1 weapon, anyway. So label it 2-12 instead of 1-11, and engrave the cone ends with cool runes or pictures of the weapon or something.

Betcha you could do one for ~$5 in plastic or ~$25 in steel via Shapeways.com.

 

[eamon]

1d4+1d6+2 has the exact same average as 2d6b1, but a slightly different distribution.

1d4+1d6+2 has:
0.21 - 8
0.17 - 9
0.13 - 10
0.08 - 11
0.04 - 12

2d6b1 has:
0.17 - 8
0.14 - 9
0.11 - 10
0.08 - 11
0.06 - 12

Something's wrong with those calculations - 2d6b1 has precisely a 1 in 25 chance of rolling 12; which is 0.04. How did you make these calculations?

The damage distribution of 2d6b1 is very very similar to that of 1d4+1d6+2. It's not worth splitting hairs here; you can use 1d6+1d4+2 instead of 2d6b1 without impacting game balance ;-).

 

[Elric]

1d6, brutal 1 gives the same average as "1d6, if the die roll is below 3, it becomes 3." So that's a good solution for 2d6, brutal 1.

 

[keterys]

Yeah, d6b1 -> d6 where 1s and 2s become 3s work (3 added divided by 6), as does 1d10b2 where 1-4 becomes 5 (10 added, divide by 10). But after that it's less convenient.

That artificer paragon path that makes everyone in the party start randomly acquiring brutal still makes me wish for a different solution, but ah well.

 

[Blackbrrd]

d6b1 is effectively 2,3,4,5,6 = 20 / 5 = 4 average, in other words, if you turn a 1 into a 4 you get the correct average.

2d6b1 can then be done as all 1's are turned into 4's.

1d10b2 can be done as 1's = 6's and 2's = 7's.

 

[keterys]

Heh. It'd be pretty odd to want to roll a 1 over a 3.