Skip to main content
Role-playing Games

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

4
  • \$\begingroup\$ Doesn't this basically just come down to "roll 1d4 3 times. Add each result to 12 once, subtract each result from 12 once"? \$\endgroup\$ Commented Jan 23 at 16:32
  • \$\begingroup\$ I don't believe so. I don't have enough of a mathematical grasp to understand why exactly (I haven't taken a math class since 2006, and I flunked that one), but I know it's not possible to roll an array of all 12s when rolling 3 d4s and +/- 12 from the result. And using this method, to achieve a 16 you have to roll a 6 and a 2, which I am guessing is slightly less likely than rolling a single 4 on a d4. (I can't back that up, it's just a guess.) \$\endgroup\$ Commented Jan 23 at 17:54
  • \$\begingroup\$ @yangking True. It should be 1d5-1. 0-4, not 1-4 like I initially said. Either way, you actually only need to roll three dice to accomplish this. Each roll is the inverse of another. \$\endgroup\$ Commented Jan 23 at 18:04
  • \$\begingroup\$ @MichaelW. the distribution of 1d5-1 and A-F (A=6d6kh1, F=6d6kl1) is different though. \$\endgroup\$ Commented Jan 24 at 5:56