For those of us a little slow on the math, for the interchangeable d66 table, what are the 15 common outcomes vs the 6 rare ones? I feel like this method is exactly what I need for some random tables I've been poking at, but I can't tell precisely where those ranges lie.
(With reference to this post here.)
In the variant d66 table in which the lowest (or highest) die is always assigned to the "tens" place rather than treating the two rolls as an ordered pair, the doubles (i.e., 11, 22, 33, etc.) will be rolled with 1/36 probability, while all other entries (i.e., 12, 13, 14, etc.) will be rolled with 1/18 probability. This is because each non-doubles entry has two possible rolls that correspond to it (e.g., entry "12" can be achieved by rolling 1 then 2, or by rolling 2 then 1), while the doubles are each obtainable only via a single roll.
The resulting table layout looks like this:
11 – rare (1/36) 12 – common (1/18) 13 – common (1/18) 14 – common (1/18) 15 – common (1/18) 16 – common (1/18) 22 – rare (1/36) 23 – common (1/18) 24 – common (1/18) 25 – common (1/18) 26 – common (1/18) 33 – rare (1/36) 34 – common (1/18) 35 – common (1/18) 36 – common (1/18) 44 – rare (1/36) 45 – common (1/18) 46 – common (1/18) 55 – rare (1/36) 56 – common (1/18) 66 – rare (1/36)
As you can see, collectively you can expect to receive one of the six rare results one time in six, while the other five times in six you'll get one of the fifteen common results, with each subset of results being equally weighted internally. (i.e., all six rare results are equally likely, given that you get one at all.)
