Dice 2x6D Action / Attack / Run

You could always have a standard Difficulty Value (DV) based on the difficulty of the action which players need to beat in order to succeed.

Jumping to grab a ledge might have a DV of 5, but if that same ledge was coated in ice, the DV could become a 9 for instance.

On 2d6, you have 11 different sums (2 to 12) and 36 discrete die roll combinations (1,1; 1,2; 1,3; ... , 6,5; 6,6).

Knowing this, I suggest that you start by figuring out how often you want each of your three results to occur in a given situation.

Example: I want A to happen 50% of the time, B to happen 25% of the time, and C to happen 25% of the time.

Note that if you want to ensure that *something* will happen, the sum of your probabilities needs to add up to 100%.

Once you have odds for your three events, match those odds to the result odds on 2d6.

Example, continued: The odds of rolling a five or less (or a nine or greater) on 2d6 are 28%. That's close enough to 25% for me. This means that B will happen if the result is two, three, four, or five; C will happen if the result is nine, ten, 11, or 12. Any other result (six, seven, eight) will cause A to happen.

Sidetrack: You might be wondering, "How do you know what the odds are?" It's all about probability, and would take me a bit of time to type and explain. However, I Googled for "the odds of rolling a given number on 2d6" and found this article on a D&D forum: http://forums.relicnews.com/showthread.php?t=151889 – It should be helpful.

Note that I'm assuming that you never want a "no result" result *and* that the results are mutually exclusive (You can't have A and B; B and C; C and A, or A, B, and C happen at the same time.). It's easy enough to do either of these things, though.

Example: I want A to happen 50% of the time, B to happen 40% of the time, and C to happen 25% of the time. 15% of the time, I want A and B to happen.

The odds of rolling a six or less on 2d6 is 42%, which is close enough to 40% for me. From the previous example, we know that the odds of rolling a six, seven, or eight come close to 50% (It's ~44%). Those ranges overlap at six, which has a 14% chance of occurring – darn close to 15%. We already know that the odds of rolling a nine or greater on 2d6 are 28%. That's close enough to 25% for me.

That gives us A on a six, seven, or eight; B on a six or less; and C on a nine or greater. Note that a six gives us both A and B.

One more note: You don't have to make your "result zones" contiguous. You could say that a two, three, seven, or 11 causes an A; a four, five, nine, or ten causes a B; and a six, eight, or 12 causes a C. (FYI: That's a 31% chance of A, 39% chance of B, and 30% chance of C – all mutually exclusive.)

If this is hard to understand, I strongly recommend spending some time learning about (and playing with) the odds of any given result on 2d6.

I wish you the best of luck in your design. I hope it comes out the way you want it to!