Solving 6=7, Dammit, I wish I thought of this in 2004 Options

[hyzmarca]

6=7, many people complained about it. A TN 6 is equal to a TN 7. There is no possible chance of getting less than a 1 on a die. It is important when adding up modifiers, as many has pointed out, since 6+1 is different from 7+1, which is true, and it is the best argument to support keeping 6=7. I'm not here to discuss the merits of 6=7, rather I intend to propose a novel solution to the problem for those who want it.

One proposed solution was a 1s never count system, but that only created a 7=8 problem. But what if, instead, we subtract 1 from the value of all sides of out dice, so that 1 becomes 0, 2 becomes 1, and so on, and we subtract 1 from all TNs so that the minimum TN becomes 1. This gives us a 0-5 die with average with the same probabilities on the first 5 TNs. The Rule of 6 becomes the Rule of 5. A TN of 6 (Formerly 7) is reached if one rolls a 5 (Formerly 6) and a 1 (Formerly 2), thus removing the 6=7. Shifting all values back one also means that a TN of 7 (Formerly 8) one must roll a 5 and a 2 (Formerly 3), and so on. The 7=8 problem is avoided.

I don't know why this would be particularly useful, I just thought of it and felt like posting it.