Quote from: hellsreach;282377*snip*

Quote from: King of Old School;282437*snip*

These were not rules related so I responded to them here. While I may or may not be responsible for the tangent, I'm sure as hell going to be the one responsible for rectifying it.

Quote from: Blue Seraph;283145But what happens if I don't roll any ones?

We can assume a successful hit does one level of damage + the number of 1's rolled. So if (6, 6, 5, 2) was a successful hit, then the multiplier would be x1, and (6, 6, 1, 1) would be x3 (2 + default).

Quote from: Blue Seraph;283145And why use 1's? It creates a strange paradox where rolling high is better to hit, but rolling low is better for damage: A roll of 24 (6,6,6,6) does less damage than a 14 (6,6,1,1). I'd expect the higher roll to do more damage, since it was a better roll. Making 6's count for damage instead of 1's makes more sense.

If multipliers are based on 6s instead of 1s, it becomes more likely that maximum damage will be done on a successful hit. For example, lets assume a target rolls a dodge result of 18:

   (1, 1, 1, 1, 1) : (1s = 5 / 6s = 0) : miss
(6, 1, 1, 1, 1) : (1s = 4 / 6s = 1) : miss
(6, 6, 1, 1, 1) : (1s = 3 / 6s = 2) : miss
(6, 6, 6, 1, 1) : (1s = 2 / 6s = 3) : hit
(6, 6, 6, 6, 1) : (1s = 1 / 6s = 4) : hit
(6, 6, 6, 6, 6) : (1s = 0 / 6s = 5) : hit

Of course there's nothing inherently wrong with your way either.

Ultimately though this dichotomy between 'if' and 'how much' shouldn't even exist in the first place, and if you transport this system to D6 Legend, it won't.

Quote from: Soylent Green;283255One way to simply D6 damage could be to use static values for Strength rolls, this may well have appeared as an optional rule in one of the D6 books I am not quite sure.

It does sound familiar.