Tuesday, May 17, 2011

Mechanics and Efficiency

I really like the math behind my proposed core mechanic (in short: roll a pool of D6s, retain the highest, boxcars = 7). It works great for PCs and singleton foes. However, in a playtest I came up against a problem: it doesn't work well for hordes of mooks.

In traditional D&D, if you have a half dozen goblin archers you just roll six D20s to see if they hit. Likewise, if the wizard hits a squad of orcs with a fireball, you just roll a fistful of D20s to see if they save. With the dice pool mechanic, you generally have to roll one monster at a time unless they have a pool of "one;" if you color code your D6s, perhaps 2-3 at a time.

One work around would be to treat the dice pool as a "reroll." For example, say each monster has a dice pool of 2D. You could roll 1d6 for each of them and then reroll for any that failed on the first check. That is probably faster than setting up a color coded dice pool but still requires two steps. Another "solution" would be to limit situational bonuses or modifiers given to mooks; maybe part of being a mook is that you don't get modifiers to your dice pool.

I also recently thought of another simple dice mechanic that uses D6s but gives a bit more fidelity at the top of the scale: Exploding dice.
  • Roll a D6 vs. a TN.
  • If you roll a 6, roll again. If you get a 5 or a 6, then add 1 to your result. If you get another 6, roll again.
That gives an approximately 1/20 chance of getting a 7, and a 1% of getting an 8. That is, setting a TN of "7" is basically like saying, "You need a hail-mary natural 20 to hit this TN."

Distribution:
1 (16%)
2 (16%)
3 (16%)
4 (16%)
5 (16%)
6 (10%)
7 (5%)
8 (1%)

The problem is that it doesn't scale well with modifiers. A straight +1 modifier is fairly huge, as it doubles the chances of getting a 7 and quintuples the chance of getting an 8. You could reduce the number at which a highly skilled individual's dice "explode." For example, if you get to roll again on a 5 or a 6 then the distribution looks like this:

1 (16%)
2 (16%)
3 (16%)
4 (16%)
5 (10%)
6 (15%)
7 (5%)
8 (1%)

That leads to a somewhat wonky distribution, though, where 6 is more likely than 5.

Math
Odds of getting a 6: .166
Odds of getting a 5 or a 6: .333
.166 * .333 = 5.5%

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