Critical - Multiply or Roll Again?

[RyanL]

Err, statistically there should be no difference between multiplying and rolling again over the course of sufficiently many die rolls. I don't see how some people are concluding otherwise??

-Ryan

 

[RyanL]

Oh, and to answer the question: I roll. It's the way you're supposed to do it in 3e, and it's more fun to boot! More dice = more fun. Except maybe vehicle mounted canons in Dragonstar. That got tedious.

-Ryan

 

[AuraSeer]

Err, statistically there should be no difference between multiplying and rolling again over the course of sufficiently many die rolls. I don't see how some people are concluding otherwise??

There's no difference in the average damage, but the distribution looks very different. A single die roll, doubled, has a flat graph. The roll of two dice, added together, generates a bell curve.

Take a greatclub for example (because it makes the math easy). It does 1d10 damage per hit, doubled on a crit.

If you simply multiply the die roll, then each number is still equally likely to show up. You have a 10% chance of doing 2 points, 10% chance of doing 8 points, 10% chance of doing 20 points.

However, if you roll the die multiple times and add, a high roll on one will usually coincide with low or average rolls on the others. You only have a 1% chance of doing 2 points (double 1), and a 1% chance of doing 20 points (double 0). 43% of the time, you'll roll within 2 points of average.

Statistically speaking, they're very different functions. Multiplying makes you far more likely to do max critical damage, and thus makes critical hits more of a factor in the game. In general, this makes things a bit harder for the PCs, just like any other increase in randomness.

(Incidentally, the higher the critical multiplier, the more pronounced the difference. As you roll greater numbers of dice, your results will cluster more and more strongly around the average.)

 

[BlackBart]

Err, statistically there should be no difference between multiplying and rolling again over the course of sufficiently many die rolls. I don't see how some people are concluding otherwise??

-Ryan

Huh, the distribution of a single die cast many times as independent trials is uniform. Since all you are doing is multiplying, the distribution will remain uniform because you still have an equal probability of achieving each value over the range. In this case "average" damage really means nothing because the distribution isn't Gaussian.

Conversely, the distribution of many die casts added together as an independent trial will yield a different distribution, such that as the number of dice increases the distribution approaches Gaussian and "average" begins to have proper meaning.

dead_radish's explanation should be a sufficient example of how the probabilities differ between the two methods(0.08% v 16.7% for max damage, or min damage).