Empower Spell calculations.... [Archive] - Wizards Community

Empower Spell calculations.... [Archive] - Wizards Community

Post/Author/DateTime Post
Jay_Ibero_911
03-11-07, 01:00 PM I'm a little confused on this...do you roll your normal variables and then actually multiply the end result, or do you just roll and adittional half of the variables? For example, 10th level wizard casts empowered fireball. Does he roll 10d6 and multiply the rolled ammount by 1.5? Or does he roll 15d6? The rules for multiplying damage seem to indicate the latter is correct, but most people I run into using it use the former....
Witch
03-11-07, 01:06 PM It's the former.
Roland the Crownbreaker
03-11-07, 02:33 PM Technically you are supposed to multiply the result by 1.5, but personally my gaming group allows to roll the extra dice if the player desires. It just feels good to toss that 1 d handfull every now and again.
ShinAkuma2
03-11-07, 02:55 PM Indeed, the feat provides an "increase by one-half", rather than a multiplication by 1.5.

Mathematically the same, but in D&D, it requires no extra dice.
StevenO
03-12-07, 02:51 PM (10d6)x1.5 = 15d6 for most real gaming applications.

Assuming your dice are all fair (random) you may save more time rolling fewer dice and using the average for the rest. In most cases rolling 4-5 dice and averaging the rest should give you a good enough distribution.
runestar
03-13-07, 02:19 AM (10d6)x1.5 = 15d6 for most real gaming applications.
I think there is a slight statistical difference in the standard deviation of results, but I guess no one really cares...:P
jokotoko
03-13-07, 06:52 AM I think there is a slight statistical difference in the standard deviation of results, but I guess no one really cares...:P

There indeed is. In the most general sense possible, rolling fewer dice and multiplying the outcome yields a higher chance of either very high or very low results. As an extreme example, compare rolling 1d6 x 15 with rolling 15d6. In the first case, the change of rolling the maximum (or minimum for that matter) is 1/6, in the latter case it is 2.13*10^-12 (just very low).
As a result, the standard deviation of the first method is much higher.

Post/Author/DateTime	Post
Jay_Ibero_911 03-11-07, 01:00 PM	I'm a little confused on this...do you roll your normal variables and then actually multiply the end result, or do you just roll and adittional half of the variables? For example, 10th level wizard casts empowered fireball. Does he roll 10d6 and multiply the rolled ammount by 1.5? Or does he roll 15d6? The rules for multiplying damage seem to indicate the latter is correct, but most people I run into using it use the former....
Witch 03-11-07, 01:06 PM	It's the former.
Roland the Crownbreaker 03-11-07, 02:33 PM	Technically you are supposed to multiply the result by 1.5, but personally my gaming group allows to roll the extra dice if the player desires. It just feels good to toss that 1 d handfull every now and again.
ShinAkuma2 03-11-07, 02:55 PM	Indeed, the feat provides an "increase by one-half", rather than a multiplication by 1.5. Mathematically the same, but in D&D, it requires no extra dice.
StevenO 03-12-07, 02:51 PM	(10d6)x1.5 = 15d6 for most real gaming applications. Assuming your dice are all fair (random) you may save more time rolling fewer dice and using the average for the rest. In most cases rolling 4-5 dice and averaging the rest should give you a good enough distribution.
runestar 03-13-07, 02:19 AM	(10d6)x1.5 = 15d6 for most real gaming applications. I think there is a slight statistical difference in the standard deviation of results, but I guess no one really cares...:P
jokotoko 03-13-07, 06:52 AM	I think there is a slight statistical difference in the standard deviation of results, but I guess no one really cares...:P There indeed is. In the most general sense possible, rolling fewer dice and multiplying the outcome yields a higher chance of either very high or very low results. As an extreme example, compare rolling 1d6 x 15 with rolling 15d6. In the first case, the change of rolling the maximum (or minimum for that matter) is 1/6, in the latter case it is 2.13*10^-12 (just very low). As a result, the standard deviation of the first method is much higher.