Maximized and Empowered... How does that work again?


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Can you give me a link to that part? Please? Thank you.

And I do realize I'm in the minority on this one. I'm even willing to accept me being wrong. I also feel it's faster and time at my table is better spent.


Cavall wrote:
Gauss wrote:

Cavall, it is the result that is multiplied by 0.5 and added again. Not the dice rolled (which is pre-result).

The rules disagree with you flat out. For them to agree with you it would have to say 'half of the dice' or something similar.

It says 'half of the normally rolled result'.
You have to roll the dice, determine the normal result, then cut it in half.

How this works:
Empowered Fireball:
1) Roll Xd6
2) Total the dice roll
3) Multiply the total by 0.5.
4) Add the total from step 2 the number from step 3.

(Note: for expediency you can combined steps 3 and 4 by multiplying the total from step 2 by 1.5.)
This is effectively: (Xd6)*1.5

Empowered and Maximized Fireball:
1) Roll Xd6
2) Total the dice roll
3) Multiply the total by 0.5
4) Multiply the number of dice by the size of the dice to get the maximized value.
5) Add the number from step 3 to the number from step 4.

This is effectively: ((Xd6)*0.5) + (X*6)

Once again, I understand what you're all saying. We disagree, and for my games and for simplicity it would be roll 5d6 not 10 and do math.

I'm not PFS so that's me and my preference, which I still feel is correct. But I'm not going to continue a circular conversation about something that would make it more tedious for me and my players.

Thank you.

And Paul? You moved on from the conversation so the conversation is allowed to move on from you. It's not an argument and no one's bickering. It's just a natural flow.

How do you resolve an odd number of dice? Your 'simplicity' has two side effects.

1) It alters the distribution of the results.
2) It makes things more complicated any time you have an odd number of dice.

Dark Archive

An empowered fireball cast by a 7th level wizard is easy. You roll 7d6. Let's say you get 26. You add half of that number to the total to get the final empowered damage roll. So it would be 39. Save for half. Not that difficult. 26 x 1.5 gives you the same number. No big deal.


Pathfinder Roleplaying Game Superscriber; Pathfinder Starfinder Roleplaying Game Subscriber
Cavall wrote:

Can you give me a link to that part? Please? Thank you.

And I do realize I'm in the minority on this one. I'm even willing to accept me being wrong. I also feel it's faster and time at my table is better spent.

For starters, it's more or less stated in the FAQ (though I admit that it could be clearer).

As for developer quotes, here's what I could find on short notice:

James Jacobs wrote:
Roshan wrote:

I would like a final, hands down ruling on how Empower spell works.

Do you roll 50% more dice (i.e 3d8 on an empowered cure moderate), or do you add 50% (i.e 2d8*1.5 on that same cure moderate) to the dice total rolled.

Roll the dice normally, as you would for the non-empowered version of the spell. Once you have that result, multiply it by 1.5 and drop any fractions.

For a 7th level cleric casting an Empowered cure moderate wounds spell, you'd roll 2d8+7, then multiply what you got by 1.5. So if you rolled 10 points of healing, the empower spell version would be 15 points of healing.

This is generally faster in play (since you're adding up fewer die rolls), but it's also more logical since you'll never have to decide how to roll half a die (which you would whenever you'd end up rolling an odd number of dice).

[LINK]

(It's worth noting that Jason Bulmahn has said differently in the distant past, but must have changed his mind, as it was later clarified differently in the official FAQ (linked above).


Cory Stafford 29 wrote:
An empowered fireball cast by a 7th level wizard is easy. You roll 7d6. Let's say you get 26. You add half of that number to the total to get the final empowered damage roll. So it would be 39. Save for half. Not that difficult. 26 x 1.5 gives you the same number. No big deal.

Yes, I, and most of the other people in this thread, know that. But Cavall is claiming he rolls half the dice in the case of 10d6 to get the empowered section of an empowered, maximized, fireball. So what does he do for half of 7d6?

Dark Archive

And as we have explained, that is not how you calculate the damage per the rules as written and FAQ's answered by the developers. You work with the numbers generated by the dice roll, not the number of d6's. If you decided to house rule it to work that way, 7d6 empowered would be 10d6 (7d6 plus 3.5 d6 rounded down to 3d6). This oddity is most likely the reason you use the number from the dice rolls instead of the number of dice to calculate empowered damage.


Cory, I think you are missing the point of my comments addressed to Cavall.

I know how it works, you know how it works, most of the posters in this thread know how it works.

The point of my last post to Cavall is that I am asking Cavall to explain how he would do odd numbers of dice since he seems insistent on modifying the number of dice rather than multiplying the total.

My last post to you was me explaining what I was asking Cavall. I wasn't asking you a question, I stated what I was asking Cavall. I still want to know how Cavall plans on running an Empowered spell with an odd number dice.

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