D&D 5E GWF vs. TWF Fighting styles

[Xeviat]

Shots fired.

Okay, so, it is accepted that the average of d6 is 3.5, yes? That's 1+2+3+4+5+6=21, 21/6 is 3.5

So, if a roll of d6 is 3.5, then we can replace 1 and 2 with 3.5 to get our 4.16666666... Where is 3.75 coming from? I'm not saying your friend can't be right, but I'm not understanding that equation in the slightest.

Our equation comes from this:

If you roll a 1, roll again, averaging 3.5.
If you roll a 2, roll again, you can get 1 to 6, averaging 3.5.
If you roll a 3 ...

There's six possible rolls on your first roll. If you roll a 1 or a 2, you roll again.

Put average((average(1,2,3,4,5,6),average(1,2,3,4,5,6),3,4,5,6) in excel.

 

[Kryx]

I used anydice and calculated it for you. See http://anydice.com/program/6d9e (it only rerolls once).
1d6 is indeed worth 4.17 as I said.

On GWF's worth: It is between a 7-18% DPR boost over Defensive style depending on class, level, etc (scales from higher to lower).

 

[bid]

apparently better than yours too (again, I'm just the messenger, don't shoot me )

Oh man, are you trolling?

If you reroll 1s and 2s on any die, let's say a d6. You pool together all the results and then divide them by their number, which means 1+2+3+4+5+6+1+2+3+4+5+6+3+4+5+6 divided by 6+6+4.

100% wrong.
You probably did not explain properly as it is impossible to come up with a different answer to this simple problem. 10/10 middle-school teachers would agree.


Without rerolls:
- (1+2+3+4+5+6)/6 = 3.5

Except it's not 1, you replace it with reroll. Same thing for 2:
- ((1+2+3+4+5+6)/6+(1+2+3+4+5+6)/6+3+4+5+6)/6
- (3.5+3.5+3+4+5+6)/6 = 4.17

 

[psychophipps]

But won't a rerolled 2 occasionally roll a 1 or a 2? I'm assuming that you get one reroll, not keep going until it's a 3 or better.

 

[Kryx]

But won't a rerolled 2 occasionally roll a 1 or a 2? I'm assuming that you get one reroll, not keep going until it's a 3 or better.

Yes it will. It will average 3.5 as normal as stated above.

 

[juggerulez]

Shots fired.
Okay, so, it is accepted that the average of d6 is 3.5, yes? That's 1+2+3+4+5+6=21, 21/6 is 3.5

So, if a roll of d6 is 3.5, then we can replace 1 and 2 with 3.5 to get our 4.16666666... Where is 3.75 coming from? I'm not saying your friend can't be right, but I'm not understanding that equation in the slightest.
Put average((average(1,2,3,4,5,6),average(1,2,3,4,5,6),3,4,5,6) in excel.

Shots fired indeed, but I'm on the receiving end myself!

I just know that to make an average you add up all the values and divide them by their quantity. And we all agree on this one.

He says that you can't average an averaged number without approximation, thus since we're in the small numbers field, your approximations are reducing your precision by a lot.

I used anydice and calculated it for you. See http://anydice.com/program/6d9e (it only rerolls once).
1d6 is indeed worth 4.17 as I said.

On GWF's worth: It is between a 7-18% DPR boost over Defensive style depending on class, level, etc (scales from higher to lower).

honestly, programs are compiled by people with a given knowledge. If this knowledge is wrong, the program will be wrong (or my excel would have told me that i was, when i was averaging 1+3.5 )

I really don't know what to say, when he was explaining how to average the dice, I was feeling like my father does when he asks me how to send a whatsapp message...

Oh man, are you trolling?

no, i'm not here to troll, as i've written in my reply above.

100% wrong.
You probably did not explain properly as it is impossible to come up with a different answer to this simple problem. 10/10 middle-school teachers would agree.

if you insist.

I now know that (1+2+3+4+5+6+1+2+3+4+5+6+3+4+5+6)/16 does 3.75, which translates to "all the possible outcomes summed up and divided by their number" which is precisely how you average something.

But won't a rerolled 2 occasionally roll a 1 or a 2? I'm assuming that you get one reroll, not keep going until it's a 3 or better.

It does. You average all the possible outcomes of a die roll with a reroll over 1 and 2, thus you have twice the chance that a '1' and a '2' show up, that's why you average like this: (1+2+3+4+5+6+1+2+3+4+5+6+3+4+5+6)/16 = 3.75

 

[psychophipps]

Hey, an extra .25 damage per die is better than a swift kick to the groin. I'll take it!

 

[Kryx]

I didn't write that formula, but it does make sense to me.

To explain:
R is the roll
N is what is passed in (3) in this case.

It says if the result of the roll (d6) is less than 3 (meaning 1 or 2) then roll d6 and return that. If the original roll wasn't 1 or 2 then use the original roll which is d6.

If that isn't convincing enough then nothing else will be. Your friend is wrong.

 

[psychophipps]

Since we're on the topic, one of the players in our group is assuming that the rerolls are also available with associated Smite dice. Is this the case?

 

[TwoSix]

I now know that (1+2+3+4+5+6+1+2+3+4+5+6+3+4+5+6)/16 does 3.75, which translates to "all the possible outcomes summed up and divided by their number" which is precisely how you average something.

Except that those different outcomes have different probabilities of occurring. Each possibility (1 through 6) has a 1 in 6 chance of occurring. For 3, 4 ,5 and 6, that's the final result. If you roll a 1 or a 2, you roll the dice again. No matter what that roll is, you keep it. The probability of each of those values is now 1 in 36 (36 being 6 squared, the 1 in 6 chance for the 1 or 2 times the 1 in 6 chance for each result that could come up on the second roll.)

So the actual average of the probability distribution (which is what you need to average, the probability distribution, not merely the sum of all possible outcomes) is

1/36+2/36+3/36+4/36+5/36+6/36 + 1/36+2/36+3/36+4/36+5/36+6/36 + 3/6 + 4/6 + 5/6 + 6/6 = 4.17

Sorry, man, but I don't think you explained the problem to your mathematician friend properly. Kryx is right, and his spreadsheet demonstrates his math-fu is pretty damn solid.