Simplified Rolls

[wlmartin]

Myself and a friend where having a discussion about using a simplified roll for Execution Axe.

He says that since it is Brutal 2 it does a minimum of 3 damage and a maximum of 12

He says that it is the same range as 3d4 and was thinking about asking his DM if he could change his roll to a 3d4 as it would mean he wouldnt need to reroll Brutal any more.

I told him that it sounded like a smart idea but was unsure if his DM would go for it or if there is something inherent in the rules that says you can't.

In theory the chances of him rolling any number are identical with 1d12 rerolling anything over a 2 and 3d4 but I am unsure if you are allowed to simply swap out the dice if you feel like it.

Of course if there was something that permitted him to reroll damage dice in a different fashion (such as the Avenger Daily that lets you reroll avenger power dice) then he would need to go back to using 1d12 but it seems a lot neater to roll 3d4 (he has many dice so it would be one clean roll as well)

Anyone tried something like this before or have any thoughts?

 

[keterys]

1d10+2 is a more accurate representation of 1d12b2 (in that it's identical, but _much faster_). I've played that before, and DMed for that as well. I've also allowed just about any other combination that worked with brutal, because I think it's a horrible mechanic

3d4 is probably close enough, though it's decidedly more likely to get its average (7.5) than its lowest or highest. Most players and DMs shouldn't really care about that difference, though. I'd just find it annoying from a "Okay, 5W attack... that's 15d4" standpoint.

 

[wlmartin]

1d10+2 is a more accurate representation of 1d12b2 (in that it's identical, but _much faster_). I've played that before, and DMed for that as well. I've also allowed just about any other combination that worked with brutal, because I think it's a horrible mechanic

3d4 is probably close enough, though it's decidedly more likely to get its average (7.5) than its lowest or highest. Most players and DMs shouldn't really care about that difference, though. I'd just find it annoying from a "Okay, 5W attack... that's 15d4" standpoint.

I don't get what you mean in saying that is likely to get the average more.

How is that logically possible?

(Ignoring the time it takes to get the result)
You are just as likely to get any result from 3-12 on a 1d12 as on 3d4

It reminds me of the conversation I had with my friend (a different one) about the chances of tossing a coin 99 times and each time it comes up heads, what is the chance of the 100th one coming up heads... he argued (wrongly) after sitting down and doing the math on it that it was some crazy high "in the millions" chance however he didnt realize that I wasnt asking what was the chance that I would flip a coin 100 times and each time it would come up heads... as his answer may have been valid.

I was discussing the chance for the 100th coin to come up heads and it is always and will always be 50/50.

So unless I am missing something, how could you be more likely to roll an average on a 3d4 than a 1d12 (ignoring 1,2) since the probability of getting each result is 100% exact?

(ie you can only get 3-12 and whilst rolling a 1 or a 2 causes a reroll that some may get into their head has more chance of returning a higher result, since the original roll that was made that the result was 1 or 2 could have equally been 11 or 12, the reroll doesnt come into it in determining the chance of rolling something other than the range of 3-12)

 

[Mengu]

(Ignoring the time it takes to get the result)
You are just as likely to get any result from 3-12 on a 1d12 as on 3d4

False. keterys is correct. 1d12[b2] has the same flat distribution as 1d10+2. 3d4 is a bell curve with more results around the middle (7-8), than the ends (3 or 12).

 

[Locutus Zero]

To expand on what's been said:

With 1d12b2(or 1d10+2) your odds of rolling any given number 3-12 is 10%. Your average number is 7.5. Your standard deviation is 3.02765.

With 3d4 your odds of rolling each number are as follows:
3 1.5625%
4 4.6875%
5 9.375%
6 15.625&
7 18.75%
8 18.75%
9 15.625&
10 9.375%
11 4.6875%
12 1.5625%

Your average is still 7.5. Your standard deviation is 1.9518.

This means, as keterys said, you will get number closer to 7.5 more often with 3d4. If you must change I recomend 1d10+2, just keep in mind that the real number is 1d12 any time something changes your rolls, other than adding a W or a flat number.

 

[Locutus Zero]

I don't get what you mean in saying that is likely to get the average more.

There are 12 possible outcomes of rolling a d12. If you reroll every time you get a 1 or 2, there are only 10 possible outcomes. Adding 2 to a d10 gives you the same odds.

With 3d4 there are 64 possible outcomes. 444 gives you 12, that's 1/64. 443, 434, and 344 give you 11, that's 3/64. 442, 424, 244, 433, 343, and 334 give you 10, that's 6/64. 7 and 8 each happen 12/64 of the time. Hopefully that gives you an idea of what we are talking about.

 

[Redbadge]

Expanding once more, hopefully in even simpler terms, here are some possible results for 3d4:

1, 1, 1
1, 1, 2
1, 1, 2
1, 1, 4
1, 2, 1
etc.

If you completed this chart all the way out to 4, 4, 4 you would see that there are 64 different possibilities (whereas 1d12 only has 12). The explanation of Locotus Zero's percentages is because there is only 1 possible result for a 3 (1,1,1). 1/64 = 1.5625%.

Similarly,there are 12 possible results for a 7 (1-2-4) (1-4-2) (3-1-3) (4-1-2) (4-2-1) (2-1-4) (2-4-1) (1-3-3) (3-3-1) (3-2-2) (2-3-2) (2-2-3)

12/64 = 18.75%

 

[wlmartin]

False. keterys is correct. 1d12[b2] has the same flat distribution as 1d10+2. 3d4 is a bell curve with more results around the middle (7-8), than the ends (3 or 12).

The maths you are talking about is going over my head...

But are you saying that somewhere in the maths side of things there is an unequal chance of scoring differently even when the base probability is the same?

ie if I wanted to use an even numbered die to calculate a 50/50 split
1d2 (coin) heads / tails
1d4 (1/2) 50% / (3/4) 50%
1d6 (1/2/3) 50% / (4/5/6) 50%
1d8 (1/2/3/4) 50% / (5/6/7/8) 50%
1d10 (1/2/3/4/5) 50% / (6/7/8/9/10) 50%

(etc)

As far as my maths was taught at school (and I appreciate that the maths gets more complex as the study of it goes on, way over my head) a probability of something happening is a probability no matter range is considered as long as the ratio remains constant since the results cancel each other out

If I am wrong just tell me I am wrong... and I wont bother with it any more
It just seems ... well icky, like someone is going to come up to me as a grown man and say

"You know what, 2+2 is normally 4 however if you stand on one leg and jiggle your hand about a bit, you can actually make it equal 5"

 

[wlmartin]

There are 12 possible outcomes of rolling a d12. If you reroll every time you get a 1 or 2, there are only 10 possible outcomes. Adding 2 to a d10 gives you the same odds.

With 3d4 there are 64 possible outcomes. 444 gives you 12, that's 1/64. 443, 434, and 344 give you 11, that's 3/64. 442, 424, 244, 433, 343, and 334 give you 10, that's 6/64. 7 and 8 each happen 12/64 of the time. Hopefully that gives you an idea of what we are talking about.

I think I get it now

The dumb way to explain this to me should be

The average will never change, that is just maths.

The chance of getting that average changes because of the type of dice you use.

I still think that all things being equal the extra combinations should cancel each other out to work out as the same result but I imagine it doesnt happen exactly like that due to the way whole numbers and fractions work with the maths involved... to me if that is the reason it works the way it does then that makes a little more sense.

Otherwise I think ill just stick my head in the toilet since it is quite significantly mashed after this discussion!

Anything else you try and explain will just go over my head

 

[Redbadge]

I think I get it now

The dumb way to explain this to me should be

The average will never change, that is just maths.

The chance of getting that average changes because of the type of dice you use.

I still think that all things being equal the extra combinations should cancel each other out to work out as the same result but I imagine it doesnt happen exactly like that due to the way whole numbers and fractions work with the maths involved... to me if that is the reason it works the way it does then that makes a little more sense.

Otherwise I think ill just stick my head in the toilet since it is quite significantly mashed after this discussion!

Anything else you try and explain will just go over my head

Just one more thing, maybe to get the lightbulb to click?

Roll 3d4 100 times and record each total. Then count the number of 7s and 8s compared to the number of 1s and 12s. You'll see that the extra combinations don't cancel out, quite logically, no higher math involved.