D&D 5E Dark Fairytale OOC [Closed]

Check the first post of this thread for chargen. You can also choose one uncommon magical item. As a couple of my players have picked up a few.
I was asking about differences between 1st post chargen and 5th level chargen. So 1st post + one uncommon (presumably minor) magic item.

No pressure, create character that you want to play. BUT. If you make something with healing, you'd be better appreciated :)

I was considering the Celestial Warlock as I've played the UA Undying Light Warlock and I'm curious to see the difference. But I only got to 5th level with it so it wouldn't be much of a comparison. OTOH, I have not played a 5e cleric or bard yet. Hmmmm.
 

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Azurewraith

Explorer
I was asking about differences between 1st post chargen and 5th level chargen. So 1st post + one uncommon (presumably minor) magic item.



I was considering the Celestial Warlock as I've played the UA Undying Light Warlock and I'm curious to see the difference. But I only got to 5th level with it so it wouldn't be much of a comparison. OTOH, I have not played a 5e cleric or bard yet. Hmmmm.
Yarp that's correct I'll vet the item on a item by item basis as some are more impactful than others.
 

Okay, I need ability scores before I can go much further. Reroll 1s and 2s changes 1d6 to 1d4+2. So best 3 of 4d4 + 6 is a range of 9-18 where each of the 3 rolls range from 3-6 so that's how I'll roll this.

_:
4D4.HIGH(3)+6 = [4, 4, 4, 4]+6 = 18
4D4.HIGH(3)+6 = [4, 2, 3, 2]+6 = 15
4D4.HIGH(3)+6 = [4, 3, 2, 2]+6 = 15
4D4.HIGH(3)+6 = [4, 2, 4, 3]+6 = 17
4D4.HIGH(3)+6 = [2, 3, 2, 3]+6 = 14
4D4.HIGH(3)+6 = [4, 2, 3, 2]+6 = 15


Wow. I might need to multiclass :) or maybe a bard skills character with all the base mods being +2 or better.
 


Neurotic

I plan on living forever. Or die trying.
Okay, I need ability scores before I can go much further. Reroll 1s and 2s changes 1d6 to 1d4+2. So best 3 of 4d4 + 6 is a range of 9-18 where each of the 3 rolls range from 3-6 so that's how I'll roll this.

_:
4D4.HIGH(3)+6 = [4, 4, 4, 4]+6 = 18
4D4.HIGH(3)+6 = [4, 2, 3, 2]+6 = 15
4D4.HIGH(3)+6 = [4, 3, 2, 2]+6 = 15
4D4.HIGH(3)+6 = [4, 2, 4, 3]+6 = 17
4D4.HIGH(3)+6 = [2, 3, 2, 3]+6 = 14
4D4.HIGH(3)+6 = [4, 2, 3, 2]+6 = 15


Wow. I might need to multiclass :) or maybe a bard skills character with all the base mods being +2 or better.

Average is the same, but distribution isn't. You scewed it far toward high end scores with no chance of low rolls. As you can see, I'm resident engineer. I don't mind high scores especially with Searinox rolls. But roll fairly as everyone else. Unless our DM allows it. Azure, I will not object or make trouble after you rule either way. Mathematically these rolls aren't same as 4d6
 

Azurewraith

Explorer
Average is the same, but distribution isn't. You scewed it far toward high end scores with no chance of low rolls. As you can see, I'm resident engineer. I don't mind high scores especially with Searinox rolls. But roll fairly as everyone else. Unless our DM allows it. Azure, I will not object or make trouble after you rule either way. Mathematically these rolls aren't same as 4d6
I was just looking into the numbers and I failed maths twice and refused a 3rd sitting. Was giving me a head ache.

If you wouldn't going with 3d6 it just makes things the same for those us that can't math.
 

Average is the same, but distribution isn't. You scewed it far toward high end scores with no chance of low rolls. As you can see, I'm resident engineer. I don't mind high scores especially with Searinox rolls. But roll fairly as everyone else. Unless our DM allows it. Azure, I will not object or make trouble after you rule either way. Mathematically these rolls aren't same as 4d6
Neurotic, there is 0% chance of rolling a 1 or 2. That's the point of reroll 1s and 2s. You are incorrect. Roll a bunch of die and only count final rolls, you will find the distribution of 1d6 drop 1s and 2s to be 1 = 0%, 2 = 0%, 3 = 25%, 4 = 25%, 5 = 25%, 6 = 25%. And that looks like 1d4+2. 1d4+2 is EXACTLY the same as 1d6 drop 1s and 2s.

So why would "3d4+6" and 3d6 drop 1s and 2s" be different? Why would "(best 3 of 4d4)+6" be different from "best 3 of 4d6, reroll 1s and 2s"?

If you go to coyote roller and use searinox's 4d6.Min(3), you will never see 1s or 2s. Yet you say his rolls are "valid" even though they too can never produce a 1 or 2.

Care to reevalute my method?
 

I was just looking into the numbers and I failed maths twice and refused a 3rd sitting. Was giving me a head ache.

If you wouldn't going with 3d6 it just makes things the same for those us that can't math.
This is not difficult math. 1d4+2 is the same as 1d6 drop 1s and 2s.

I've used this method in other games without any question before. 1d5+1 is 1d6 drop 1s, for example.

Neurotic would be right if we were doing: 1d6 reroll 1s and 2s ONCE. Then my method does not work. But usually, reroll 1s and 2s means, as long as the result is 1 or 2, reroll. And thus the smaller die size works fine.
 

Charwoman Gene

Adventurer
This is not difficult math.
Yes, this is difficulty parsing English.
Neurotic would be right if we were doing: 1d6 reroll 1s and 2s ONCE.

Which Ironically is NOT the method Neurotic used. Nor Vlad the Destroyer.

I, however rerolled 1s and 2s once. Therefore I protest unless I get to reroll a few 1s and twos and rearrange my ability scores.
I also miscopied a 16 as a 15 at some point.

[sblock=rolls]
6 6 3 3
15
6436
16
---2536
---14
---5651
---16
3364
13
---6245
---15


6 6 3 3
15
6436
16
***5536
***16
***5653
***16
3364
13
***6645
***17
scores: 1D6.MIN(3) = [5] = 5
1D6.MIN(3) = [3] = 3
1D6.MIN(3) = [6] = 6

[/sblock]

Not updating RG unless I get the OK
 

Yes, you did reroll ONCE. That's unfortunate. I hope you get those new stats.

When I said, I didn't have a problem before doing it my way here, I forgot there actually was a discussion of whether "rerolls 1s" means "reroll ONCE" or "reroll 1s until there's no 1s". Once we figured out it was the latter, my method of rolling was just fine.
 

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