D&D 5E 4d6 Drop the Lowest Etiquette

[Miladoon]

@Yunru, why do we need maths when we have guns?

I let them choose, and offer a different way o rolling stats. I like to expose the players early to rules as guidelines.

My current option is to let them roll pairs of stats six times, in order. Then they choose a high stat from the pair and then they have to choose a low stat from another pair.

Example, Green is a high choice, red is a low choice

STR 15,16
DEX 10,11
CON 13,9
INT 13,14
WIS 12,17
CHA 9,12

So the player would wind up with:

STR 16
DEX 10
CON 13
INT 13
WIS 17
CHA 9

 

[lowkey13]

*Deleted by user*

 

[Yunru]

I'm not predetermining them: I'm ordering them. If any three of the dice are 6s, you get 18, no exceptions. If two of the four are 6s, and one's a 5, you've a 5 in 6 chance of getting 17.

Edit: Okay I guess I'm predetermining two sixes, but that's irrelevant because if there's not two sixes, you can't get either 18 or 17.

 

[lowkey13]

*Deleted by user*

 

[Yunru]

Sure it is. For one thing it's called Maths

No but seriously: if there are three 6s you are assured an 18, true or false?
If there are two 6s and a 5, you have a 1 in 6 chance at 18 and a 5 in 6 chance at 17, true or false?
There are no other combinations that give either 18 or 16, true or false?

Going to admit now, Statistics is a complete mystery. Don't understand it at all. Which is why I'm simply mapping the outcomes instead.

 

[Halivar]

If you order them, there are only 6 combinations that yield 18. (all 6's in first 3, any in last place)
There are 15 combinations that yield 17. (again, you order them, so 6-6-5, 6-5-6, and 5-6-6 plus any non-6 in last place yields 15)

 

[Yunru]

If you order them, there are only 6 combinations that yield 18. (all 6's in first 3, any in last place)
There are 15 combinations that yield 17. (again, you order them, so 6-6-5, 6-5-6, and 5-6-6 plus any non-6 in last place yields 15)

Hey're idependent, order doesn't matter. I was ordering them from highest to lowest for convienience.

 

[Miladoon]

yeah, keep explaining it to each other because you might realize it doesn't matter after 400 posts.

 

[lowkey13]

*Deleted by user*

 

[Yunru]

No, but seriously, you are wrong. You aren't looking at this correctly. You don't roll three dice, and then determine the outcome *based on the roll of the fourth die*. You roll the four dice together.

Which gives you the eleven possible outcomes I listed before.