# Dice Averaging Dilemna

1. Sep 6, 2007

### Monoculus

This is for a PnP RPG I am creating... I've looked around for places on the internet, but haven't really found anything so I figured I'd go to the source. ;)

QUESTION: I'm rolling 3d8, (3 eight-sided dice), seven times. What is the average of each roll? Because these are dice, I figured each average would be different because the rolls would be random with each roll.

If someone could help me with this, that would be great. I know it's probably just a simple question, but I can't seem to figure out how to solve it.

2. Sep 6, 2007

### nicktacik

The average for each die would be 4.5 . So for 3 dice it would be 13.5 . So for 7 times it would be 94.5 .

3. Sep 7, 2007

### Monoculus

That's not quite the answer I was looking for, but still helps in my creative process.

I think I was just over-thinking a bit.

When you roll 3d8 seven times, what would the results be for each roll? I now realize that it would be different each time, obviously. I tend to over-analyze things, I think that's what happened LOL

Appreciate your time though, thanks again.

4. Sep 7, 2007

### HallsofIvy

The outcome will be different but the average of all the outcomes will be, as nicktacit said, 4.5.

5. Sep 7, 2007

### CRGreathouse

Tell us what you are trying to do. Are you dropping some dice and summing the others? Are you looking for the likelihoods of the highest, second highest, ... dice being certain values?

For example, the highest roll is 24 with probability 1.36%, 23 with probability 3.98%, 22 with probability 7.55%, ....

6. Sep 7, 2007

### DaveC426913

Also, there are 512 possible unique results, distributed as a bell curve across the numbers 3 though 24.

You want to build a matrix like this:
Code (Text):

sum  rolls                        chances   percent
3   111                           1/512     0.2%
4   112 121 211                   3/512     0.6%
5   113 122 131 212 221 311       6/512     1.2%
...
23   788 878 887                   3/512     0.6%
24   888                           1/512     0.2%

Last edited: Sep 7, 2007
7. Sep 7, 2007

### DaveC426913

There is only one way to roll 24, thus the chances are 1 in (8^3) = 1/512 = 0.195%.

8. Sep 7, 2007

### CRGreathouse

Quite sure. I'm reporting the probabilities across all 7 sets -- the probabilities for a single roll of 3 8-sided dice is easy (and symmetric).

9. Sep 7, 2007

### DaveC426913

Oh I see. So, 0.195 x 7. Gotcha.

Last edited: Sep 7, 2007
10. Sep 7, 2007

### CRGreathouse

Well, subtracting off the double-counting, yes. At least one 24 out of the 7 sets has probability 1-(511/512)^7. The others are a little trickier, but not bad.

11. Sep 7, 2007

### DaveC426913

Waitaminute. Won't your numbers add up to 700%? You've only listed 3 numbers and you're already covering off 13% of the results.

12. Sep 7, 2007

### Monoculus

I have seven stats, (Strength, Dexterity, etc). For each stat, you roll 3d8 to get your stat score.

What I was trying to do was figure out what numbers would be rolled if you initially rolled 3d8 seven times.

I would then take my results and list them in the book as an example for the reader. I was also going to use the results to create a point-buy system.

13. Sep 7, 2007

### CRGreathouse

The numbers will add to exactly 100%. Consider how unlikely it is to have one's highest roll be a dozen or less -- 0.18%, within rounding.

(Did you see post #10? We posted at nearly the same time.)

Last edited: Sep 7, 2007
14. Sep 7, 2007

### CRGreathouse

What's the seventh, luck? Comeliness? As for point-buy, the probabilities won't help you -- you need to decide how much each is worth in the game. Inverse cumulative probabilities would be silly -- you wouldn't want a 24 to be worth four times what a 23 is worth, would you?

15. Sep 7, 2007

### Monoculus

This is my end result:

I took the average of a 3d8 roll (13.5) and rounded it to 14.

14 x 7 stats = 98 points for stat distribution

I added 2 more points to give the player a very small edge, so I'm at 100 points total now.

My 100 points were decreased to 79 because all stats are automatically defaulted to 3 when using the point-buy system, (7 stats x 3 = 21 points initially used). I then dropped the 79 down to 75 so I would have a more generalized number.

If a player chooses to roll for their stats, they will more than likely get the better results. If they want more control over their stat totals, then they sacrifice the possibility of getting better dice rolls. I'm satisfied with the way this works, for now anyway; all things are subject to change as we explore more game mechanics.

The seventh stat is Charisma. The game is skill-based, so skills will be dependent upon the character's attributes.

Last edited: Sep 7, 2007
16. Sep 7, 2007

### CRGreathouse

Random sampling (100,000 sets of 7x3d8) shows the average roll is (19, 17, 15, 13, 12, 10, 8). To two decimal places (I'm not confident the others are accurate), that is
(18.82, 16.59, 14.95, 13.50, 12.05, 10.41, 8.15)

17. Sep 7, 2007

### Monoculus

Do you have a software tool for calculating that? If so, can I have the link? :D

And thank you, I'll use those numbers instead of what I rolled LOL

18. Sep 7, 2007

### CRGreathouse

I was just wondering which stats you has that differed from the D&D-inspired Str, Dex, Con, Int, Wis, Cha.

The average point buy is of course 73.5. 46% of random rolls will have at least a 75 point buy; the remainder will have less. Of course players who get to choose their own stats will make them effectively better by careful placement.

19. Sep 7, 2007

### CRGreathouse

I just made a short PARI/gp script.

This table is the number of rolls (out of 512) that are at least 1, 2, 3, ..., 24.
rtable = [0, 0, 1, 4, 10, 20, 35, 56, 84, 120, 162, 208, 256, 304, 350, 392, 428, 456, 477, 492, 502, 508, 511, 512]

This rolls the 3d8 randomly:
roll() = local(r); r=random(512);for(i=3,24,if(r<rtable,return(i)))

This takes seven rolls and sorts them:
rseven() = vecsort([roll(),roll(),roll(),roll(),roll(),roll(),roll()])

This takes the average of the lowest-of-7 of 100,000 rolls:
sum(i=1,10000,rseven()[1])/10000.

To get the average of the second-lowest, use the above line with a [2] instead of a [1].

If you want to download PARI, you can just copy/paste what I wrote.