Question: What is the average of seven rolls of 3d8 in a PnP RPG?

  • Thread starter Monoculus
  • Start date
  • Tags
    Dice
In summary, the conversation is about creating a PnP RPG and the question is about the average of rolling 3d8 seven times, with the conclusion being that the average for each die is 4.5 and for 7 rolls it would be 94.5. The conversation also discusses the likelihoods and probabilities of different outcomes when rolling 3d8 seven times, as well as the process of creating a point-buy system for the game.
  • #1
Monoculus
5
0
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.
 
Mathematics news on Phys.org
  • #2
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
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
The outcome will be different but the average of all the outcomes will be, as nicktacit said, 4.5.
 
  • #5
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
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:
 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:
  • #7
CRGreathouse said:
For example, the highest roll is 24 with probability 1.36%, 23 with probability 3.98%, 22 with probability 7.55%, ...
You sure about these numbers?

There is only one way to roll 24, thus the chances are 1 in (8^3) = 1/512 = 0.195%.
 
  • #8
DaveC426913 said:
You sure about these numbers?

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

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
CRGreathouse said:
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).
Oh I see. So, 0.195 x 7. Gotcha.
 
Last edited:
  • #10
DaveC426913 said:
Oh I see. So, 0.195 x 7. Gotcha.

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
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
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
DaveC426913 said:
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.

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:
  • #14
Monoculus said:
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.

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
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:
  • #16
Monoculus said:
What I was trying to do was figure out what numbers would be rolled if you initially rolled 3d8 seven times.

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

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

Monoculus said:
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.

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
Monoculus said:
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

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.
 

1. What is the Dice Averaging Dilemma?

The Dice Averaging Dilemma is a statistical paradox that involves finding the average of a large number of dice rolls. It demonstrates that in certain situations, the average of multiple dice rolls can actually be lower than the average of a single dice roll.

2. How does the Dice Averaging Dilemma work?

The Dice Averaging Dilemma works by using a large number of dice rolls to calculate the average. However, the more dice that are rolled, the more likely it is that the average will be lower than the average of a single dice roll. This is due to the way probabilities work and is a counterintuitive result.

3. Why is the Dice Averaging Dilemma important?

The Dice Averaging Dilemma is important because it highlights the limitations of using averages in certain scenarios. It also demonstrates the importance of understanding probability and how it can lead to unexpected results.

4. How can the Dice Averaging Dilemma be applied in real life?

The Dice Averaging Dilemma can be applied in real life situations where averages are commonly used. For example, in sports statistics, it can help explain why a team's average score may be lower than the average score of their individual players.

5. Is there a solution to the Dice Averaging Dilemma?

There is no definitive solution to the Dice Averaging Dilemma, as it is a paradox that challenges our understanding of probabilities and averages. However, understanding the concept and its implications can help in making more informed decisions in situations where averages are used.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
16
Views
2K
  • Precalculus Mathematics Homework Help
2
Replies
53
Views
5K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
  • Set Theory, Logic, Probability, Statistics
2
Replies
41
Views
3K
  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
Replies
11
Views
2K
Replies
4
Views
26K
Replies
17
Views
7K
Replies
4
Views
2K
Back
Top