Dice – Ensuring Fairness in Dice Rolling Methods

cheatingdicestatistics

Recently, in D&D sessions I have been rolling poorly (nothing over 10), and thus tried ways to improve my rolling. I was using a plastic solo cup last session, and got really good results compared to normal. I would place the die with the 20 facing down while holding the cup upright. I then spin the cup in a circular motion, slowly tilting it to about horizontal, then slamming it down on the table to get the result(kind of like Yahtzee). I don't personally think this has any real advantage to rolling normally, but one of my friends said that rolling in a consistent method would essentially make the rolls less random, which would be considered cheating. I know some people will not let others touch their die, or start all their rolls with a 20 facing up, but I thought that was more superstition than a tangible change. Recently I found this article that rolls would not be completly random but it doesn't notate any specific methods used for the motion of the dice. It does imply that changing the situation in which the dice are rolled can affect the outcome or probability.

So my questions are:

Does the physical rolling method change the odds of the dice in a significant way?
Is simply seeking for a method to improve rolling considered cheating?

Best Answer

This is a classic clustering fallacy

Wikipedia defines the clustering fallacy as:

...the tendency to erroneously consider the inevitable "streaks" or "clusters" arising in small samples from random distributions to be non-random.

Basically, a perfectly random die will have long (potentially very long) steaks of seemingly non-random behavior, either very high or very low. Therefore, your observation of low rolls in multiple sessions could simply be true random variation. Streaks, both high and low, might not be common, but you'll definitely notice them a lot more than more varied behavior. After all, a sequence of exactly 19,20,19 will be a lot more noticeable than a 3, 16, 9, even though the exact probability of both sequences is the same.

The only way to truly know if any method or die is biased or not is to do dozens of identical rolls and look at the resulting distribution. Additionally, everything from the die itself to the rolling surface to your throwing technique is going to affect the roll.

How random does it need to be?

Moreover, how random is random enough? If you read the paper that your article is citing, they say that for a 6-sided die that bounces 4-5 times, the probability that the die lands on the face that's the lowest at the beginning is about 20%, whereas the expected random value is about 16%. Would you notice a 4 percentage point difference on a d6 roll?

For a more concrete personal example, I have a gold d20 that that has a very strong bias toward rolling 20s, according to the saltwater flotation method. However, when I use this "golden" d20, I don't actually notice a greater proportion of 20s showing up in my rolls.

Therefore, the answer to your first question is yes: rolling methods do affect the roll, but unless you're using actual loaded dice, the difference is too small to matter. This means that the answer to your second question is also no, as well.

Related Solutions

Dice Rolling Programs – How Reliable Are They?

Any well written dice roller will give you perfect random rolls (well, pseudo-random — the difference doesn't matter for gaming).

Some dice rollers are not well written, or depend on the underlying OS/language's source of random numbers, which itself can be well written or not.

The vast majority of the time you will find dice rollers more random than actual dice, which, being physical objects, will be flawed.

I know of one product that had to change the library it used to generate rolls because the OS implementation was poor, but that is a rare case.

[RPG] Dice probabilities when rolling more dice than needed

We'll pretend we have k s-sided dice and we are trying to find the probability to get n specific different results. The first die roll we have an n/s chance of getting a number we need and an (s-n)/s chance we get a number we don't need. When we get a number we need we are looking to get n-1 numbers we need on k-1 different dice. When we get a number we don't need we are still looking to get n numbers but with k-1 dice left to roll. The probability of getting n = 0 results is always 1 regardless of the number of dice. The probability of getting any more results when there aren't any dice left (k = 0) is 0.

prob s k 0 = 1
prob s 0 n = 0
prob s k n = (n*(prob s (k-1) (n-1)) + (s-n)*(prob s (k-1) n))/s

We can make a table to calculate these for s = 6 sided dice.

  0   1        2        3       4      5 6 n
0 1   0        0        0       0      0 0
1 1   1/6      0        0       0      0 0
2 1  11/36     2/36     0       0      0 0
3 1  91/216   30/216    6/216   0      0 0 
4 1 671/1296 302/1296 108/1296 24/1296 0 0
.
.
.
k

The problems you are interested in are getting n=2 specific different results with k=2 dice which has a probability of 2/36, getting n=2 specific different results with k=3 dice, which has a probability of 30/216, and getting n=3 specific different results with k=4 dice, which has a probability of 108/1296.

There is probably a simpler non-recursive formula based on the factorial function.

We can check the above formula with counting arguments.

To get a 3 and a 4 rolling 2d6 there's 1 combination of dice that will work. We don't care if we roll the 3 first or the 4 first, so there are 2! = 2*1 = 2 orders we can roll the combination in. 2 of the 6^2 possible rolls result in a 3 and a 4. The probability of rolling a 3 and a 4 on 2d6 is 2/36 or 1/18.

We can make a table of all the ways to roll a 3 and a 4 on 2 dice and count them to make sure this is correct

3 4    1 way
4 3    1 way  
       2 ways

2 ways out of 36 roll a 3 and a 4.

There are more ways to roll a 3 and a 4 than there are ways to roll two 3s. Theres' only one way to roll two 3s. If we switch the order we roll the dice in it doesn't add another way to get two 3s. This will matter when we add extra dice.

3 3    1 way
3 3    0 ways! We already counted this.
       1 way

Let's write down all the ways we can roll a 3 and a 4 on 3d6. We'll use * to denote any result that is neither a 3 or a 4 (there are 4 possible such results: 1, 2, 5, and 6)

3 4 *    4 ways
3 4 3    1 way
3 4 4    1 way
4 3 *    4 ways
4 3 3    1 way
4 3 4    1 way
3 * 4    4 ways
3 3 4    1 way
3 4 4    0 ways! We already counted this.
4 * 3    4 ways
4 3 3    0 ways! We already counted this.
4 4 3    1 way
* 3 4    4 ways
3 3 4    0 ways! We already counted this.
4 3 4    0 ways! We already counted this.
* 4 3    4 ways
3 4 3    0 ways! We already counted this.
4 4 3    0 ways! We already counted this.
         30 ways

When one of the results isn't a 3 or 4 it's easy, there's 4 combinations that can be in 3! = 3*2*1 = 6 orders. When all of the results have a 3 or 4 then one of the numbers will be repeated. Instead of having 2 combinations that can be in 3! = 3*2*1 = 6, half of the orders are duplicates, so there are only 30=6*4+2*3 ways to roll a 3 and a 4. The probability of rolling a 3 and a 4 on 3d6 is therefore 30/6^3 or 5/36.

Counting results to check larger problems is tedious. This line of thinking is leading to a recursive formula involving factorial which is no simpler than the recursive formula we started with.