Probability help on rolling two D4 with all doubles counting as zero except double 4s, which count as 4

Question

So I have devised a game system involving two D4 being rolled to determine movement on a chart. It works by tabulating the difference between the dice pips on each die, e.g., Die 1=rolls '1' vs Die 2=rolls '4' resulting in 4-1=3 movement. The point is to allow for '0' movement when doubles are rolled. But I also wanted to allow for a roll of '4' so I set a roll of double 4s to equal 4 movement, while all other doubles cancel, or zero movement. I have tried calculating this myself but I would appreciate some help to check my work and offer any corrections or suggestions. So....

1. Five possible outcomes on two D4: 0,1,2,3,4
2. The probabilities for each option are not equal: 4 is least likely at 1/16 (0.0625)
3. 0 requires (1|1), (2|2) or (3|3), so that is 3x more likely than 1/16 so 1/16 + 1/16 + 1/16 = 3/16 (0.1875) correct? The rest I will do based on "differences" between the dice
4. Difference of 1: (1|2), (2|1), (2|3), (3|2), (3|4) and (4|3). So would my probability for a "1" be 6/16 (0.375)?
5. Difference of 2: (3|1), (4|2), (2|4) and (1|3). So 2 is 4/16 (0.25)?
6. Difference of 3: (4|1), (1|4). So 3 is 2/16 (0.125)?
7. Summary from highest probability to lowest: 1 (38%), 2 (25%), 0 (19%), 3 (13%), 4 (6%)

Did I get that right?

If so, then my last question concerns the mean value for everything? Definitely need some help here. The zeros are screwing with my brain here. I presume the mean value should be greater than 1 but less than 2.

Answer 1

The probabilities are right, but beware of approximations.

You got the right probabilities in terms of fractions, but when you approximated them with percentages you get a sum of 101%. You are adding 0.5% in items 4 and 6.

A further confirm can be obtained by simulating the rolls and computing the distances via your rules. The histogram below depicts the simulation of 1 million rolls (blue bars) and compares the empirical distribution with the theoretical one (orange dots).

For the average movement you can use the formula for the expected value.

The formula for the expected value \${\rm E}[X]\$ of a discrete random variable \$X\$ that can take the values \$x_1, x_2,\dots,x_n\$ is given by $$ {\rm E}[X] = \sum_{i=1}^n x_i P(x_i) $$

where \$P(x_i)\$ is the probability of each value. This formula takes into account the fact that the values have different probabilities: this is a sort of weighted mean, where the weights are given by the probabilities. The higher the probability of a value is, the higher its contribution to the average is.

In this case you have $$ {\rm E}[X] = 0 \frac{3}{16} + 1\frac{6}{16} + 2\frac{4}{16} +3\frac{2}{16} +4\frac{1}{16} = \frac{24}{16} = 1.5 $$

Your intuition that the average movement should be between 1 and 2 is correct.

Answer 2

Eddymage's answer is correct, but I thought I would provide a supplementary answer to introduce one of RPG.SE's favorite tools: AnyDice. (We even have a tag for it: anydice.)

After wrestling with some error messages for a while, I was able to implement your rolling strategy using two functions:

• movehelper: takes two numbers (i.e. the results of your rolls) and returns the difference between the higher and the lower, unless both numbers are 4, in which case it returns 4.
• move: feeds two 1d4s into movehelper and returns the result. (This is technically unnecessary—you could call output [movehelper 1d4 1d4] directly— but I liked the idea of having a neat wrapper.)

The program is here: https://anydice.com/program/3162d

As you can see, AnyDice shows you the probabilities of each result.

• 0: 18.75% chance
• 1: 37.5% chance
• 2: 25% chance
• 3: 12.5% chance
• 4: 6.25% chance

The two numbers in parentheses after "output 1" are the mean/average result (1.5) and the standard deviation (1.12).

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You mentioned in a comment that you might want 0 and 3 to be equally likely. Under your current scheme, 0 has a 3/16 chance of occurring, while 3 has a 2/16 chance.

The simplest way to implement this would be to say that a double-1 roll is not a valid result, i.e. roll again. This would change the chances of both 0's and 3's to 2/15.

AnyDice program for that strategy: https://anydice.com/program/31632

This raises your mean result to 1.6.