I'm having a little difficulty recalling the procedure for calculating probability of dice results. My question is, using 4 Fudge d6 dice, what is the probability of getting three of any result in one roll? (+++,---,ooo)

  • 7
    \$\begingroup\$ Does "++++" count for this purpose as containing three of any result, or does it have to be exactly and only three of the same? There is an Atomic Robo stunt I wonder if you're referring to, which permits "+++0", but excludes "++++". \$\endgroup\$ Oct 16, 2014 at 0:12

3 Answers 3


There are 81 possible permutations of this roll. Of these, 3 have all 4 results the same, and 24 have 3 results the same. This gives us a probability of 27/81 or 33.33%, for at least 3 results the same, and 24/81 or 29.63% for exactly 3 results the same.

  • \$\begingroup\$ Could you post a mathematical formula for the percentage chance? \$\endgroup\$
    – Sandwich
    Oct 16, 2014 at 3:54
  • \$\begingroup\$ I know the percentage is correct, but for the entire computation, out of curiosity. Normally to determine the percentage of a certain digit coming up on a d6 you use (1/6^n) for the number of dice, Since Fudge dice are technically d3's, You'd use (1/3^n), but that only covers the odds of getting a single result, and it doesn't compute for an additional die, I'd just like to know, for future reference. \$\endgroup\$
    – Sandwich
    Oct 16, 2014 at 4:07
  • \$\begingroup\$ "Where did you get the 81 integer?", is what I'm trying to ask. \$\endgroup\$
    – Sandwich
    Oct 16, 2014 at 4:35
  • \$\begingroup\$ Ah nevermind, I figured it out, 3 different branches, one for +, One for -, One for o, each containing 9 different results. 9*3*3. 81. I'm dumb. \$\endgroup\$
    – Sandwich
    Oct 16, 2014 at 4:40
  • \$\begingroup\$ You have 4 dice, each has 3 possible outcomes (evenly distributed I assume). That makes 3^4 different permutations, or 3*3*3*3 = 81. Note that the result +000 and 0+00 are different permutations, although they show the same totals of pluses, dashes and zeroes. \$\endgroup\$
    – fgysin
    Oct 16, 2014 at 6:49

The odds of getting 3 of the same result out of four are 33%.

For each set of Four Fudge dice there are a total of 27 different roll results with 3 alternate outcomes per set of rolls, +, -, and o. For a total of 81 possible results. Those outcomes are below:

++++ ++-+ ++o+
+++- ++-- ++o-
+++o ++-o ++oo

+-++ +--+ +-o+
+-+- +--- +-o-
+-+o +--o +-oo

+o++ +o-+ +oo-
+o+- +o-o +oo+
+o+o +o-- +ooo

This is the set of outcomes for only the "+" set of rolls, There are sets for the "-" and the "o" as well. As you can see, 33% of the outcomes above have three or more of the same result. This includes the rolls beginning in "-" and "o" as well.

  • \$\begingroup\$ Since there's an Atomic Robo RPG stunt which asks for exactly three, and the question doesn't use "at least" phrasing, I'm not sure this is complete without the probability of a result of exactly three (and without it, this answer appears functionally identical to WaxEagle's). But then, that's why I voted to close the question as unclear. \$\endgroup\$
    – BESW
    Oct 16, 2014 at 4:53
  • 3
    \$\begingroup\$ Awww. But mine has cool formatting. \$\endgroup\$
    – Sandwich
    Oct 16, 2014 at 4:55

The probability of 3 of any one result is 33.33%. Here's why:

The basic probability of getting at least 3 of any one outcome of fudge dice is 11.11%. This is 1/9. The fact that you have 3 possibile results from a set where the outcome of any 3 dice is 11% means that the overall probability is 33.33%.

Showing my work. This program utilizes the fact that Fate dice are essentially d3s, and takes into account the fact that you're rolling 4 dice. The "at Least" option from the following command:

 output [count 1 in 4d3]

Gives us the output of 11.11% to show the odds of getting at least 3 of any one number. Since these 3 outcomes are equally likely, and are independent. The probabilities add allowing us to say that on 33.33% of rolls at least 3 dice will show the same face.

With a small number of possibilities like this it is easiest to simply count the number of desirable outcomes in the possible results and use those. However, there are mathematical formulas that allow us to generalize this to any number of results, across any number of rolls with any sized die.

The two numbers that matter here are the number of results that we care about (in this case 9) and the number of total results (in this case 81). The number of total results is easy to derive. That's simply the number of results in the dice (3) raised to the power of the number of times you're rolling in this case 4. (3^4 = 81).

The simple thing for determining the permutations is to think about the number of possible dice rolls that would come up on our target value if we rolled 3d3. 3d3 show the 3 of the same face a single time. So if we use 4d3, we get 3 for the fourth die, and then we also get 2 for each other die since the other 3 dice can show the same face. This gives us 2+2+2+3 = 9 permutations. This obviously doesn't scale and the math for how to scale this is currently escaping me (it's basically the same math I'm showing here, but reducing it to a mathematical form is beyond me right now)

So that means that our formula for three of any 1 die is

 (2+2+2+3)/81 = 9/81 = 11.11%

Multiplying this by 3 gives us 33.33%.

The other thing we need to pay attention to (and since it's not clear from the question which is desired), is that sometimes we may want to know the probability of only getting 3 of a kind. In this case we can take our previous form:

 27/81 (that's 3x9/81)

and subtract 3 permutations from it (one each for ----, ++++ and oooo) Giving us

 24/81 = 29.62%

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .