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)
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.
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.
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%