Probability spread on 3d6

Question

I tried to calculate the number of unique combinations to get the probability of specific results on a 3d6 roll and I am getting wildly different numbers from what I see posted online, and I would like to see where I am going wrong.

There are 216 possible combinations, as you know. However for all intents and purposes a (2,1,1) is the same as a (1,2,1) or a (1,1,2) for example, in terms of the result to a human player.

Combining duplicate results I arrive at 56 possible combinations (please excuse the tedious iteration of variants, I include them in case they have relevance):

1 combination for 3:
(1,1,1)

1 combination for 4:
(1,1,2 or 1,2,1 or 2,1,1)

2 combinations for 5:
(1,1,3 or 1,3,1 or 3,1,1)
(1,2,2 or 2,1,2 or 2,2,1)

3 combinations for 6:
(1,1,4 or 1,4,1 or 4,1,1)
(1,2,3 or 1,3,2 or 2,1,3 or 2,3,1 or 3,1,2 or 3,2,1)
(2,2,2)

4 combinations for 7:
(1,1,5 or 1,5,1 or 5,1,1)
(1,2,4 or 1,4,2 or 2,1,4 or 2,4,1 or 4,1,2 or 4,2,1)
(1,3,3 or 3,1,3 or 3,3,1)
(2,2,3 or 3,2,2 or 2,3,2)

5 combinations for 8:
(1,1,6 or 1,6,1 or 6,1,1)
(1,2,5 or 1,5,2 or 2,1,5 or 2,5,1 or 5,1,2 or 5,2,1)
(1,3,4 or 1,4,3 or 3,1,4 or 3,4,1 or 4,1,3 or 4,3,1)
(2,3,3 or 3,2,3 or 3,3,2)
(2,2,4 or 2,4,2 or 4,2,2)

6 combinations for 9:
(1,2,6 or 1,6,2 or 2,1,6 or 2,6,1 or 6,1,2 or 6,2,1)
(1,3,5 or 1,5,3 or 3,1,5 or 3,5,1 or 5,1,3 or 5,3,1)
(1,4,4 or 4,1,4 or 4,4,1)
(2,2,5 or 2,5,2 or 5,2,2)
(2,3,4 or 2,4,3 or 3,2,4 or 3,4,2 or 4,2,3 or 4,3,2)
(3,3,3)

6 combinations for 10:
(1,3,6 or 1,6,3 or 3,1,6 or 3,6,1 or 6,1,3 or 6,3,1)
(1,4,5 or 1,5,4 or 4,1,5 or 4,5,1 or 5,1,4 or 5,4,1)
(2,2,6 or 2,6,2 or 6,2,2)
(2,3,5 or 2,5,3 or 3,2,5 or 3,5,2 or 5,2,3 or 5,3,2)
(2,4,4 or 4,2,4 or 4,4,2)
(3,3,4 or 3,4,3 or 4,3,3)

6 combinations for 11:
(1,4,6 or 1,6,4 or 4,1,6 or 4,6,1 or 6,1,4 or 6,4,1)
(1,5,5 or 5,1,5 or 5,5,1)
(2,3,6 or 2,6,3 or 3,2,6 or 3,6,2 or 6,2,3 or 6,3,2)
(2,4,5 or 2,5,4 or 4,2,5 or 4,5,2 or 5,2,4 or 5,4,2)
(3,3,5 or 3,5,3 or 5,3,3)
(3,4,4 or 4,3,4 or 4,4,3)

6 combinations for 12:
(1,5,6 or 1,6,5 or 5,1,6 or 5,6,1 or 6,1,5 or 6,5,1)
(2,4,6 or 2,6,4 or 4,2,6 or 4,6,2 or 6,2,4 or 6,4,2)
(2,5,5 or 5,2,5 or 5,5,2)
(3,3,6 or 3,6,3 or 6,3,3)
(3,4,5 or 3,5,4 or 4,3,5 or 4,5,3 or 5,3,4 or 5,4,3)
(4,4,4)

5 combinations for 13:
(1,6,6 or 6,1,6 or 6,6,1)
(2,5,6 or 2,6,5 or 5,2,6 or 5,6,2 or 6,2,5 or 6,5,2)
(3,4,6 or 3,6,4 or 4,3,6 or 3,6,4 or 6,3,4 or 6,4,3)
(3,5,5 or 5,3,5 or 5,5,3)
(4,4,5 or 4,5,4 or 5,4,4)

4 combinations for 14:
(2,6,6 or 6,2,6 or 6,6,2)
(3,5,6 or 3,6,5 or 5,3,6 or 5,6,3 or 6,3,5 or 6,5,3)
(4,4,6 or 4,6,4 or 6,4,4)
(4,5,5 or 5,4,5 or 5,5,4)

3 combinations for 15:
(3,6,6 or 6,3,6 or 6,6,3)
(4,5,6 or 4,6,5 or 5,4,6 or 5,6,4 or 6,4,5 or 6,5,4)
(5,5,5)

2 combinations for 16:
(4,6,6 or 6,4,6 or 6,6,4)
(5,5,6 or 5,6,5 or 6,5,5)

1 combination for 17:
(5,6,6 or 5,6,5 or 6,5,5)

1 combination for 18:
(6,6,6)

So then:

(1x4)+(2x2)+(3x2)+(4x2)+(5x2)+(6x4)= 56 Combinations (taking all variants of the same roll into account).

My next step was to divide 100 by 56 to get the probability that any one of these 56 combinations would show up:

100/56 = 1,7857142857

Thus:
3,4,17 & 18 have a 1.79% (rounded up) chance of showing up since each only has 1 combination
5 & 16 have a 3.57% chance since they each have 2 combinations
6 & 15 have a 5.36% chance (3 combinations for each number)
7 & 14 have a 7.14% chance (4 combinations for each number)
8 & 13 have a 8.93% chance (5 combinations for each number)
9,10,11 &12 have a 10.71% chance (6 combinations for each number)

Are my calculations correct and if so why don't they match the probabilities I see listed online (including slightly higher probabilities for 10 and 11, which I find strange since 9,10,11 and 12 all have the same number of combinations.)

My apologies for any misuse of symbols. I am not a mathematician. Thanks to anyone who troubles to read through all this.

Answer 1

To calculate the probabilities of different values, you can see how many times the value occurs divided by $216$. To conceptualise each result, you could use a 3D point $(D_1, D_2, D_3)$, or a septimal (base $7$) number without 0. While the result isn't equal to this number, it gives us a meaningful way of talking about the first option, the second option, the last option, and so on.

There is only one way to get three (111), meaning that it has a $\frac{1}{216}$ chance of occuring. There are three ways to get four (112, 121, 211), meaning it has a $\frac{3}{216}$. There are six ways to get five, ten to get six, fifteen to get seven, and so on.

Overall, there are [1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1] ways to get [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18].

Answer 2

You can only compute the probabilities by counting if each case is equally likely. This means your error is in assuming "for all intents and purposes a (2,1,1) is the same as a (1,2,1) or a (1,1,2) for example, in terms of the result to a human player" implies they can be combined and still have equal weight in the probability as any other combination.

For an extreme case that exhibits how absurd this thought process can be, consider the lottery. For all intents and purposes, one losing ticket is the same as any other to the player, so really there are two results: you win or you lose. Does this mean that you have an equal chance of winning or losing? Of course not!

In the same manner, even though a dice throw of (2,1,1) looks the same as (1,2,1) or (1,1,2) to the player, this is a superficial similarity as far as the probability computation goes. We cannot simply combine them and assume that they have the same likelihood of occurring as (1,1,1).