Rolling 6 dice and 3 on the same side

Question

I found similar answer for 'Rolling 6 dice and 4 on the same side'. Will exactly the same rule apply for 3 dice?

Instead of 15000 (according to the above link), I found somewhere 14700.

$${6 \choose 2} \times {6 \choose 3} + 6 {6 \choose 3} \times (5^3-5) = 15 \times 20 + 6 \times 20 \times 120 = 14700$$ So there are 14700 ways to roll three of a kind.

Answer 1

I apologize for the error in my first answer, which I have since deleted—I was intent on proving one thing right and another wrong, and failed to notice that the truth was contrary to my expectation.

Your web link, with this justification:

Three of a kind can be formed either as two three of a kinds or as three of a kind and three other dice from one of the other five values that aren't all three the same value.

Is correct, as is the value 14700.

What is wrong with doing it in the way given by the other question you linked? Suppose we choose a number (6 options) and a group of three for that number ($6 \choose 3$ options) and then three additional numbers that are distinct ($5^3$ options). Then, we are double-counting combinations such as this: (1, 1, 1, 2, 2, 2)! Notice that we count this once, by choosing the roll $1$, the combination $\{1, 2, 3\}$, and the other numbers $(2, 2, 2)$. Then we count it again, by choosing the roll $2$, the combination ${4, 5, 6}$, and the other numbers $(1, 1, 1)$.

So, the method that works for exactly four of a kind, does not work for three of a kind, because of doubly counting certain rolls.

To avoid this double-counting we shall consider two cases:

Case 1: There are two three-of-a-kinds combined in a single roll.

In this case, there are ${6 \choose 2}{6 \choose 3}$ options, as correctly mentioned in your post. The $6 \choose 2$ represents the choice of the two numbers that are rolled, each three times. The $6 \choose 3$ represents the number of ways to distribute those two numbers.

Case 2: There is only one three-of-a-kind in a single roll.

We solve this case similarly to the linked answer. First we choose a number to be repeated thrice ($6$ options), then a group of three rolls for that number to appear ($6 \choose 3$ options), then finally the values for the other three. We must however be careful not to include cases where the other three values are identical. Supposing this restriction did not exist, we would have $5^3$ choices, but because it does, we must remove the $5$ cases where all three other values are identical. Hence, there are $5^3 - 5$ options for the other cases.

Multiplication of our independent terms gives $6 {6 \choose 3} (5^3 - 5)$.

Total

Combining our two cases gives your equation:

$${6 \choose 2} {6 \choose 3} + 6 {6 \choose 3} (5^3 - 5) = 14700$$