Probability of two distinct numbers on three d6s I have a question that asks:

Three fair dice are rolled. Find the probability that exactly two different numbers appear on the three dice.

I have come up with my own solution that is incorrect, and I don't understand why. 
When I get problems like this, I usually split up the "possibilities" into their own "slots". This problem will have 3 slots because there are 3 different dice. So, the first dice gets its own "slot", and that slot has 6 different possibilities since the first result can be anything that dice can output.
The second slot has 5 different possibilities because there need to be two different numbers, so the second probability must be a different number - it cannot be the same as the previous number, hence the five, instead of six, possibilities.
The third slot has 2 different possibilities because it can equal either the first slot or the second slot. 
That gives me $\frac{6×5×2}{6^3}=\frac5{18}$.
However, the final solution gives me $\frac5{12}$. There is an explanation as to how they got there ($1-\frac{6×5×4}{6^3}-\frac{6×1×1}{6^3}$), and I understand how they got there and why that solution makes sense.
I just don't understand why my solution doesn't. Any assistance in understanding that would help...
Thanks.
 A: In your calculation you missed out the possibility that the first two slots are the same. 
I would approach this problem in the following way: Choose two numbers out of six, then choose which one of the two numbers appears on two dices, then choose which two of the three dices have this number. Hence, $$P=\frac{{6\choose 2}{2\choose 1}{3\choose 2}}{6^3} = \frac{5}{12}$$
A: The problem is that the second die can still show the same number as the first, as long as the third die then shows a different number. This case is not counted in the attempt leading to $\frac5{18}$, and has probability
$$\frac66×\frac16×\frac56=\frac5{36}$$
Adding this to $\frac5{18}$ produces the correct answer of $\frac5{12}$.
A: Another way to approach this is to consider that you have a first die, a second die, and a third die, and that two must show the same number while one shows a different number.
Now ask first which two dice show the same number.
Among three dice, there are three ways to choose two dice: first and second, first and third, or second and third.
Now, for each choice of which two dice are the pair, consider how those two dice can be matched but different from the other two.
To describe any of the "pair and a singleton" outcomes, you can name the number shown on the pair and then the number shown on the singleton, or you can name the singleton first and then the pair, but either way you have $6$ choices for the first number named and $5$ for the second.
So we have altogether $3 \times 6 \times 5$ outcomes out of the $6^3$ equally likely possible outcomes of rolling three dice:
$$
\frac{3 \times 6 \times 5}{6^3} = \frac{5}{12}.
$$

The way to get $\frac{5}{12}$ as shown above is one combinatoric argument.
Alternatively, after the point where you "consider how those two dice can be matched but different from the other two," 
you could find the probability by constructing a probability tree as in this answer (your original $\frac{5}{18}$ includes two of the three choices of pairs, and that answer shows how to do the tree for the third pair.)
Yet another alternative another combinatoric method in which you first ask what numbers come up (a pair and a singleton) and then ask which dice the pair came up on.
That's the method in this answer.
Any of these methods is OK as long as they account for the three possible pairings of the same-numbered dice. That's why I recommend doing at least a little combinatoric thinking about any problem like this, even if you end up doing a probability tree in the end, just to make sure you don't miss a branch of the tree.
