Probability of two pair of dice in 5 rolls I was browsing this question:
Probability of two pairs of poker dice and letter arrangement.

Poker dice is played by simultaneously rolling $5$ dice. The probability of two pairs is approximately $0.2315$. 

and when I tried to solve it in my head I came up with the answer
$$\frac{\binom{6}{2}\binom{4}{1}}{6^5}$$ 
because we need to choose a value for each pair of dice (from 1 to 6) then we can choose from the remaining values for the last die. In other words I would assume that all die are indistinguishable but clearly that is not most people's interpretation. The answer given in the question is:
$$\frac{\binom{6}{2}\binom{5}{2}\binom{3}{2}\binom{4}{1}}{6^5}$$
which is the same as mine, except it includes the factor 
$$\binom{5}{2}\binom{3}{2}$$
which is equivalent to saying we can choose two of the dice to be a pair, and choose two more of the remaining dice to be a pair. However, I think of dice as all the same so we are over-counting. How can I learn to read these questions properly?
Edit:
A couple more examples of different answers from different interpretations, 
How many ways can we pick two teams of 4 people from 20 people
 A: I would suggest you try gambling on those odds and see how quickly you lose money :).
Truth is the dice are distinguishable. You could imagine you have 5 different colored dice if you'd like, or that you roll each die at a prescribed spot on the table. 
I'll use the second example: imagine you roll each die one by one, in a line from left to right. A roll that comes in $1,2,3,4,6$ from left to right is a different (and equally likely) roll from one that comes in $6,3,4,1,2$ from left to right.
In answer to the question about people, this is open to two interpretations. If there's a red team and a blue team, the answer is ${20\choose 4}{16\choose 4}$ where you can interpret the first factor as the choice for the red team and the second as the choice for the blue team. However, if the teams are 'indistinguishable' then there is only half that number. I would tend to interpret the question as written in the second way, although I would ask for clarification if possible, since oftentimes people mean the first way.
A: The word "indistiguishable" is tricky here, and I don't know precisely what you mean by it. I interpret this problem to be a question about a five-letter word over an alphabet of six letters. The fact that you've indicated that you have a space of $6^5$ outcomes indicates that you're thinking about it along the same lines, where you have five spaces in some order, and you must fill them with numbers $1$ through $6$. Thus we consider $(1,1,1,1,2)$ to be a different outcome than $(1,1,1,2,1)$. So if you roll two pairs, it's not enough to say what the values are. You have to say where the values appear in the sequence, for it to really be one of your possible $6^5$ outcomes.
Suppose you roll two dice and you want to know the probability of getting exactly one $6$. There may be nothing to distinguish the dice, but you still have to consider dice individually to really determine the answer. You could say it's 
$$\frac{\binom{5}{1}}{6^2} = \frac{5}{36}$$
because one die must be $6$ and then you have $5$ choices of value for the remaining die, but you can verify just by counting that the answer should be $\frac{10}{36}$. That's because we need to impose an arbitrary order on the dice, and then say which one is the $6$.
