<s>[INST] Probability that $2$ dice selected from $3$ rolled dice will have sum $7$ If I roll $3$ dice ($6$-sided) and then proceed to choose $2$ of them (discarding the 3rd), what is the probability of the total of those $2$ dice totalling $7$?
 [/INST] Let's say the dice are blue, green, and red.  Each outcome can be represented by the ordered triple $(b, g, r)$.  Since there are six choices for each entry, there are $6^3 = 216$ possible outcomes.  Assuming the dice are fair, they are equally likely to occur.
Method 1:  The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.  
There are $\binom{3}{2} = 3$ ways to select a pair of dice.  Say we select the blue die and the green die.  There are six ways for that pair of dice to have a sum of $7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$, where each ordered pair is of the form $(b, g)$.  For each such pair, there are six possible outcomes for the red die (or, in general, the other die), giving $3 \cdot 6 \cdot 6$ favorable outcomes.
However, we have counted each outcome in which there are two pairs of dice that have sum $7$ twice, once for each way we could have designated one of the pairs as the pair that has sum $7$.  For this to occur, two of the dice must show the same number and the third die must show the difference between $7$ and the number that appears on the other two dice.  There are $\binom{3}{2} = 3$ ways for exactly two dice to display the same number and $6$ numbers these two dice could display.  For any such outcome, there is only one possible outcome for the third die.  Hence, there are $3 \cdot 6$ such outcomes.
Thus, the number of favorable cases is $3 \cdot 6 \cdot 6 - 3 \cdot 6 = 108 - 18 = 90$.
Therefore, the probability that when three dice are rolled that a pair of dice may be selected with sum $7$ is 
$$\frac{3 \cdot 6 \cdot 6 - 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$
as InterstellarProbe found.
Method 2:  The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.
There are two possibilities.  Either each die shows a different outcomes or two of the three dice show the same outcome.  Since $7$ is not a multiple of $3$, it is not possible for all three dice to show the same outcome.
Each die shows a different outcome:  There are $\binom{3}{2} = 3$ ways for two of the dice to have sum $7$ and $6$ ways for those two dice to have sum $7$.  There are four possible outcomes for the other die.  Hence, there are $3 \cdot 6 \cdot 4$ such outcomes.
Two of the dice show the same outcome:  There are $\binom{3}{2} = 3$ ways for exactly two of the three dice to show the same outcome and six possible outcomes those two dice could show.  The other die must show the difference between $7$ and the outcome on those two dice.  Hence, there are $3 \cdot 6$ such outcomes.
That yields a total of $3 \cdot 6 \cdot 4 + 3 \cdot 6 = 90$ favorable outcomes.
Hence, the probability that a pair of the three dice has sum $7$ is 
$$\frac{3 \cdot 6 \cdot 4 + 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$
in agreement with the answer we obtained above.
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