Given a set of 4 dice, what is the probability on a given roll that any 2 of them will have a sum of 7?
(Assume that the two dice are selected in favor of getting a $7$, such that if it's possible to make a sum of $7$ from any two dice in the roll, it counts.)
Given a set of $4$ dice, there are $5$ potential outcome sets for the roll:
Useful facts to know beforehand:
Because $7$ is not divisible by $2$, there is no way for any pair of identical dice to add to $7$. Since the only possible sum of two dice in a four-of-a-kind will be even, we know that there are 0 outcomes in this category that can add to $7$.
There are $4$ ways this outcome can be rolled, one for each die to be the unique one of the set. Since there are $6$ seven-pairs that can be thrown, we know that there are $6 * 4 = 24$ outcomes.
There are actually $6$ ways to roll a particular two-pair, but we don't want to count sets like $\{1,1,6,6\}$ for both the $\{1,6\}$ and $\{6,1\}$ seven-pairs. You can equivalently think of this category as asking "Given the result of the first die, how many ways can the remaining three dice be rolled to be two of the missing seven-pair value and one that matches the first." Similar to category $2$, we know that there is one outcome for each die to be the unique one, so there are $3$ outcomes per seven-pair in this category.
Therefore, there are $6 * 3 = 18$ outcomes.
We can break this category into $2$ distinct cases: those where the repeated number is used in the seven-pair (4a), and those where the unique numbers are a seven-pair (4b).
4a. We know that there are $6$ possible pairs that can be rolled. Additionally, we know that one of the other dice's value is fixed to be the corresponding number in the seven-pair. For the fourth die to be unique then, it cannot be either of those two values, leaving $4$ valid values remaining. We can get the number of ways this can be rolled by taking ${}_4 \mathrm{ P } _4 = 24$ and dividing by the factorial of the number of duplicate dice $(2! = 2)$, giving us $12$.
$6$ pairs that can be rolled $*$ $1$ fixed value for the seven-pair $*$ $4$ remaining values for the fourth die $*$ $12$ ways to roll $=$ $288$ outcomes.
4b. Again, we'll have $6$ possible pairs that can be rolled. However, this time, we'll have $4$ of the seven-pairs as the values for the other dice (this is the number of seven-pairs that don't use the value of the rolled pair). We can get the number of ways these can be permuted using the same method as in 4a, except this time we want to reuse the caveat from category 3 to avoid double counting sets like $\{1,1,3,4\}$ for both $\{3,4\}$ and $\{4,3\}$, so we halve it.
$6$ pairs that can be rolled $*$ $4$ values for our third die $*$ $1$ corresponding value for the third die's seven-pair $*$ $6$ ways to roll $=$ $144$ outcomes.
Intuitively, we have $6$ seven-pairs, $4$ remaining unique values for die number three, and $3$ remaining unique values for die number four. However, one unique caveat here is that if dice three and four are a seven-pair, they would be double counted! To address this, we can break this category into $2$ cases: rolls with a single seven-pair (5a), and rolls with two seven-pairs (5b).
5a. We have our $6$ seven-pairs, $4$ remaining unique dice values for die number three, and $2$ remaining unique, non-seven-pair values for die number four. Of the $24$ ways these dice can be permuted, we want to avoid double counting our seven-pairs, as well as instances where the values of dice three and four swap (as they would represent the same pair). This means that we want to consider $24 / (2 * 2) = 6$ ways to roll.
Our product is then $6 * 4 * 2 * 6 = 288$ outcomes.
5b. We have our $6$ seven-pairs, $4$ remaining unique dice values for die number three, and just $1$ value for die number four to make a seven-pair with die number three. We want to avoid double counting the same permutations as in 5a, but with the additional restriction of not wanting to double count the pairs of seven-pairs whether they are made of the set of dice 1 and 2 or the set of dice 3 and 4. (In 5a, this was unnecessary, as we knew that the dice 3 and 4 could never be the same pair as the dice ever appearing in 1 and 2, thus giving us unique permutations.) This is solved by again halving the number of ways to roll from $6$ to $3$.
Our product is then $6 * 4 * 3 = 72$ outcomes.
Finally, we add up our outcomes for each category and arrive at our total: $$24 + 18 + 288 + 144 + 288 + 72 = 834$$
And for the final probability we divide our desired outcomes by the number of possible combinations and arrive at: $$\frac{834}{1296} \approx 64.35\%$$