<s>[INST] $4$ dice sum probability What is the probability that, when $4$ dice are rolled, we can choose two of them such that the sum of the numbers on the upper faces is $7$?
I get $139 \over 216$ but I'm not sure I'm correct.
 [/INST] It is convenient to assume that the dice were rolled one at a time, and that we recorded the results as they came in. So there are $6^4$ equally likely outcomes. We count the favourables.
First we count the number of strings that have  $6$ and a $1$.  It is easier to count the strings that don't. There are $5^4$ strings that don't have a $1$, and $5^4$ that don't have a $6$, and $4^4$ that are missing both. So the number of strings that have a $6$ and a $1$ is $6^4-2\cdot 5^4+4^4$. Call this number $a$.
There are also $a$ strings with a $2$ and a $5$, and $a$ with a $3$ and a $4$. So our first estimate for the number of favourables is $3a$.
However, we have double-counted the strings that have, for example, both $1$ and $6$ and $2$ and $5$ (or the other $2$ possibilities). 
For both $1$ and $6$ and $2$ and $5$, we have $\binom{4}{2}\binom{2}{1}$. Multiply by $3$. We conclude that the number of favourables is $3a-3\binom{4}{2}\binom{2}{1}$.
We leave it to you to put the pieces together.
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