Fix $X$ as a value between $1$ and $10.$ Consider a game of rolling three six-sided dice, where the highest valued die rolled is removed. If the sum of the resulting two dice is equal to or less than $X$, do nothing. If the sum of the resulting two dice is greater than $X$, reroll all three dice and apply the afore noted removal process again.
How would you calculate the probability of obtaining a value less than or equal to a chosen $X$ in this game, with the minimisation and reroll?
I wonder why you've excluded $X=11$ and $X=12$.
If the lower two dice have the same value $k$, and thus a sum of $2k$, there is $1$ result in which the highest die is also $k$ and $3(6-k)$ results in which the highest die is higher, for a total of $19-3k$.
If the lower two dice have different values $k\lt l$, and thus a sum of $k+l$, there are $3$ results in which the highest die is also $l$ and $6(6-l)$ results in which the highest die is higher, for a total of $39-6l$.
Thus, for odd $n\le 7$, the number of ways to get a sum of $n$ from the lower two dice is
\begin{eqnarray} \sum_{k=1}^{(n-1)/2}(39-6(n-k)) &=& \frac{(39-6n)(n-1)}2+6\cdot\frac12\cdot\frac{n-1}2\cdot\frac{n+1}2 \\ &=& \frac{2(-6n^2+45n-39)+3(n^2-1)}4\;, \\ &=& \frac94(9-n)(n-1) \end{eqnarray}
whereas for $n=9$ it's $39-6\cdot6+39-6\cdot5=12$ and for $n=11$ it's $39-6\cdot6=3$.
For even $n\le6$, the number of ways to get a sum of $n$ from the lower two dice is
\begin{eqnarray} 19-3\cdot\frac n2+\sum_{k=1}^{n/2-1}(39-6(n-k)) &=& 19-3\cdot\frac n2+(39-6n)\left(\frac n2-1\right)+6\cdot\frac12\cdot\left(\frac n2-1\right)\cdot\frac n2 \\ &=& \frac{4\cdot19-6n+(39-6n)(2n-4)+3(n-2)n}4\;, \\ &=& \frac{-9n^2+90n-80}4\;, \end{eqnarray}
whereas for even $n\ge8$ the number of ways to get a sum of $n$ from the lower two dice is
\begin{eqnarray} 19-3\cdot\frac n2+\sum_{k=n-6}^{n/2-1}(39-6(n-k)) &=& \frac{-9n^2+90n-80}4-\sum_{k=1}^{n-7}(39-6(n-k)) \\ &=& \frac{-9n^2+90n-80}4-(39-6n)(n-7)-6\cdot\frac12(n-7)(n-6) \\ &=& \frac{-9n^2+90n-80-4(39-6n)(n-7)-12(n-7)(n-6)}4 \\ &=&\frac{3n^2-78n+508}4 \end{eqnarray}
To summarize, we have the following numbers $a_n$ of ways of getting a sum of $n$ in the lower two dice:
\begin{array}{c|c} n&a_n\\\hline 2&16\\ 3&27\\ 4&34\\ 5&36\\ 6&34\\ 7&27\\ 8&19\\ 9&12\\ 10&7\\ 11&3\\ 12&1 \end{array}
The sum is $6^3=216$, as it should be. Now add up the numbers $a_n$ for $n\le X$ and divide by $216$ to obtain the probabilities $p_X$ for obtaining a sum less than or equal to $X$ with the lower two dice:
\begin{array}{c|c} X&p_X\\\hline 2&\frac{16}{216}\\ 3&\frac{43}{216}\\ 4&\frac{77}{216}\\ 5&\frac{113}{216}\\ 6&\frac{147}{216}\\ 7&\frac{174}{216}\\ 8&\frac{193}{216}\\ 9&\frac{205}{216}\\ 10&\frac{212}{216}\\ 11&\frac{215}{216}\\ 12&\frac{216}{216} \end{array}
As noted by @kccu in a comment, you effectively have two tries to get $X$ or less, so the probability to succeed is $1-(1-p_X)^2=2p_X-p_X^2$.