throwing a two-dice 7 with in total 6 dice We have been stuck on the following questions regarding the game Qwixx:
What is the probability that there is a combination of two dice which add up to exactly 7 when throwing with a total of 6 dice once?
Thank you in advance!

For $k=1,2,3$ let's say that we are in status $S_k$ if exactly $k$ distinct faces have shown up and no pair exists that adds up to $7$.
Now throw a die.
It does not matter what outcome it has and we have landed in status $S_1$.
From here $5$ steps are made giving a total of $6^5=7776$ possible routes. 
Note that: 


*

*there is $1$ possible step $S_1\to S_1$ and $4$ possible steps $S_1\to S_2$.

*there are $2$ possible steps $S_2\to S_2$ and $2$ possible steps $S_2\to S_3$ 

*there are $3$ possible steps $S_3\to S_3$.


This results in the following scheme:
$$\begin{array}{ccccc}
S1\\
1\uparrow\\
S1 & \stackrel{4}{\to} & S2\\
1\uparrow &  & 2\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow &  & 2\uparrow &  & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow &  & 2\uparrow &  & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3\\
1\uparrow &  & 2\uparrow &  & 3\uparrow\\
S1 & \stackrel{4}{\to} & S2 & \stackrel{2}{\to} & S3
\end{array}$$
That gives:


*

*$1^5=1$ route that ends in $S_1$

*$4\times[1^42^0+1^32^1+1^22^2+1^12^3+1^02^4]=124$ routes that end in $S_2$

*$4\times2\times[1^3+1^22^1+1^23^1+1^12^2+1^12^13^1+1^13^2+2^3+2^23^1+2^13^2+3^3]=720$ routes ending in $S_3$


That gives probability $\frac{845}{7776}$ that we end up in $S_1$, $S_2$ or $S_3$ hence probability $$\frac{6931}{7776}$$ that we do not end in $S_1$, $S_2$ or $S_3$.
