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!
 A: Note:  what follows has several cases each of which is somewhat error prone.  The calculation should be checked carefully.
The pairs that add to $7$ are $(1,6), (2,5), (3,4)$.  We'll say that a pair is "hit" if at least one member of the pair comes up in your six tosses.  We'll work from the complement...that is we will consider the probability that no two values sum to $7$ and then subtract from $1$.
Case I: only one pair is hit.  In this case only one value can be hit, so the probability is $\frac 1{6^6}$.  As there are $6$ possible values, the probability in this case is $$\frac 1{6^5}$$
Case II:  exactly two pairs are hit.  There are $3$ ways to choose the two pairs.  Then $4$ ways to choose the values in those two pairs which are hit.  Given those choices, we need every toss to be one of those two values (but not all the same), the probability of which is $\frac 1{3^6}-2\times \frac 1{6^6}=\frac {31}{23328}$  Combining we see that the probability is $$3\times 4 \times \frac {31}{23328}=\frac {84}{15552}=\frac {31}{1944}$$
Case III: all three pairs are hit.  Then there are $8$ ways to choose the three values that occur.  Given those choices we need the probability that all the values we get are contained in those three, and that we hit each of those three at least once.  The probability of that is $\frac 1{2^6}-3\times \frac 1{3^6}+3\times \frac 1{6^6}=\frac {5}{432}$.  Combining we get $$8\times \frac {5}{432}=\frac {5}{54}$$
It follows that the desired result is $$1-\frac 1{6^5}-\frac {31}{1944}-\frac {5}{54}=\boxed {\frac {6931}{7776}}=0.8913323\cdots$$
A: 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 $S_3$.
