Four tetrahedron dice probability We have four dice in the shape of a tetrahedron. Each dice has faces numbered 2, 0, 1, and 7. If we roll all four dice simultaneously, what is the probability that we can compose the number 2017 using one of the three visible numbers from each dice?
Here is what I tried:
$\frac{3}{4}$x$\frac{3}{4}$x$\frac{3}{4}$x$\frac{3}{4}$=$\frac{81}{256}$
The real answer is $\frac{63}{64}$.
Please help me with this answer!
 A: As mentioned in above answers, $2017$ is impossible if all four bottom numbers are the same. There are $4$ such outcomes:
$$2222; \ 1111; \ 0000; \ 7777.$$
There are $4^4$ outcomes in total (in sample space). Hence the probability of getting $2017$ is:
$$P(2017)=1-P(\text{not} \ 2017)=1-\frac{4}{4^4}=1-\frac{1}{4^3}=\frac{63}{64}.$$
A: As LKM says, unless all four dice have the same number at the bottom, you can always compose 2017.
OK, but why exactly is that true? That is not immediately obvious, so here is a proof:
If all four dice have a different number at the bottom, then to compose 2017 we can do: pick the $2$ from the die that has the $1$ at the bottom, pick the $0$ from the one that has the $7$ at the bottom, the $1$ from the one with the $0$ at the bttom, and the $7$ from the one with a $2$ at the bttom. Schematically, we can write this as:
$1 \rightarrow 2$
$7 \rightarrow 0$
$0 \rightarrow 1$
$2 \rightarrow 7$
OK, that was easy, but what if you have two or three dice with the same number at the bottom? Can we still do it?
Well, suppose we have two $0$'S, a $1$, and a $2$ at the bottom. Then we can do:
$0 \rightarrow 1$
$0 \rightarrow 2$
$1 \rightarrow 0$
$2 \rightarrow 7$
This, of course, generalizes to any case where you have three different numbers at the bottom. That is, if the numbers/faces at the bottom are two $A$'s, a $B$, and a $C$, then we can do:
$A \rightarrow B$
$A \rightarrow C$
$B \rightarrow D$
$C \rightarrow A$
OK, what if we have two $A$'s and two $B$'s at the bottom? then we can do:
$A \rightarrow B$
$A \rightarrow C$
$B \rightarrow D$
$B \rightarrow A$
finally, with three $A$'s and one $B$ we can do:
$A \rightarrow B$
$A \rightarrow C$
$A \rightarrow D$
$B \rightarrow A$
A: Here's a more "high-tech" solution showing that "all dice hide the same number" is the only bad case, and therefore $1 - 4 \cdot (\frac14)^4 = \frac{63}{64}$ is the right answer.
We can define a bipartite graph with the dice on one side, their faces on the other, and an edge from each die to each of its visible faces, as shown below:

Picking a different number from each die is equivalent to finding a perfect matching in this graph. To see if this is possible, we need to check Hall's condition.
Here, it says that for each subset of dice, there must be at least as many visible numbers on them as the number of dice in the subset. This is automatically true if we pick a subset of $3$ or fewer dice, since even one of the dice will show $3$ faces. So the only case to worry about is the set of all $4$ dice: we need them to show all $4$ faces, which happens unless all the dice are showing the same three faces.
This generalizes to the case of $n$ dice with $n$ sides, where we need to pick a face from each die so that all faces chosen are different. Again, having all dice hide the same number is the only case where Hall's condition fails.
A: Out of the visible parts of the dice, 3 parts are visible, the worst case is that all four dice show same visible part, otherwise, $2017$ can be composed.
So let's say there are four numbers the first dice not showing, the remaining three must not be showing the same number, just as the first dice do, which its probability is $[ \frac{1}{4} ]^3$, so the answer is $1-[ \frac{1}{4} ]^3 = \frac{63}{64}$
