# Probability on a die facing two directions

Say, I have a die. When we roll a die we know that one side will face up and four sides will face north, west, east and south (assuming the die is always in the right position).

We also know that the sum of the sides is always 7. For example if one side is 6 then the opposite side will be 1. That means the probability of sum of east + west or north + south is 7 is always 1.

My questions are:

1. How to compute the probability of getting 4 on the north given that six is on top?
2. What if there are two dice, what is the probability of getting 3 facing West in one die and getting 4 facing East in the other die given one of the die has 6 facing the top?

So far, my approach is:

The sample size is, the possible number on the top x the remaining possible number facing a direction x the matched number x the remaining possible number x the matched number.

For example:

There are 6 possible numbers on top, so 6. There are 5 possible number facing North, so 5. There are only one possible number for South, so 1. There are 3 possible number facing West, so 4. There are one possible number facing East, so 1.

S = 6 x 5 x 1 x 3 x 1 x 24 (because there are 24 combinations of North, South, West and East).

Am I in the right direction?

Thank you.

1. Each direction is equally likely: (North) \ (North + West + East + South) = $\frac 14$
2. If $3$ is West $4$ is necessarily East: P($3$ West | $6$ Top)P($3$ West)= $\frac1 4 \frac1 6 = \frac1{24}$
The probability of having a $3$ on West is $\frac 16$ because each number is as likely to land on West. But given that the dice landed such that $6$ is on top only $4$ equally likely numbers are competing for the West spot: $(2,3,4,5)$ therefore the probability is $\frac 14$.
There are many fewer combinations than that. The numbers on the faces of a die don’t change, so once you’ve chosen a number to be on top and one for the north face, the other four faces are also determined. Also, once you’ve chosen $k$ for the top face, there are only four choices for north since $7-k$ must be on the bottom.