# Probability of three dice falling in the same quadrant of a box

This is surely really basic for most people here but it's tripping me up.

You get a box and draw lines to split it up into 4 parts. I got asked what the probability was that when rolling three dice, all three dices would end up in the same quadrant.

My first take on this was

• a 1/4 chance of die 1 in quadrant x
• a 1/4 chance of die 2 in same quadrant x
• a 1/4 chance of die 3 in same quadrant x => 1/4*1/4*1/4 = 1/64 chance

My second take on this was that the first die doesn't matter at all so all that's left is

• a 1/4 chance of die 2 in same quadrant
• a 1/4 chance of die 3 in same quadrant => 1/4*1/4 = 1/16 chance

But I have been given a solution where all possible combinations are drawn out and as there are 20 possible combinations, the odds are 1/20.

What is correct (if any) and why?

• It's actually more complicated than that, because dice have non-zero size. The locations of the dice aren't independent if they physically cannot be closer together than the length of each edge. But since it's not possible to address this with the information given, I assume the problem deliberately ignores it for simplicity, and is intended to be equivalent to selecting numbers with replacement from the set $\{1,2,3,4\}$ representing the four quadrants. Apr 14, 2016 at 13:51
• It didn't even come to mind but for sake of simplicity, yes, you can ignore the fact that they cannot be closer together than the length of each edge. Apr 15, 2016 at 6:19

The possible combinations are not equi-probable. For instance it is more probable to have 3 dice in 3 known different quadrants than in a single one. You can not get the probability of a combination by taking the inverse of the number of combinations. So your result $\frac{1}{16}$ is correct.
The first take and the second take are the same. The point is , in take 1, what happens is you are inherently fixing the quadrant $$x$$ in which you want the dice to fall. In truth the dice could fall in any of the four quadrants, but they all have to fall in the same one. Thus, $$\frac{1}{64}*4 =\frac{1}{16}$$ is the right answer without doubt.
As for the third answer, you may tell the solution giver: It's quite simple. At the end of the roll, let $$x_i$$ be the number of dice present in quadrant $$i$$, $$i=1,2,3,4$$. In the end $$x_1+x_2+x_3+x_4=3$$, and each of these numbers $$x_i$$ is between $$0$$ and $$3$$. How many combinations of $$x_i$$ are possible?$$\binom{6}{3}=20$$. But the combinations are not equiprobable : in fact they are distributed multinomially.
• I upvoted this, then realized seven minutes later (too late to change the vote--oops!) that there are not $64$ different ways to assign $x_i$ from $\{0,1,2,3\}$ so that $x_1+x_2+x_3+x_4=3$. For example, $1+1+0+2$ is one of the $64$ ways to make a sum $x_1+x_2+x_3+x_4$ from that set, but the sum is $4$, not $3$. In fact there are exactly $20$ ways to make the sum equal to $3$, i.e. there are exactly $20$ combinations counted this way; but as noted already, they are not equiprobable. Apr 14, 2016 at 12:26