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?
 A: Assuming equal size of each box (more precisely equal probability of ending up in each of the boxes) the solution is 1/16. Your second take is correct. Alternatively take your first take (which gives the solution for a specific box out of the four boxes) and multiply by four.
A: 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.
A: 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.
