3-dimensional light up cube, # of rows/cols/diags in/on a 4 × 4 × 4 cube 
Imagine a 3-dimensional cube (much like a 4 × 4 × 4 Rubik's cube) except the planes of the cube cannot be twisted individually and instead of faces with different colors, it is clear (see thru) but with a single light in each of the $64$ little cubes contained within the $1$ big cube.
The $64$ lights are connected to $64$ switches, all of them initially turned off.  We now turn on a light by choosing a switch uniformly at random, toggling it (so that the light turns on if it was off, and turns off if it was on), and repeat this lighting process for a total of $32$ times.
At the end, we therefore have at most half of the $64$ lights lit — but the actual number is likely to be less, since some lights may be toggled several times.
I'd like to confirm the answers to the following five questions:
$1$. What will be the average number of lights on after all $32$ random numbers are processed sequentially, meaning that the corresponding lights were toggled properly from them, possibly multiple times each?
$2$. At the end of the lighting process, we want to know what is the probability that only $1$ complete row, column, or diagonal will be illuminated.  That is, $4$ adjacent/connected lights that complete one of those.  It may help if you think about and answer question 3 first.
$3$. How many possible complete unique/distinct rows, columns, and diagonals are there in/on the cube?  Remember they can go thru the cube, not just along the faces.  Also, they can be on the inner planes such as planes $2$ and $3$ (of $4$). They must be collinear.  You cannot go around a corner for example and get $4$ in a row.  All $4$ lights must lie in a line segment of length $4$ with length (in this context) meaning # of adjacent lights illuminated in a collinear fashion. 
$4$. What is the ideal number of lights on that will give us the highest probability of having exactly $1$ complete row/col/diag?
$5$. How many random lights (minimum) need to be on to guarantee that at least $2$ complete rows/cols/diags are illuminated?  That is, the probability of exactly $1$ complete row,col,diag becomes $0$.

I used computer simulation and got the following results so far:


*

*On average about $20.4$ lights will be on.  

*I get about $31.89$%.  The result is somewhat surprising to me because of all the possible ways to get $1$ complete row/col/diag, you would think out of $20$ or so lights (on average), you would get multiple of those more often.  My simulation program is even telling me that getting $8$ complete is even possible, but very unlikely.  I don't know what the max is, maybe $20$ or so if all $32$ lights stay lit and they just happen to be perfectly placed for maximum lit rows/cols/diags.

*I counted $76$ distinct complete rows, columns, and diagonals (so far) of length $4$.

*It appears that the ideal # of lights on is $23$, which gives about a $39.5$% chance of exactly $1$ complete row/col/diag. $22$ was a close 2nd with a $39.3$% chance. $24$ came in 3rd with a $38$% chance.

*I think the minimum is $47$.  The algorithm I used is very simple and very fast. Obviously you cannot have any $4$ in a row, col, or planar diagonal, so I simply started with $48$ (of $64$) lights on, making sure to turn off $1$ in each row, col, and planar diagonal (but not checking 3D diagonals).  My simulation program told me I had $3$ complete "lines" at that point.  I then checked to see if I could get only $1$ line lit by testing each of the $48$ on lights by turning them off one at a time then putting them back on before the next test.  That dropped it to $2$ lines illuminated.  I then looped thru combinations of $2$ on lights (there are $48 \choose 2$ of them which is only $1128$).  That dropped it to only $1$ line lit.  F.Y.I. my program told me there were $27$ two light combos that (when turned off) dropped it to only $1$ line lit.  So with that info, we just take the penultimate distinct # of lit lights we tried (which was $47$) and that is our answer.
The correct answer for Question $2$ depends on getting Question $3$ correct, so that is why I am asking for some help, because I want to get this exactly right.
For the diagonals, I counted $2$ per face, so that is $12$ facial diagonals (imagine an X shape on each face), $12$ inner planar diagonals which are Xs also but on planes $2$ and $3$, twisting the cube in all $3$ dimensions.  I also count $4$ inner "3D" diagonals that go thru the "heart" of the cube.
Can someone please help me confirm these numbers?
 A: Question 1
This is a simple Markov chain.
Let $o_i$ be the number of lights ON at step $i$, then
$$o_{i+1}=\begin{cases}
o_i-1,&p=\frac{o_i}{64}\\
o_i+1,&p=1-\frac{o_i}{64}\\
\end{cases}$$
And $o_0=0$
You can build the matrix for this (I'm not going to) and solve for 32 steps.
Question 2
For 4 specific lights in row, calculate the chance that each of them is "hit" an odd number of times and note how many of the 32 shots each combination uses up (hint: it can be $4, 6, \dots,32$).
Then for all other rows, calculate the chance that at least 1 of the bulbs is hit an even number of times (including 0) and allowing for the fact that some bulbs are common with the illuminated row.
Question 3
From each face there are 16 rows/columns. There are 3 independent faces. Therefore there are 48 rows/columns. 
Diagonals that are 4 long can only start from the corners. There are 2 in each 1x4 block. There are $4+4+4$ of these. 24.
Plus there are 4 diagonals that go from corner of the cube through the "guts" to the other corner.
$48+24+4=76$
