As a suggestion, look for extreme cases. For instance, $Z=2\implies (A,C,D,E)=(1,1,1,1)$ which would imply that $X=2$, so $(X,Z)$ are dependent. It also implies that $W=B+1<6$ so $(W,Z)$ are dependent. Continue in that spirit.

I think Z cannot equal 1 because it is forced to be (1+1)(1+1).

Yes, you are correct. That was a typo, now repaired. I meant $Z=2$.

Minimal value of Z has to be 4.

Right, sorry, I keep making the same mistake. I just wanted the minimum case $Z=(1+1)(1+1)$ which, as you say, is $4$. The numerical value isn't important (though it's idiotic that I can't seem to get it right)...the point is that knowing $Z$ is minimal determines all of $(A,C,D,E)$ uniquely.

No, problem I just wanted to make sure that Im understanding it correctly :D. I will post shortly my findings of those extrem cases here so I hope someone can check it.

Note that, while it can be easy to show dependence this way, if the variables are independent you can't prove anything just by looking at the extreme cases. Still, those cases might be very suggestive...if you have conditional independence in all extreme cases, maybe the variables really are independent.

For the hard case, $(Y,Z)$ I suggest proving that, for all values of $(A,C,D,E)$, we have $Y\in \{0,1,2,3\}$ with equal probability.

You are right, but if you wanna find independence you have to check all cases just checking Z with another letter there are at least $4^4$ options. Thats why I think there has to be some trick to this problem.

Since my claim covers all values of $(A,C,D,E)$, we certainly get all values of $Z$.

Write it out. Fix some value of $Z$, say, $Z=24$, and compute the conditional $P(Y=0\,|\,Z=24)$. Since every specification of $(A,C,D,E)$ that gives $Z=24$ gets you $Y=0$ with probability $\frac 14$, the conditional probability is $\frac 14$.

So I fixed P(Z = 64) and P(Y = 0) so I got P(Z = 64)*P(Y = 0) = $\frac{1}{4^5}$

Well, in the extreme case $Z=64$ we know we have $(A,C,D,E)=(4,4,4,4)$ so it's obvious that $Y$ is uniform in that case.

Well I think Z and Y are independent because Z is forcing only A and C and we can get any number for Y with same probability so I think it is independent.

they are, and what I said amounts to a proof, but I agree it needs to be written out. Sketch: For any fixed value of $Z$ there is some finite list of "good" $(A,C,D,E)$, let's say $k$ of them. Each of these has probability $p=\frac 1{4^4}$. Then the conditional probability that $Y=0$, conditioned on $Z$ having that fixed value is $\frac {p\times k\times .25}{p\times k}=.25$ as desired. But you should write it out more carefully.

yes I will, thank you, and I would like to ask you about expected value if I calculated it correctly check Edit.

Just did it quickly and I got $46.75$ but I might have made an error. Individually, I get $E[W]=6.5, E[Z]=25, E[X]=13.75$. $E[Y]$ might be a bit ambiguous. I assume $Y\in \{0,1,2,3\}$ uniformly, so $E[Y]=1.5$ .

Hmm I got all same as you but only $E[Y]$ I got 3.5

I see that we agree on $X,W,Z$. For $Y$, I say the answer is $1.5$ for the reason I gave. Again, I interpret $\pmod 4$ to mean the remainder on division by $4$. Hence the possible values of $Y$ are $\{0,1,2,3\}$.

may I see how you calculated it? I calculated it like $\cfrac{3(1+2+3+4)}{4}$ mod $4$

That doesn't make sense. $Y$ can only take the values $\{0,1,2,3\}$. It can't have expectation $>3$. My computation can be done mentally: $Y$ is uniform on $\{0,1,2,3\}$ so $E[Y]=\frac {0+1+2+3}4=\frac 64=1.5$

so is it $\cfrac{3(0+1+2+3)}{4}$ mod $4$ ?

But you are calculating only 1 letter and there 3 of them so it should be 1.5 * 3 mod 4 and its 0.5 but Im not sure if it is allowed something like that.

Again, $Y$ is uniform on $\{0,1,2,3\}$.

oh, yes I'm blind thank you for your help and attention. I think triples I will do on my own so you do not have to waste time :D.

@Mr.S You can compute covariances first. Pairs or triples whose covariance matrix are not $0$ can be ruled as dependent, while other can be investigated further.

Thanks Kakashi but my knowledge of covariance matrix are unfortunately 0.