Dice probability problem collection I tag these questions as homework, as they are older exam questions and every year
Can you try to solve/explain how to solve with a method some of these?
If something is answered in an old post please post the link.

1. We toss 2 dice. Using probability-generating functions compute the probability that their sum is 4.

2. We roll two different dice 10 times. What's the probability the first die to show a number greater than the number of the second die for 3 three times*?
*I think they mean that this happens three times (and they don't care about the precedence of those 3 times out of 10)

3. Consider a random experiment in which we roll consecutively two dice until "we get a sum of 10 for the first time". We denote the random variable $X$ which countsthe number of repetitions until the end of the experiment. Compute the probability-generating function of $X$.

To community: I'm familiar with many terms of probability theory so with a good analysis I think I might understand your answers.
Reminder:Their diffictulty may seem easy to medium but not everyone is great at this.
 A: 1:
The probability generating function for the total of a single die is
$$
G(z) = \frac16 (z+z^2+\cdots+z^6)
$$
The probability generating function of a sum is just the product of the two probability generating functions. So, the probability generating function for the sum of two dice will be
$$
G_2(z)=G(z)\cdot G(z)=\left[G(z)\right]^2 =
\left[\frac16 (z+z^2+\cdots+z^6)\right]^2
$$
The answer to your question will be the coefficient of $z^4$.
2:
If we roll two fair dice, there are three possibilities: the two results are equal, the first is greater than the second, or the second is greater than the first.  The sum of the probabilities of each of these events is $1$, and the probabilities of the second two events are equal. The probability that two dice are equal is $\frac16$, which means that there's a $\frac56$ chance that one of the other two occur, which means that the latter two events each have a $\frac5{12}$ chance of occurring.
The probability that the first one is greater than the other is then simply $p=\frac5{12}$. The probability that this occurs $3$ times out of $10$ trials total is
$$
\binom{10}{3}\left(p\right)^{3}\left(1-p\right)^{7}
$$
3:
First of all, we need the probability that two dice sum to $10$. This, as it ends up, is $p=\frac{3}{36}=\frac{1}{12}$.  The probability of getting a sum of $10$ on the first try is then
$$
p = \frac{1}{12}
$$
The probability of getting the first sum of $10$ on the second try is the probability of not getting $10$ the first try times the probability of getting $10$ on the second, which is
$$
(1-p)p = \frac{1}{12}\times\frac{11}{12}=\frac{11}{12^2}
$$
The probability of getting the first sum of $10$ on the $n^{th}$ try (extending the above logic) is
$$
(1-p)^{n-1}p=\frac1{12}\times\left(\frac{11}{12}\right)^{n-1}
$$
So, our probability generating function is simply
$$
\begin{align}
G(z) &= \sum_{n=1}^\infty P(X=n)z^n \\&=
\sum_{n=1}^\infty \frac1{12}\cdot\left(\frac{11}{12}\right)^{n-1} z^n \\&=
\frac{z}{12} \sum_{n=1}^\infty \left(\frac{11z}{12}\right)^{n-1}
\end{align}
$$
This expression can be simplified by noting that this is a geometric series.
