For a particular die roll the cumulative probability is $ P(X_i \leq x ) = x/6 $, for $x=1,...,6$. So, if the die rolls are independent,
$$ P(\max \{ X_1, ..., X_n \} \leq m) = P(X_1 \leq m, ..., X_n \leq m) = \prod_{i=1}^{n} P(X_i \leq m ) = \left( \frac{m}{6} \right)^n $$
for $m=1,...,6$. When $m > 6$ this probability is clearly $1$ and $0$ if $m < 1$.
From this it's simple to deduce that $$P(\max \{ X_1, ..., X_n \} = m) = \frac{m^n - (m-1)^n}{6^n} $$
(I've suppressed the indicator that $m \in \{1,...,6\}$). Note that to generalize this to a $k$-sided die, just replace $6$ everywhere with $k$.
Suppose players $A$ and $B$ throw the die $n_A$,$n_B$ times with maximum rolls $M_A, M_B$, respectively. By the description above, player $A$ wins if $M_A > M_B$. Using the law of total probability,
\begin{align*}
P(M_A > M_B) &= E_{m} \Big( P(M_A > M_B | M_B = m) \Big) \\
&= E_{m} \Big(1 - \Big( \frac{m}{6} \Big)^{n_A} \Big) \\
&= \frac{1}{6^{n_A + n_B}} \sum_{m=1}^{6} \left(6^{n_A} - m^{n_A} \right)
\left(m^{n_B} - (m-1)^{n_B} \right)
\end{align*}
If $n_A = n_B = n$, this simplifies to
$$ \frac{1}{6^{2n}}\sum_{m=1}^{6} \left(6^n - m^n \right) \left(m^n - (m-1)^n \right) $$
Below this is plotted as a function of $n$. In this example, it's intuitive that the probability of $A$ winning quickly goes to zero as $n$ increases since their maximums become increasingly likely to both be six, in which case $B$ wins.
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I'll outline some things but avoid explicit solution. The point is for you to try to work it out.
At each step, you will have some amount of points, $M$. If you stand for the next roll you will either win $k$ more points (giving you $M+k$) or you will lose $M$ (leaving you with $0$).
The expected value-maximizing strategy would say that any time a $5/6$ chance of $k$ is worth more than a $1/6$ chance of $M$ -- i.e. when the expected gain from playing another round would be positive -- you benefit (on average) from continuing. A little mental arithmetic lets you see when to sit down for a variety of similar games of this type (as long as its not a game that becomes much, much more profitable later, there's no need to sum series, you can just stay standing until the net gain on the next roll is no longer positive).
Imagine, for example, you played the first round and got $100 $points. Now in the second round you would get $200$ points if you win ($5/6$ chance) and lose $100$ points if you got a $5$ ($1/6$ chance). The expected winnings is $200\times 5/6$ and the expected loss is $100 \times 1/6$, for an expected return of $900/6 = 150$; on average you stand to win much more than lose.
However, if you are trying to be the person with the most points of all at the end of many such games of Greed, maximizing expected value each turn is not the optimum strategy to be the eventual winner!
You may be better taking a smaller chance of a larger return, or banking smaller, surer gains, depending on where you are placed relative to other players.
Imagine you're coming second but the person coming first is a fair way ahead of you. If you sit down at the point that would maximize your expected return for that single round, you may simply make it certain you don't win. Similarly if you're far ahead in the last round, you will be better off sitting down much earlier than would maximize your expected return that round -- you may be unnecessarily risking a near-certain win if you continue.
If you were to be playing many more rounds you would (approximately) seek to maximize expected value per turn, but as you get toward the last few rounds the winning strategy changes: if you're behind, you should take more risk, if you're ahead, less risk. When you're getting close to the end, it's not just the current round of points that matter if you want to maximize the chance of being ahead at the end.
You can work out the exact strategy mathematically, but my guess is that the teacher is trying to get you to undertake the simpler task to maximize your expected winnings (which works quite well in the early to mid stages of the iterated game).
You might like to look at articles on strategy for the dice game Pig (of which this is a variant). There's lots of information on strategy to be found for various versions of this game online, but its more fun to work it out in detail yourself. Here is one document that also makes the point about the expectation-maximizing strategy not being optimal. Your game is a little different (your payoff increases and everyone gets the same rolls) but the basic ideas are the same.
Best Answer
$$P(Z=2) = P( (X=1) \cap (Y=1)) = P(X=1) \times P(Y=1) = \dfrac{1}{6} \times 0 = 0$$
$$P(Z=3) = P( (X=1) \cap (Y=2)) + P((X=2) \cap (Y=1)) = \dfrac{1}{6}\times\dfrac{1}{4} + \dfrac{1}{6}\times0 = \dfrac{1}{24}$$
and so on...
This would be $P(Z=11) + P(Z=7)$. You will need to compute them as I have done above. Note that you can make 7 and 11 in two ways though, which is why $P(Z=3)$ is the sum of two probabilities.
Given the answer to 2, what do you think?
This is more interesting. Suppose you roll $x$ and $P(Z=x)= p$. You can now win in a couple of ways.
a) You roll an $x$ on your second roll. If $Z_i$ is the $i^{th}$ roll, then the probability you win on your next roll is $p$ since rolls are independent.
However assume you roll neither an $x$ nor a $7$. We know the probability this happens (hint: its 1 - something). Let's call it $\theta$ for now. The probability you win on the third roll is $\theta \times p$. The probability of a win on the fourth roll is $\theta^2 \times p$, and so on. The probability of winning going forward involves these quantities. You should be able to write out the probability of winning and spot a geometric series, and so the sum (since probabilities are less than 1) can be written down analytically.