Three people throwing. Two die, one after the other Three players are throwing two dice. One is marked 1 ,2,3,4,5,6 and the other is marked 11,12,13,14,15,16. They take turns throwing the dice one after the other and mark the score.
At the end of one round the person who has the highest aggregated score wins.
Following are the questions
a) What is the sample space 
b) What is the probability that one of them would get a 13 
c) what is the possibility of one of them not scoring a 9 or scoring 16 
d) What is the probability of one of them scoring a 14 or scoring 20 
I am new to probability so I am not sure how to go about doing this problem. Any help is greatly appreciated.
 A: The thing you need to answer is a). Once you have that, everything else is simply counting. To see that (I hope you saw it recently in class) let $\Omega$ be the sample space and $A \subseteq \Omega$ the set of outcomes you are interested in (e.g. for b) all outcomes such that the sum is $13$). Then the probability of this set is 
$$\frac{|A|}{|\Omega |}$$
and thus you only need to count the size of $A$ and $\Omega$.
For the sample space you need to find a way to write down every possible outcome.
If you have no idea for that (and you didn't do anything helpful in class at all...), try to think about how the players would write down their scores: They through a dice, they write something down, they throw the second one. In which way would you take these notes, to later be able to read off the results and e.g. say "well, I scored neither a 9 nor a 16 this round". If you find a nice way for that, this will most likely give you your sample space.
For a little example consider another experiment: A player is throwing two coins, one after the other. Both coins have a head side H and a tail side T. Then the sample space (or rather a possible sample space) would be $\{(H,H),(H,T),(T,H),(T,T)\}$, as this are all possible outcomes.
A: (a) If rolling of dice is independent, sample space is simply $A$ x $ B$ where $A$ is the set of outcomes of first dice and $B $ of the second. So, S={(1,11), (1, 12), (1, 13),..., (2, 11), (2, 12),..., (6, 15), (6,16)}. There will be 6 x 6 = 36 elements.
(b) Probability that one of them will get a 13: ${3\over 18} $ =${1\over 6} $
(c) Probability of neither 9 nor 16 of one of them is ${36 - 11\over 36} $ = ${25\over 36} $
(d) Probability of 14 or 20 of one of them is ${11\over 36} $
Here, the meaning of one of them is a little ambiguous. Just confirm whether I have understood right.
