Understanding of the probability I have a problem with understanding some of my statistics homework. I hope that some of you could help me understand. 
In summary the question is as follows:
There are 30 people in a group, which are split up in 2 groups of 15. The first group we call group1 and the other group2.  Group1 has been placed in an order from 1 till 15 and group 2 is ordered 16 till 30.  In the question we need to find two persons. We do this by removing each time the 3rd person till we only have 2 people left. 
So for example:
1 2 3 4 5 6 7 8 9 10 11.. 30; We start with number 4, first number 7 will be removed. Then number 10 will be removed. Next will be number 13. Etc, And after number 28 number 1 will be removed. till there are only 2 people left.
The start position is random, so it could be 1 till 30. 
I need to answer the following questions:


*

*What is the probability that there are 0 persons left from group1?  

*What is the probability that there is 1 person left from group1?  

*What is the probability that the two persons, who are still there are both from group1?


Someone told me the answers, but I do not understand them. So I hope that someone could explain it to me. I got the following answers:
1) 4/30
2) 22/30
3) 4/30
 A: First, we look at an example. If we start at number $1$, we'll remove elements from the sequence
$$
(1, 2, 3, \dots, 19, 20, 21, \dots, 28, 29, 30)
$$ 
until we end up with persons $20$ and $30$ (please correct me if I'm wrong).
When we start with person $2$, we could use the same sequence and redo the calculation. Instead, we modify the order as follows:
$$
(2,3,4, \dots, 20, \underset{\underset{20\text{th}}{\uparrow}}{21}, 22, \dots, 29, 30, \underset{\underset{30\text{th}}{\uparrow}}{1})
$$
After removing elements from this sequence, we observe that we have discarded the same positions as in the last example. In particular, the persons on the $20$th and $30$th place would remain (numbers $21$ and $1$). Especially, we conclude that the two positions have a constant distance of $10$.
If we start with number $n \in \{1, \dots, 30 \}$, we obtain the persons $p_1, p_2 \in \{1, \dots, 30\}$ from the $20$th and $30$th position, respectively. 
$$
(n, \dots, \underset{\underset{20\text{th}}{\uparrow}}{p_1}, \dots, \underset{\underset{30\text{th}}{\uparrow}}{p_2})
$$
There are $15$ different choices for $n$, s.t. $p_1 \leq 15$. Then, we have $p_2 = p_1 + 10$ (because of the distance of the indices). Consequently, there are only $5$ possibilities for our starting number $n$, s.t. both persons are in group #$1$. It follows, that the answer to question 3 is $5/30$.
Because of symmetry, the answer to question 1 is $5/30$, too. The remaining probability (question 2) has to be $1- 5/30 - 5/30= 20 / 30$.
