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:

  1. What is the probability that there are 0 persons left from group1?
  2. What is the probability that there is 1 person left from group1?
  3. 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

  • $\begingroup$ "We start with number 4, first number 7 will be removed. Then number 10 will be removed. Next will be number 13..." Who is removed after number 28? $\endgroup$ – Mike Pierce May 4 '15 at 20:03
  • $\begingroup$ Thank for the reply. Number 1 will be removed. $\endgroup$ – Danique May 4 '15 at 20:06

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 $1- 5/30 - 5/30= 20 / 30$.

  • $\begingroup$ I think your right. Thanks for the clear explanation! $\endgroup$ – Danique May 5 '15 at 15:41

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