The probability of being paired with someone? This question was presented to me last summer. It goes as follows:
There are 8 people in a bus. All of them will be attending a camp, and will all be placed into rooms with one other person. There are 50 people total who will be attending this camp. What's the probability of at least one person from the bus to be paired with another person on the bus? How many people on the bus can expect to be roommates? Will it on average be no pairs, one pair, two pairs, etc.?
I honestly have no idea how to approach this problem. I have limited experience in statistics -- however I'm very curious in how the math works for a question like this! Am I overthinking it or is it actually a complex problem?
 A: We solve a small part of the problem, the expected number of pairs from the bus. Call the people on the bus $P_1$ to $P_8$. For any $i$ from $1$ to $8$, define random variable $X_i$ by $X_i=1$ if $P_i$ is paired with someone from the bus, and by $X_i=0$ otherwise. Then the number $Y$ of paired people from the bus is given by $Y=\frac{X_1+\cdots+X_8}{2}$. 
By the linearity of expectation we have $E(Y)=\frac{1}{2}\left(E(X_1)+\cdots+E(X_8)\right)$.
We have $E(X_i)=\Pr(X_i=1)=\frac{7}{49}$, and therefore $E(Y)=\frac{28}{49}$.
A: Hint 1: For the first question, it is easier to compute the probability of the opposite happening: what is the probability that each of the people on the bus are in separate rooms?
Hint 2: for problems like this, the usual approach is to enumerate all outcomes. To find the probability of some event occurring, you take the number of outcomes that satisfy the event and divide by the total number of outcomes. There are multiple ways you could enumerate outcomes (do you consider the rooms to be different or indistinguishable?) but as long as you are consistent in your counting, you should get the right answer.

One possible enumeration is the following: each of the 50 people are assigned to 50 labeled beds, and the first room has beds $1$ and $2$, the second room has beds $3$ and $4$, and so on. Counting in this manner, there are $50!$ possible bed assignments.
From the first hint, we should count the number of assignments where the $8$ people on the bus are in separate rooms. If we assign these eight people first, there are $25!/17! \cdot 2^8$ ways to give them beds in separate rooms (there are $25 \cdot 24 \cdots 18 \cdot 17$ ways to put them in separate rooms, and then each person can choose one of the two beds in each room). Then, there are $42!$ ways to assign the remaining $42$ people. Thus dividing this by $50!$, we see that the probability of the eight people being in separate rooms is.
$$\frac{25!/17! \cdot 2^8 \cdot 42!}{50!} = 2^8 \frac{25 \cdot 24 \cdots 18 \cdot 17}{50 \cdot 49 \cdots 43 \cdot 42} = \frac{50 \cdot 48 \cdots 36 \cdot 34}{50 \cdot 49 \cdots 43 \cdot 42}.$$
Remark: the last expression suggests another way to arrive at the same answer. If we sequentially assign beds to the eight people in order, the first person can go to any $50$ of the $50$ beds, the second person can go to $48$ of the remaining $49$ beds, the third person can go to $46$ of the remaining $48$ beds, and so on. So the answer is $\frac{50}{50} \cdot \frac{48}{49} \cdots \frac{36}{43} \cdot \frac{34}{42}$.
To get the probability of at least one pairing on the bus, just subtract this probability from $1$.
A: 
There are 8 people in a bus. All of them will be attending a camp, and will all be placed into rooms with one other person. There are 50 people total who will be attending this camp.
What's the probability of at least one person from the bus to be paired with another person on the bus?

There are so many ways to arrange 50 people into pairs.   Count ways to arrange 50 people, divided by ways to arrange 25 pairs and people within each pair.
There are so many ways to arrange the 42 other people so 8 pair with the bus people and the remainder form 17 pairs.    Count ways to arrange the 42 people, divided by ways to arrange the later 34 into 17 pairs and people within each of these pairs.   (The earlier 8 are paired to bus people, already counted as part of the arrangement of 42 people.)
Dividing the second count by the first will give the probability that no people from the bus share a room.
The probability that some bus people share a room is then found by the Law of Complements.

How many people on the bus can expect to be roommates? Will it on average be no pairs, one pair, two pairs, etc.?

There are so many possible pairs.   The probability that any pair will share a room is so much.   The expected number of pairs sharing a room is found by using the Linearity of Expectation.
