Expected value of the number of teams

Question

Suppose there are 32 people, and we want to create a championship consisted of 16 teams (i.e., 2 people in each team). A system will pick these teams at random.

Further, suppose 4 of these 32 people are sick, and any team containing any sick person can't play in the championship.

Question is what is the expected number of teams remaining in the championship?

My Approach

I think we only have 3 possible outcomes here: 14 teams, 13 teams, and 12 teams based on how these sick people get distributed in the teams.

So the expected value would be $$ (P(14)*14) + (P(13)*13) + (P(12)*12) $$ where P(n) is the probability that n teams can participate in the championship.

My Question

I'm not sure if my intuition is correct or not. Also, I got stuck with finding P(n)

Answer 1

Let $$X_i = \begin{cases} 1 &\text{if team i includes at least one sick person} \\ 0 &\text{otherwise} \end{cases}$$ for $i = 1,2,3 \dots ,16$. Then $$P(X_i = 0) = \frac{28}{32} \cdot \frac{27}{31}$$ so $$E(X_i) = P(X_i = 1) = 1 - \frac{28}{32} \cdot \frac{27}{31}$$ for $i = 1,2,3, \dots ,16$. Therefore the expected number of teams which contain at least one sick person is $$E \left( \sum_{i=1}^{16} X_i \right) = \sum_{i=1}^{16} E(X_i) = 16 \left( 1 - \frac{28}{32} \cdot \frac{27}{31} \right) = 3.806$$

So the expected number of teams with no sick persons is $$16 - 3.806 = \boxed{12.193}$$

Answer 2

Consider that there are $4$ sick people. Let's consider them "identical" for the purpose of this explanation. Let's distribute them into positions $a_1,b_1,a_2,b_2,...,a_{16},b_{16}$. whare the team $t_1$ consists of $a_1$ and $b_1$, the team $t_2$ consists of $a_2$ and $b_2$, etc.

There are $\binom{32}4=35960$ ways to distribute $4$ people into $32$ positions.

To find the number of ways to get $14$ teams participating, we choose $2$ teams (out of $16$) and put the sick people in both positions of each. There are $\binom{16}2=120$ possibilities.

To find the number of ways to get $13$ teams participating, we choose $3$ teams (out of $16$), then pick $1$ of those $3$ teams to contain two sick people, then for each of the $2$ remaining teams pick a position (out of $2$) in the team to contain a sick person. There are $\binom{16}3\cdot\binom31\cdot2^2=6720$ possibilities.

To find the number of ways to get $12$ teams participating, we choose $4$ teams (out of $16$), then pick a position (out of $2$) in each of those $4$ teams. There are $\binom{16}4\cdot2^4=29120$ possibilities.

A quick inspection verifies that the number of possibilities of the cases ($120,6720,29120$) add up to $35960$.

Now let's get the probabilities:

$$P(14)=\frac{120}{35960}\\ P(13)=\frac{6720}{35960}\\ P(12)=\frac{29120}{35960}$$

Substituting, the expected number of teams participating is

$$\begin{align} &14\cdot\frac{120}{35960}+13\cdot\frac{6720}{35960}+12\cdot\frac{29120}{35960}\\ =&\frac{1680+87360+349440}{35960}\\ =&\frac{438480}{35960}\\ =&\frac{378}{31}\\ =&12\frac6{31} \end{align}$$

Hey, the sick people are different!

Consider that each possibility mentioned above can be turned to $4!=24$ "the sick people are different" (TSPAD) possibilities. It doesn't have any effect on the result. To see that, try multiplying each the nominators and denominators in the probability equations by $24$. It doesn't change the probability.