Probability of 4 drivers in the same group

Question

Yesterday I asked a question that a colleague quickly answered @mjw but some things were not clear and the question was closed. I will reformulate it and count on everyone's support again.

The problem is this: I have a go-kart group in which 32 drivers participate. In each stage, the drivers are divided into 2 groups (GroupA and GroupB). The names of the 32 drivers (D1, D2, ..., D32) are drawn one by one and distributed in the two groups according to the example below:

1º - I take a name randomly from a box and place it in the "A" group;

2º - I take a name randomly from a box and place it in the "B" group;

3º - I take a name randomly from a box and place it in the "A" group;

...

I will perform these steps until all the drivers are distributed in the groups as shown. Each group will have 16 drivers.

Question: What are the chances of drivers "D1", "D2", "D3", and "D4" being in the same group (GroupA or GroupB) considering that all drivers can be chosen equally?

I tried to think about solving this problem using the inclusion-exclusion theory, or conditional probability, but I failed. If I'm not wrong, the number of possible combinations is

$\binom{32}{16}$

But how can I consider the division into 2 groups?

Answer 1

As you say, there are $\binom{32}{16}$ equiprobable ways to divide the drivers into two groups of $16.$

The number of ways that $4$ specified drivers can be in group A is $\binom{28}{12}$, since the remaining $12$ drivers in the group can be any of the other $28$. Of course, there are the same number of ways that the $4$ can all end up in group B.

The probability is $$\frac{2\binom{28}{12}}{\binom{32}{16}}=\frac{91}{899}$$

Answer 2

Let $D_1$ be in any of the two groups. Then the probability that

• $D_2$ is in the same group as $D_1$ is $\displaystyle \frac{15}{31}$. This is because there are still $15$ spots in the group and $31$ remaining drivers to choose from.

Similarly,

• that $D_3$ is also in the same group is $\displaystyle \frac{14}{30}$.
• that $D_4$ is also in the same group is $\displaystyle \frac{13}{29}$.

So probability $P$ that $D_1, D_2, D_3, D_4$ are all in the same group is

$ = \displaystyle \frac{15}{31} \cdot \frac{14}{30} \cdot \frac{13}{29} = \frac{91}{899}$

Or you can also state it as $P = \displaystyle \frac{2 \cdot {28 \choose 12}} {32 \choose 16} = \frac{91}{899}$

(the numerator is number of ways of having $D_1, D_2, D_3, D_4$ drivers together in either group $A$ or group $B$ and rest $12$ drivers in the group chosen from remaining $28$ drivers).