Combinatorics with colored beans

Question

I have some difficulties with the following exercise in combinations:

There are $8$ beans in the box: $6$ white beans, $2$ green beans. Two players one by one pick $2$ beans; first player one picks $2$ beans, after that player two picks $2$ beans. For every green bean that player picks he gets $5$ points.

What's the expected number of points for player one? What's the probability that player two picks only one green bean?

Solution:

$$E(\text{points of player one}) = 4 \cdot 5 \cdot \frac{2}{8} \cdot \frac{6}{7} + 10 \cdot 2 \cdot \frac{2}{8} \cdot \frac{1}{7}$$

Unfortunately I didn't find any good way to fir the binomial distribution here.

I don't get any idea how to answer the second question.

Answer 1

For question 1 you have to take the sum of the value of each outcome multiplied by the probability of the outcome. This is how I would do it:

2 beans: $\underbrace{\frac{1}{8\cdot7}}_\text{probability}\cdot\underbrace{10}_{\text{value of outcome}}$

1 bean: $\frac{7}{8\cdot7}\cdot5$

0 beans: value is 0 so probabilty won't matter.

Add them all up to get $\frac{15}{14}\approx 1.071$

For the second question break in to cases:

case 1: player 1 took 1 green bean. Probability:$\frac{2\cdot6\cdot5}{8\cdot7\cdot6}$

assuming this happens the probability player 2 gets the remaining green is $\frac{6\cdot5}{6\cdot5\cdot4}$ So the probability of both happening is $\frac{2\cdot6\cdot5}{8\cdot7\cdot6}\cdot\frac{6\cdot5}{6\cdot5\cdot4}=\frac{5}{112}\approx 0.044$

case 2: player 1 took no green balls: probability $\frac{6\cdot5\cdot4}{8\cdot7\cdot6}$

assuming this happens the probability player 2 takes exactly 1 green ball is $\frac{2\cdot6\cdot5}{6\cdot5\cdot4}$. So the probability both happen is $\frac{6\cdot5\cdot4}{8\cdot7\cdot6}\cdot\frac{2\cdot6\cdot5}{6\cdot5\cdot4}=\frac{5}{28}\approx.178$

So the total probvability is the sum of the probabilities of the two cases which is $\frac{5}{28}+\frac{5}{112}\approx 0.2232$