World Cup Group prediction probability We have an office world cup bet where each person guesses the team that finishes 1st and 2nd from their qualifying group. E.g.


*

*A1: Brazil

*A2: Mexico

*B1: Netherlands etc...


You get a point for every correct guess. The person with the most points wins.
My question is, if you selected each guess at random what is the expected number of points you would get?
There are 8 groups, each with 4 teams; 2 guesses per team -> so the maximum points is 16. You can't guess the same team for 2 guesses. I.e. you couldn't guess Brazil to come 1st and 2nd in their group.
 A: For each group you have $24$ possible guesses, and the number of points that you can get is $0$-$2$:


*

*For guessing $1234$ you get $2$ points

*For guessing $1243$ you get $2$ points

*For guessing $1324$ you get $1$ point

*For guessing $1342$ you get $1$ point

*For guessing $1423$ you get $1$ point

*For guessing $1432$ you get $1$ point

*For guessing $2134$ you get $0$ points

*For guessing $2143$ you get $0$ points

*For guessing $2314$ you get $0$ points

*For guessing $2341$ you get $0$ points

*For guessing $2413$ you get $0$ points

*For guessing $2431$ you get $0$ points

*For guessing $3124$ you get $0$ points

*For guessing $3142$ you get $0$ points

*For guessing $3214$ you get $1$ point

*For guessing $3241$ you get $1$ point

*For guessing $3412$ you get $0$ points

*For guessing $3421$ you get $0$ points

*For guessing $4123$ you get $0$ points

*For guessing $4132$ you get $0$ points

*For guessing $4213$ you get $1$ point

*For guessing $4231$ you get $1$ point

*For guessing $4312$ you get $0$ points

*For guessing $4321$ you get $0$ points


So the expected (average) number of points that you will get for each group is:
$$0\cdot\frac{14}{24}+1\cdot\frac{8}{24}+2\cdot\frac{2}{24}=\frac{12}{24}=\frac{1}{2}$$
Given $8$ independent groups, the expected number of points that you will get is:
$$8\cdot\frac{1}{2}=4$$
A: There are $4! = 24$ ways to arrange the ordering of the groups. For any winner/runner-up pairing, there are two ways it may occur. This means that at random, you have $\frac{2}{24} = \frac{1}{12}$ probability of getting it right.
Another way of looking at it is there are 4 choices for the winner, and consequently 3 choices for the runner-up (or vice-versa). Therefore there is a $\frac14 \cdot \frac13 = \frac{1}{12}$  chance you will pick both right randomly.
But, we must also independently consider the cases where you get only one answer right.
In this case, suppose you pick the winner right. There are 6 such ways to do that, but two of them, as we have noticed, also involve picking the runner-up correctly. As such, there are 4 ways to pick the winner right but not the runner-up, and vice versa. So this has a $\frac{4}{24} = \frac{1}{6}$ chance of happening.
The expected value of a single group is computed using the expected value formula:
$$\begin{align*}E[G_i] &= \underbrace{\frac{1}{12} \cdot 2}_{\textrm{picking both correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only winner correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only runner-up correctly}} \\
&= \frac16 + \frac16 + \frac16 \\
&= \frac12 \end{align*}$$
Since expectation is linear,
$$E\left[ \sum_{i=i}^8 G_i\right] = \sum_{i=1}^8 E[G_i] = \sum_{i=1}^8 \frac12 = 8\cdot \frac12 = 4.$$

Suppose we made the pool more interesting, and we gave 2 points for picking the winner, and 1 for the runner up. Then we can use this approach with only trivial modification:
$$\begin{align*}E[G_i] &= \underbrace{\frac{1}{12} \cdot 3}_{\textrm{picking both correctly}} + \underbrace{\frac{1}{6} \cdot 2}_{\textrm{picking only winner correctly}} + \underbrace{\frac{1}{6} \cdot 1}_{\textrm{picking only runner-up correctly}} \\
&= \frac14 + \frac13 +\frac16 \\
&= \frac34\\
E\left[\sum G_i\right] &= 8\cdot \frac34 \\
&= 6.
\end{align*}$$
A: If the group competition is modeled by picking two balls without replacement from an urn with four balls, then each pick you make has 1/4 chance of a point. So you'd expect to earn 4 points.
A: Observe that you have a 1 in 4 chance of correctly guessing team A1, since there are 4 teams to choose from. Similarly, you have a 1 in 4 chance of correctly guessing A2. In other words, you expect your guesses for A1 and A2 to each contribute $1 \over 4$ points to your total points.
By linearity of expectation, your expected number of points is then ${1 \over 4} +  {1 \over 4} = {1 \over 2}$. For 8 groups in total, you can expect $8 \cdot {1 \over 2} = 4$ points.
You don't need to make the calculation more complicated than this. For instance, it isn't necessary to consider the cases where you get A1 right but A2 wrong, or both right, etc. This is precisely because expectation is linear and it doesn't matter whether the events in question are dependent or independent.
There is a well-known puzzle similar to yours: $n$ ladies take their umbrellas to a party, and then take one at random upon leaving. What is the expected number of ladies who end up with their own umbrella? By linearity of expectation, $n \cdot {1 \over n} = 1$.
