Climbing the mountains of Tour de FiveThirtyEight

You and three other, equally talented bicyclists are competing to get up the mountain to accrue points: 5 for first, 3 for second, 2 for third and 1 for last places, respectively.
Since you are all equally talented, the relative rankings of your average speeds for this stage should all be equally likely outcomes. However, two of your competitors are on a team and through a clever use of drafting off of one another will be able to both finish at roughly the same time as the faster of the two teammates.

Given that two of your competitors are working together, what is your expected number of points on this stage?
Let's call your average speed $A,$ the average speeds of the two teammates $B$ and $B^\prime,$ and the average speed of other individual competitor $C.$ Given that there only $24$ possible, equally likely orderings of $A,$ $B,$ $B^\prime,$ and $C,$ we will go with brute force to calculate the expected number of points for this stage.

If $A > \max\{ B, B^\prime \}$, then all of the teammate's schemings are for nought, and you will get the place that you rightfully deserve. There are 6 orderrings with $A > C$ and $A > \max \{B, B^\prime\}$, where you are in first and get $5$ points.
There are another 2 orderrings where $C > A > \max\{B, B^\prime\}$, where you place 2nd and receive $3$ points.

Similarly, if $\min \{B, B^\prime\} > A,$ then the cooperative advantage is also nill from your perspective. These cases cover the 6 orderings where $C > A$ and $\min\{ B, B^\prime\} > A$ where you were in last place and receive $1$ point.

There are also 2 orderings where $\min\{B, B^\prime\} > A > C$ and you finish in 3rd place receiving $2$ points.

The teamwork only affects you when either $B > A > B^\prime$ or $B^\prime > A > B.$ There are four possible orderings with $A$ between $B$ and $B^\prime$ and $A$ the second largest number in $\{A, B, B^\prime, C\}.$ In this case, though rightfully you should have finished in second place the slower of $B$ and $B^\prime$ gets pulled ahead of you by the faster and you place in 3rd, receiving $2$ points.
Similarly, there are 4 possible orderings with $A$ between $B$ and $B^\prime$ and $A$ the third largest number in $\{A, B, B^\prime, C\}.$ In this case, though rightfully you should have finished in third place, the slower of $B$ and $B^\prime$ gets pulled ahead of you by the faster and you place last, receiving $1$ point.

Summing up all of the points, we get $$6 \cdot 5 + 2 \cdot 3 + 6 \cdot 1 + 2 \cdot 2 + 4 \cdot 2 + 4 \cdot 1 = 58 \text{points,}$$ for an average of $2.41\bar{6}$ points.

Meanwhile in the absences of the teammate shenaniganry, your expected points is the simple average of the points $\frac{5+3+2+1}{4} = 2.75,$ so teammates cost you only a third of a point.

Pickup basketball teams

The number of ballers from each of Blacksburg, Greensboro and Silver Spring who show up to a pickup basketball game each week is identically and independently distributed as the uniform distribution on $\{1, 2, 3, 4, \dots, N\},$ for some integer $N.$

What is the probability that, on any given week, it’s possible to form two equal teams with everyone playing, where two towns are pitted against the third?

Let $B,$ $G$ and $S$ be the number of players who showed up from Blacksburg, Greensboro and Silver Spring, respectively. Without loss of generality, let's assume that $S$ is the largest and we seek the number of configurations where $B+G=S.$

First note that if $S$ is the maximum, then no such configurations can occur if $S=1(=B=G).$ For any $S = 2, \dots, N,$ there are $S-1$ different combinations of integers $B$ and $G,$ such that $B+G=S;$ those being $(B,G) \in \{ (1,S-2), (2, S-3), \dots, (S-3, 2), (S-1, 1) \}.$ So the total number of games where $B+G=S$ is $$\sum_{S=2}^N (S-1) = \frac{N(N-1)}{2}.$$
From symmetry, we get three times this number (since we arbitrarily assumed that $S$ was the maximum, but it could equally likely be $B$ or $G$), so the total number of configurations where two equal teams where two towns are pitted against the third are $3 \frac{N(N-1)}{2}.$ The total number of possible configurations of players to show up on a given week is $N^3,$ so the probability is \[p_N = \frac{3(N-1)}{2N^2}.\]
In particular, if $N = 5,$ then the probability is $p_5 = 24\%.$