[Soccer Flipcup]{.underline}

Soccer Flipcup can be expressed as a 1-dimensional random walk with boundaries at 0 and $N$. In each of the intermediate states $(1,N - 1)$, there exist a pair of competing players who play "flip cup". If the player from the onward team wins, the state increases from $n$ to $n + 1$, in which case the (n+1)th players compete. Otherwise, the state decreases from $n$ to $n - 1$, and the (n-1)th players compete. This continues until the state reaches either 0 or $N$, when the gameplay stops and the corresponding team wins.

The game typically begins with state $\frac{N}{2}$ for odd players or $\frac{N}{2} \pm \frac{1}{2}$ for even players.

For each of the teams of $N - 1$ players, we consider what is the optimal assignment of states to each of the $N - 1$ players as a function of their rank in flipcup skill.

This game, for one of the teams, can be represented by the following set of equations, where $p(n)$ is the probability of victory given action is on state $n$, and where $q_{n}$ is the likelihood of winning state $n$

$$p\left( n \right) = \left{ \begin{matrix} \ 1,\ \ \& n = N \ q{n}p\left( n + 1 \right) + \left( 1 - q{n} \right)p\left( n - 1 \right),\ \ \& 0 < n < N \ 0,\ \ \& n = 0 \ \end{matrix} \right.\ $$

The non-trivial equation becomes

$$\frac{1 - q{n}}{q{n}} = \frac{p\left( n + 1 \right) - p\left( n \right)}{p\left( n \right) - p\left( n - 1 \right)}$$

Recursively,

$$p\left( n + 1 \right) - p\left( n \right) = p\left( 1 \right)\prod{m = 1}^{n}\frac{1 - q{m}}{q_{m}}\ $$

Notice we can also write $p\left( n \right)$ as

$$p\left( n \right) = \sum{m = 1}^{n}{p\left( m \right) - p\left( m - 1 \right)} = p(1)\sum{m = 1}^{n}{\prod{s = 0}^{m - 1}\frac{1 - q{s}}{q_{s}}}$$

To find $p\left( 1 \right)$, we use the fact that $p\left( N \right) = 1$,

$$p\left( 1 \right) = \left( \sum{m = 1}^{N}{\prod{s = 0}^{m - 1}\frac{1 - q{s}}{q{s}}} \right)^{- 1}$$

Now we have a closed form equation for $p(n)$

$$p\left( n \right) = \frac{\sum{m = 1}^{n}{\prod{s = 0}^{m - 1}\frac{1 - q{s}}{q{s}}}}{\sum{m = 1}^{N}{\prod{s = 0}^{m - 1}\frac{1 - q{s}}{q{s}}}}$$

[Nash Equilibrium]{.underline}

What is clear that given two exactly identical teams and perfect information, $q{n} = \frac{1}{2}$ for all $n$. If one team seeks an advantage by swapping $y$ with $z$, the other team can make the identical swap to keep $q{n} = \frac{1}{2}$ . As this process continues, both teams would end up with optimal positions and would be unable to improve their odds. This end state is a proper Nash Equilibrium state. And it would happen that $q_{n} = \frac{1}{2}$ for all $n$. In the equilibrium state, $p\left( n \right)$ simplifies to

$$p\left( n \right) = \frac{\sum{m = 1}^{n}1}{\sum{m = 1}^{N}1} = \frac{n}{N}$$

That is, for odd player games, the likelihood of winning is $\frac{1}{2}$, and for even player games, one of the teams would have an advantage of $\frac{1}{N}$.

[What is this Nash Equilibrium state?]{.underline}

To find which $n = t$ has the most importance, and therefore requires the greatest strength, we take the partial derivative of $p\left( n \right)$ with respect to $q{t}$. We assume $q{n} = \frac{1}{2}$, as per the Nash equilibrium.

$$\frac{\partial p\left( n \right)}{\partial q_{t}} \propto \text{Nmax}\left( n - t,0 \right) - n(N - t) = \text{Nmin}\left( n,t \right) - nt$$

This equation tells you that with $N - 1$ players and a starting position $n$, which positions "t" are the most important and therefore require the strongest players.

The optimal strategy is as follows:

The equation is maximized at $t = n$ (since $N > n$). That shouldn't be surprising to anyone, that the best player should be in the starting position
$\partial p\left( n \right)$ decreases both as both $t$ increases and decreases from $n$. The best players should be around the starting position
As $t$ increases, $\partial p\left( n \right)$ decreases by $n$. As $t$ decreases, $\partial p\left( n \right)$ decreases by $N - n$.
This implies that for $n = \frac{N}{2}$, which is a common starting position, there is symmetry, meaning you are indifferent between switching players between the $\frac{N}{2} + a$ and $\frac{N}{2} - a$ positions
However, say $n > \frac{N}{2}$. Then you would want to put your second-best player in the $n - 1$ spot. This is particularly relevant for games with an even number of players.

Here is a diagram for a special case of N=8.

N   8

t                            
      1   2    3    4    5    6    7

n 1 7 6 5 4 3 2 1 2 6 12 10 8 6 4 2 3 5 10 15 12 9 6 3 4 4 8 12 16 12 8 4 5 3 6 9 12 15 10 5 6 2 4 6 8 10 12 6 7 1 2 3 4 5 6 7