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 $n$

\[p(n) = \begin{cases} 1, & n=N \\ q_{n}p \left( n+1 \right) + \left( 1-q_{n} \right) p \left( n-1 \right) , & 0<n<N \\ 0, & n=0 \end{cases}\]

The non-trivial equation becomes

\[\frac{1 - q_{n}}{q_{n}} = \frac{p\left( n + 1 \right) - p\left( n \right)}{p\left( n \right) - p\left( n - 1 \right)}\]


\[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

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?

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.

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

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, where each column each row represents a $n$, and each column represents a $t$.

      1   2    3    4    5    6    7
  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