. In the $O-E$ game, a round starts with player $A$ paying $c$ cents to player $B$. Then $A$ secretly arranges the numbers $1,3,5,7,9,11,13$ in some order as a sequence $a_{1}, a_{2}, \ldots, a_{7}$, and $B$ secretly arranges $2,4,6,8,10,12,14$ as a sequence $b_{1}, b_{2}, \ldots, b_{7}$. Finally, the players show their sequences and $B$ pays $A$ one cent for each $i$ in $X=\{1,2,3,4,5,6,7\}$ such that $a_{i}<b_{i}$. This finishes the round. What number $c$ would make the game fair? (The game is fair if the total payments by $A$ to $B$ equals the total payments by $B$ to $A$ after all possible distinct rounds are played exactly once.)
I. Let $k$ be in $X$. There are $7 ! 6 !(8-k)$ choices of the sequences $a_{1}, a_{2}, \ldots, a_{7}$ and $b_{1}, b_{2}, \ldots, b_{7}$ for which $2 k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$ 's can be any of the 7 ! permutations of the 7 odd integers; then $j$ is the subscript such that $a_{j}=2 k-1$, and $b_{j}$ must be one of the $8-k$ numbers in $\{2 k, 2 k+2, \ldots, 14\}$; the remaining 6 even integers can be arranged in 6 ! ways.

The total of the payments by $B$ to $A$ for the $(7 !)^{2}$ possible rounds is then

$$
\sum_{k=1}^{7} 7 ! 6 !(8-k)=7 ! 6 !(7+6+\cdots+1)=7 ! 6 ! 28=(7 !)^{2} 4
$$

$A$ pays to $B$ a total of $(7 !)^{2} c$; so $c=4$ makes the game fair.