6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wins, satisfying the following conditions:
(1) The winner depends only on the relative order of these 200 cards: if these 200 cards are placed face down in increasing order, and the audience is informed which card belongs to which player, but not what number is written on each card, the audience can still determine who will win;
(2) If the elements of the two sets are written in increasing order as
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{100}\right\},
$$

where, for all $i \in\{1,2, \cdots, 100\}$, $a_{i}>b_{i}$, then $A$ defeats $B$;
(3) If three players each select a set of 100 cards, denoted as $A$, $B$, and $C$, and $A$ defeats $B$, and $B$ defeats $C$, then $A$ defeats $C$.
Question: How many such rules are there?
[Note] Two different rules mean that there exist two sets $A$ and $B$ such that in one rule, $A$ defeats $B$, and in the other rule, $B$ defeats $A$.
6. There are 100 ways.

To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different rules.
Next, we prove that there are no other ways to define such rules.
Assume a rule satisfies the conditions, let $k \in\{1,2$, $\cdots, n\}$ be the smallest positive integer satisfying the following property:
$$
\begin{array}{l}
A_{k}=\{1,2, \cdots, k, n+k+1, n+k+2, \cdots, 2 n\} \\
B_{k-1}$.
Thus, $U>W$ (trivial when $k=1$).
The elements in the set $V \cup W$ arranged in increasing order are the same as the elements in the set $A_{k} \cup B_{k}$ arranged in increasing order.
By the choice of $k$, $A_{k}<B_{k}$.
Thus, $V<W$.
Therefore, $X<V<W<U<Y$.
By condition (3), $X<Y$.
Hence, the conclusion holds.