6.42 12 players participated in a chess tournament. After the tournament, each of them made 12 lists. On the 1st list, they only wrote their own name; on the 2nd list, they wrote their own name and the names of all the opponents they defeated; on the 3rd list, they wrote all the names on the 2nd list and the names of all the opponents these people defeated; and so on, until on the 12th list, they wrote all the names on the 11th list and the names of all the opponents these people defeated. It is known that for each player, on their 12th list, there are names that did not appear on their 11th list. How many draws were there in the tournament?
[Solution] Since everyone's 12th list contains a new name that was not on their 11th list, it follows that everyone's 12 lists are all different. In fact, if two of someone's 12 lists were the same, then all the lists from the earlier of these two lists onward would be the same, and in particular, the 11th and 12th lists would be the same, which is a contradiction.

Since the 12 lists are all different, starting from the 2nd list, each list has at least one more name than the previous list. Therefore, the 12th list has at least 12 people. But there are only 12 people in total, so each person's each list has exactly one more person than their previous list. It is not hard to see that the one additional person on each list is precisely the opponent defeated by the one additional person on the previous list. From this, we can conclude that each of the 12 players wins exactly one match. Therefore, the number of draws in the tournament is $C_{12}^{2}-12=66-12$ $=54$ matches.