# Chess tournament problem

$12$ chess players took part in a tournament. Each played against each other exactly once. After the tournament every chess player did $12$ lists of names.

• On the first list, the player only wrote his own name.
• On the second list, they wrote their own names as well as all man they had won against.
• They proceeded to write lists: Every next list contains all names from the previous list and the new names that players from the previous list had won against.

It turned out that for all chess players, the $11$th and the $12$th list contained different amount of each name.

How many games ended draw in the tournament?

I have been thinking of the problem the hole day but I am clueless...

• It is a good question but we also do need to see your efforts in solving this. Otherwise you could move it to puzzling.stackexchange.com Commented Dec 12, 2015 at 15:39
• I have tried to solve the problem the whole day. My attempts have been to approach it to a logical way in each every player has won. There is a total of 66 games, and lets say that 54 was even. Player 1 won against 2, 2 against 3 and so on. But that didnt work and I can't figure it out... Commented Dec 12, 2015 at 15:58
• Good. At least you have tried. And the correct spelling is 'whole'. I'm sure someone will help. Commented Dec 12, 2015 at 16:00

In order for the $11$th and $12$th lists to be different, all previous lists must be different, and that means that each list only increases by one name from the previous list - or you will have run out of names to add before the $12$th list.
So $12$ matches ended in a result, and the remaining $\fbox{54}$ matches were drawn.