Congrats to John Tromp on his game with Leoni.

Only John, Frode and Jan have no losses now.

And now, only John and Frode have only one loss. David, Leoni, Bill and Jan have two losses. Which means that if John Tromp wins all 6 remaining games he will be the winner (he has won the game against Frode).

On the other hand, if it is a four way tie, the son nod will to to Leoni.

Congratulations to Leoni for winning his second chammpionship and at the same time achieving a well deserved rating of 2000!

And congrats to John for tieing in his first championship, I predict his rating will also approach 2000 after the next championship.

And congrats to Dave and Frode for some great games as well.

Looks like our next Tournement is going to be be one of the smallest (9 players) , but nonetheless one of the alltime strongest ever.

Are you sure that neither 6.2.1 nor 6.2.2 will have a tie for second place? If either does, then 7.1.1 will have more than 9 players.

Well, a first place tie would add a tenth player and a second place tie would add a tenth and eleventh player, but I don't see any.

A second-place tie would add a 10th player. For a tie for 1st to add a 10th player would require that it be a 3-way tie.

As for whether one will happen, I'll defer to you Hex judgment if you say it's unlikley, but there are still lots of ways it could happen.

Son ties are indeed 3 way.

If everybody has only one loss each, yes.

The players must have exactly the same wins(losses) with Player1 loses to Player2 loses to … PlayerN loses to Player1 where N >= 3.

It's not unlikely that there is a 3-way tie for second place in one of the 2nd leagues. If that is the case, there will be 11 players in 1st leagues.

Marius

By my calculation any tie in either league is mathematically impossible.

A 2-way tie for first is possible (in general, I'm not talking about any specific tournament) when Everyone has at least two losses. (It's easier in games with draws, of course.)

The simplest example to see is probably this: Imagine that without SON there is a 3-way tie between A, B, and C. But instead of each losing only 1 game, each player lost 2 – one to someone in the tie and one to somone who did less well. Since the second losses can be against players who have differing scores, the SONs of the three leading players can all vary. If A and B each lost to a player who had 6 points while C lost to a player who had 8 points, then A and B wwould share 1st place while C would finish 3rd.

I guess we will see whether there is a tie or not.

I'm not looking at any board positions, so I'm not commenting on any lines of play.

In 6.2.1, if Marius Halsør wins the rest of his games and agoui beats ypercube, then there will be a 3-way tie for first. Other such ties are possible, too. I'm not prediciting any will actually occur, but they are possible.

DVD in your example, A and B would almost share first place, except that if A lost to B lost to C lost to A, then A beat C while B beat A and therefore B would gain more son than A, and B would win the tournement by a nose.

Strike my example in 6.2.1. I'll cite a good example in a moment.

Bill, you're missing the effect of the second losses. Yes, B is ahead in SON if you only look at games among the top three players. But A makes up that difference by beating a player with 8 points (whom B lost to) while B only beat a player with 6 points (whom A lost to).

As for 6.2.1, if

• agoui beats Marius Halsør
• ypercube beats agoui, Marius Halsør, and Norbi
• Subtleblue beats Taral, Marius Halsør, ypercube, Norbi, and elliot

then agoui, ypercube, and Subtleblue will tie for first.

Dvd, only C, not B, lost to the player with 8 points in your example.

How about this, Bill:

Ypercube wins every match.

Taral beats me, I beat Aguoi, Agoi beats Taral.

This would mean a 3-way tie for second place.

Yes, I stand corrected, it is still mathematically possible for a first or second place tie.

And there were examples of even 5-way (!!) ties in the past.

Looks like 9 players in the next tier one.