2016 Candidates Moscow 10-29 March

[Mick Norris]

3. 7 Tie-breaks
If the top two or more players score the same points, the tie will be decided by the following criteria, in order of priority:
a) The results of the games between the players involved in the tie.
If they are still tied:
b) The total number of wins in the tournament of every player involved in the tie.
If they are still tied:
c) Sonneborn - Berger System.

This reads ambiguously to me

Why?

[Richard Bates]

This reads ambiguously to me

Why?

Who is "Every player involved in the tie" if the first tiebreak doesn't determine a single winner?

[NickFaulks]

I agree and have asked.

[Brian Towers]

Who is "Every player involved in the tie" if the first tiebreak doesn't determine a single winner?

Can you give an example where this is not clear?
It seems clear to me.

Suppose A, B and C are tied on +3.
A beat C and B beat C while A and B drew. Hence after the first tie break (results against other tied players) C is eliminated and only A and B are tied.
Suppose A has +5,=7,-2 and B has +4,=9,-1
Then B has fewer wins and is eliminated on second tie break leaving A as the overall winner.

[Mick Norris]

Then B has fewer wins and is eliminated on second tie break leaving A as the overall winner.

Or fewer losses, as Kramnik complained about at London, although interesting it hasn't been changed since

[Mick Norris]

chess.com report

includes Joal Lautier interview, and MVL analysis of Caruana - Anand

[Richard Bates]

Who is "Every player involved in the tie" if the first tiebreak doesn't determine a single winner?

Can you give an example where this is not clear?
It seems clear to me.

Suppose A, B and C are tied on +3.
A beat C and B beat C while A and B drew. Hence after the first tie break (results against other tied players) C is eliminated and only A and B are tied.
Suppose A has +5,=7,-2 and B has +4,=9,-1
Then B has fewer wins and is eliminated on second tie break leaving A as the overall winner.

I have absolutely no doubt that that is how the process is supposed to work. It is not however how it is unambiguously written above, because "the tie" is never redefined. All it has to do is make reference to players being eliminated, or even just add the word "still" - ie. every player still involved in the tie (after the first tiebreak). Just poor drafting imo.

[NickFaulks]

I have absolutely no doubt that that is how the process is supposed to work.

I have confirmed that you/we are correct, C is out. I agree that the wording is ambiguous and in fact I think the closest strict interpretation would be that C stays in.

[Brian Towers]

I have absolutely no doubt that that is how the process is supposed to work. It is not however how it is unambiguously written above, because "the tie" is never redefined. All it has to do is make reference to players being eliminated, or even just add the word "still" - ie. every player still involved in the tie (after the first tiebreak). Just poor drafting imo.

So, the "If they are still tied:" bit doesn't do it for you?
To me it seems that after first tie break in my example above A and B are "still tied" and C is not and so by implication is out.

[NickFaulks]

So, the "If they are still tied:" bit doesn't do it for you?

I would like it to be clearer who "they" are. If, for instance, the wording were "if the tie is still not resolved ..." then I think it is clear that all three would still be in. I don't like the shades of meaning being so close.

[Christopher Kreuzer]

Looking at the final four rounds (11-14), and who the four leaders are playing (I refuse to believe Giri can score 3/4 or more even though he is playing Anand and the three other tail-enders Svidler, Nakamura and Topalov), the following are the pairings (the colour in brackets for each row is the colour the player at the start of that row has against his opposition listed in that row):

Caruana: Topalov (B); Aronian (W); Svidler (W); Karjakin (B).
Karjakin: Anand (B); Topalov (W); Aronian (B); Caruana (W).
Anand: Karjakin (W); Nakamura (B); Giri (W); Svidler (B).
Aronian: Svidler (W); Caruana (B); Karjakin (W); Nakamura (B).

Some observations from the above:

- The tournament has divided into 'leaders' and 'tail enders' (four in each group).
- Karjakin and Caruana favourites due to their slender half-point advantage.
- Colour balances are even, though Caruana has successive Whites in rounds 12 and 13.
- Two of the leaders (Caruana and Aronian) play another leader twice and a tail ender twice.
- One of the leaders (Anand) plays a leader once and finishes with three games against tail enders.
- The other leader (Karjakin) plays three leaders and one tail ender (but has a half point lead over Anand).

From that alone, I would make Caruana favourite (slightly easier run in) and Karjakin second-favourite, with Anand third-favourite and Aronian fourth-favourite. The final consideration is the rest day in the middle of these final four rounds. Who manages to summon enough energy and resources on that rest day for the closing stretch might be the decisive factor. It will probably go down to the final round, which will be incredibly tense. As was seen in the London Candidates, lots of mistakes get made under such pressure.

[Mick Norris]

The final consideration is the rest day in the middle of these final four rounds. Who manages to summon enough energy and resources on that rest day for the closing stretch might be the decisive factor. It will probably go down to the final round, which will be incredibly tense. As was seen in the London Candidates, lots of mistakes get made under such pressure.

Mental and physical tiredness will be a factor, plus the pressure of having to win (or avoid defeat) - it may be decided by a tail-ender having a good day against a favourite

I hope it won't be decided on SB tiebreak by whether, say, Topalov ends up with a worse score than Svidler

[Christopher Kreuzer]

Forgot to add that the tail ender with most say in what happens is Svidler (he plays three leaders, though only one of the actual leaders, and has two Blacks and one White). Topalov plays two leaders, but both are the actual leaders (White and Black). Nakamura plays two leaders (but in both cases the 'half-a-point-behind' leaders) and has White in both cases. Giri only gets to play Anand (and Giri has Black).

[JustinHorton]

(I refuse to believe Giri can score 3/4 or more even though he is playing Anand and the three other tail-enders Svidler, Nakamura and Topalov)

I don't think he will, but it ought to be eminently possible. Now or never today, of course.

I wonder what approach Topalov will take against Caruana today. And for that matter, what approach Caruana will take against Topalov.

Odds, rounded a little for comprehensibility:

Caruana 11/8
Karjakin 8/5
Aronian 9/2
Anand 7/1
Giri 16/1
100/1 Bar

Am I right in thinking that Caruana would have to score +3 over Karjakin over the next three rounds for Karjakin not to come ahead of him in the event of a last-round win?

[JustinHorton]

(Re: last comment, I guess it would be more complicated if any tie between them was also with a third party or more. But just between the two of them, that's right, isn't it?)