## View Poll Results: Who will be the 2012 Victorian Champion?

Voters
28. You may not vote on this poll
• Zelesco

1 3.57%
• Morris

5 17.86%
• Garner

0 0%
• Stojic

2 7.14%
• Levi

0 0%
• Sandler

1 3.57%
• Stevens

1 3.57%
• Cheng

7 25.00%
• Hacche

1 3.57%
• Goldenberg

2 7.14%
• West

5 17.86%
• Dragicevic

3 10.71%

# Thread: Victorian Championship 2012

1. Rather than predicting the winner I'm going to predict the winning score: 8.5/11.

2. Originally Posted by Max Illingworth
Rather than predicting the winner I'm going to predict the winning score: 8.5/11.

I took this mathematical approach myself only as a means to examine as many factors as possible in order to reach a conclusion ie final prediction.
If I am correct your outcome of 8.5/11 (*) can only be a result of

• Eight wins, one draw and two losses
• Seven wins, three draws and one loss and
• Six wins and five draws

Not wishing to restrict my prediction to specific numerical permutations, I am going to examine other simpler as well as more complicated factors such as

• players' strength,
• players' age
• experience in this kind or tournament play,
• recent performances,
• individual encounters
• non chess related situations such as education commithments etc

Although I am not committing myself as in taking part in the poll yet, my present field of preferences includes three or four other players, two of which are Domagoj Dragicevic and Dusan Stojic (harder to separate these two, than any other probable variations.

I will then spend a couple of days before the deadline trying to separate the final pair of my preferred players before my vote!

(*) Mathematicians please help, Is it possible to establish how many equal firsts (x way tie) can we have in a 12 player RR using a specific formula, or do I have to resort to practical counting means?

3. Originally Posted by JaK
(*) Mathematicians please help, Is it possible to establish how many equal firsts (x way tie) can we have in a 12 player RR using a specific formula, or do I have to resort to practical counting means?
You can have any number from 2 to 12 players finishing equal first, and this can be trivially demonstrated by assuming that that number of players all draw with each other but beat everyone else they play.

As for the number of different player combinations that can finish equal first, I believe the number of different possible n-way ties for a given n is

12!/(n!*(12-n)!)

(! = factorial)

on which basis there are:

66 possible two way ties
220 possible three way ties
495 possible four way ties
792 possible five way ties
924 possible six way ties
792 possible seven way ties
495 possible eight way ties
220 possible nine way ties
66 possible ten way ties
12 possible eleven way ties
1 possible twelve way tie

for a total of 4083 ties. (There are also 12 possible one-way ties, ie outright firsts, and the maths suggests there should be 1 way to have a zero way tie, which I assume consists of not having the tournament at all.)

Another question might be: what are the number of different scores on which an x-way tie for first can occur for each value of x?

4. ## thanks

Thanks Kevin, that was great help, I am very tempted to start a course in Maths (starting with repeating year 12) at some stage when I retire!

5. Originally Posted by Kevin Bonham
Another question might be: what are the number of different scores on which an x-way tie for first can occur for each value of x?
I think that for x = 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 the answers are (in order) 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

An x-way tie (for x<12) can occur on any score from 6 to (11.5-x/2). x=1 also follows this pattern as there are 11 different possible outright winning scores. The case of x=12 is special because it is the only one for which 5.5 is a possible winning score.

I am ignoring the possibility of double forfeits or 1/2:0s or other such irregularities!

6. Excellent field, wish I could have played but work prevents me

Good luck to all players

7. Originally Posted by Kevin Bonham
I think that for x = 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 the answers are (in order) 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

An x-way tie (for x<12) can occur on any score from 6 to (11.5-x/2). x=1 also follows this pattern as there are 11 different possible outright winning scores. The case of x=12 is special because it is the only one for which 5.5 is a possible winning score.

I am ignoring the possibility of double forfeits or 1/2:0s or other such irregularities!
I will use the previous formula but with restricted field ie 6-7 players in order to do calculations with real players and possible outcomes. Elimination of an x number of players is based upon the assumption that the x-number has not realistic chances of winning the title but not excluding possibilities of players of that subset causing major upsets!
Main difficulty for this approach is research for players' performances in major competitions as well as individual encounters with eachother during a period that I am very busy and spend more of the time away from home.
We 'll see how it goes anyway!

8. This event starts tonight at Box Hill Chess Club!

New FIDE ratings are also out & the field has an average rating of 2242, making it one of, if not the strongest event so far in Australia this year!

Players for the event (with updated ratings - FIDE then ACF) in pairing order:

Code:
```1. Zelesco, Karl	2093	2156
2. Morris, James	2327	2356
3. Garner, David J	2171	2134
4. Stojic, Dusan	2316	2300
5. Levi, Eddy L	2120	2118
6. Sandler, Leonid	2265	2298
7. Stevens, Tristan	2178	2148
8. Cheng, Bobby	2352	2362
9. Hacche, David J	2067	2081
10. Goldenberg, Igor	2385	2414
11. West, Guy	2328	2325
12. Dragicevic, Domagoj	2303	2304```

9. ## Domagoj

Back for a sec to vote for this (*)
I just noticed that the poll for this event closes on May 3, 2012.
Won't the voters who 'd tip their favourite players by then have an advantage over the rest of us since they would know tonight's results.
Anyway, my tip for the Victorian 2012 Champion , is Domagoj Dragicevic, after a hard battle (it might go to a play off match) vs his team mate at Noble Park Chess Club Dusan Stojic!
(*)I had to hurriedly redo my calculations since for some silly reason I had included Mirko in my list!

10. [QUOTE=Kerry Stead]This event starts tonight at Box Hill Chess Club!

"New FIDE ratings are also out & the field has an average rating of 2242, making it one of, if not the strongest event so far in Australia this year!"

I'm inclined to the that Doeberl, SIO and the Australian Championships may be just a tad stronger.

Cheers,
Jammo

11. JAK, in going for a possible tie for first between Domagoj and Dusan you've stolen my prediction! An excellent pick. The only difference is that I predict it will be an 11 player tie, not just those two. And yes, my simulation used exactly the formula used in global warming predictions, so it should be taken very seriously. It also predicted Black Caviar to win its last race and later eat some carrots.

Robert is right. There are plenty of stronger tournaments in Australia, although this is a reasonably good Vic Championship with plenty of improving younger players. It's just a shame Darryl's not in it; he's the form player in Victoria this year.

If anyone thinks Gelfand is a rough chance to beat Anand you can get 4.75 at Pinnacle. Interesting... he's not a bad match player it seems.

12. Results from tonight's round:

Code:
```No Name            Rtg  Loc   Result   Name                Rtg  Loc

1 Zelesco, Karl   2093 2156    1:0    Dragicevic, Domagoj 2303 2304
2 Morris, James   2327 2356    1:0    West, Guy           2328 2325
3 Garner, David J 2171 2134    0:1    Goldenberg, Igor    2385 2414
4 Stojic, Dusan   2316 2300    1:0    Hacche, David J     2067 2081
5 Levi, Eddy L    2120 2118    P:P    Cheng, Bobby        2352 2362
6 Sandler, Leonid 2265 2298    1:0    Stevens, Tristan    2178 2148```

13. Originally Posted by Gattaca
If anyone thinks Gelfand is a rough chance to beat Anand you can get 4.75 at Pinnacle. Interesting... he's not a bad match player it seems.
Very tempting! Seem like I've only seen Anand draw since beating Topalov!

Are these VIC Champs games published anywhere?

Cheers,
Paul

14. Originally Posted by jammo
Originally Posted by Kerry Stead
This event starts tonight at Box Hill Chess Club!

"New FIDE ratings are also out & the field has an average rating of 2242, making it one of, if not the strongest event so far in Australia this year!"

I'm inclined to the that Doeberl, SIO and the Australian Championships may be just a tad stronger.

Cheers,
Jammo
Although definitely stronger at the top end, the tournament does not have the 'tail' of the Doeberl or SIO & the average rating is almost identical to that of this year's Australian Championships (in fact if you take the average of the Aust. Champs & multiply it by 12, there is just one rating point difference between the two fields).

Average for the Doeberl Cup Premier was 2182
Sydney International average was 2112
Australian Championships average was 2242

I'm still happy to stand by my statement ... though it does help the averages that Karl Zelesco's FIDE rating has gone up over 100 points already this year!