< Earlier Kibitzing · PAGE 28 OF 28 ·
|Mar-31-13|| ||Daisuki: Here are updated odds based on a 30,000-tournament (1,680,000-game, including the results of round one through thirteen games already played) simulation:|
Win frequency Win frequency change since last simulation Win frequency change since the beginning of the tournament Player
95.06% +54.94% +37.41% Carlsen
4.94% −54.93% −8.82% Kramnik
0.00% −0.02% −12.04% Aronian
0.00% =0.00% −9.10% Radjabov
0.00% =0.00% −2.35% Grischuk
0.00% =0.00% −1.84% Ivanchuk
0.00% =0.00% −2.43% Svidler
0.00% =0.00% −0.83% Gelfand
(Caissa forbids me to have proper formatting here.)
As far as winning scores go, 9.5/14 or +5 had a 66.60% frequency, 9.0/14 or +4 had a 33.00% frequency, and 8.5/14 or +3 had a 0.40% frequency. Aronian now has no possible way of winning the tournament, leaving all odds to Carlsen and Kramnik. Carlsen catching up to Kramnik (without Kramnik losing, which would hurt Carlsen's tiebreak odds) was extremely important, as he is better on the second tiebreak, number of wins. The remaining meaningful games are Carlsen-Svidler and Ivanchuk-Kramnik.
So anyway, why does he have such high odds in my simulation? Well, for one thing Ivanchuk's current erratic play is not being simulated. Carlsen is also considered much stronger by the data than Kramnik, and Svidler is not considered especially stronger than Ivanchuk. Carlsen also has white and is very good with white, and Kramnik has black. The result of all this is that Carlsen is very unlikely to lose and has significant odds of winning. If he wins then Kramnik can do nothing to win the tournament. If he draws Kramnik has to win, but Kramnik is considered weaker than Carlsen and has black, so that's not considered to have particularly good odds. If Kramnik loses (and as black this is considered to have some possibility, although not a huge one) Carlsen wins the tournament. Anyway, my simulation is sensitive to remaining individual games, so that's that. ;p
|Mar-31-13|| ||Kinghunt: <Daisuki> Can you break that up into probability of win, loss, or draw for Carlsen and Kramnik in their respective games? I'm a bit curious, because your numbers are so far off of everything I've generated.|
|Apr-01-13|| ||Tiggler: <Kinghunt> and <Daisuki> Is simulation necessary now? I think the expectations can be solved analytically. Thus for this case:|
<Kinghunt: <donjova: For <Kinghunt>: Carlsen vs. Svidler:
Ivanchuk vs. Kramnik:
Final odds for these probabilities
Carlsen - 73%
Kramnik - 27%>
There are only nine possible outcomes. Three of them have Kramnik winning:
Kramnik wins and (Carsen draws or loses):
P = 0.4*0.6 = 0.24
Kramnik draws and Carlsen loses:
P = 0.3*0.1 = 0.03
Total: 0.27 Kramnik, so Carlsen must be 0.73.
|Apr-01-13|| ||Tiggler: Upper bound on Kramnik's chances:
Assume his expectations for win/loss/draw are the same as Carsen's, even though he has black and Carlsen has white, and even though his rating difference vs Ivanchuk is less than Carlsen's versus Svidler.
Assume also that win/loss/draw are equally likely (best case for Kramnik, I believe, because the smallest sum of three squared positive numbers that add to 1 is when the three numbers are equal - ask me to explain this better if interested).
Then his win probability = 1/9 + 1/9 + 1/9 = 0.3333
Thus a win probability of ~70% for Carlsen seems extremely conservative.
|Apr-01-13|| ||Kinghunt: <Tiggler> Yes, indeed. I am solving these analytically, rather than through simulation. It's simple enough to do that anyone could do it easily from here, but I find that most people are simply too lazy to take the 10 seconds to think it through and do the math. Your solution is correct.|
|Apr-01-13|| ||Daisuki: <Kinghunt: <Daisuki> Can you break that up into probability of win, loss, or draw for Carlsen and Kramnik in their respective games? I'm a bit curious, because your numbers are so far off of everything I've generated.>|
Well, I consider it much like your far higher odds for Aronian when he was almost out. There really aren't any odds I can give you, because the relevant four players' strengths change from tournament to tournament at random.
Tiggler, yes, I can solve it that way and was thinking of it exactly like you did, but of course I lack hard percentages, which are required for that method. I think ~70% is really generous to Kramnik. If you factor in both relative rating and color differences Carlsen has a 160+ point advantage over Kramnik in this last round. So really it makes sense to me that a draw on Kramnik's part is almost like a loss; Carlsen is unlikely to lose (in theory). And of course even if Kramnik wins Carlsen has to not win. I hope Svidler folds to Carlsen, though. ;p
|Apr-01-13|| ||Tiggler: <Daisuki>:<Tiggler, yes, I can solve it that way and was thinking of it exactly like you did, but of course I lack hard percentages, which are required for that method.>|
This method is no more dependent on the assumed probabilities than any other. And a 'hard' percentage, what does that mean? Since all these players have played 13 games in the last two weeks in London, one could argue that converged performance ratings from those 52 games are the most relevant ones to use. If that is done, Carlsen's advantage in rating over Kramnik disappears, and Svidler is much stronger than Ivanchuk. Also the white advantage almost disappears, except that Svidler is more likely to lose with black.
The likliest outcome is two draws anyway, except that the tournament situation is a unique disturbing influence that we have no way to evaluate.
|Apr-01-13|| ||Kinghunt: And Carlsen wins! What an exciting round it was, too. I'm very much looking forward to Carlsen-Anand in the fall.|
|Apr-01-13|| ||Daisuki: Tiggler, it is dependent in the sense that I can't give you one set of odds covering the two relevant games. Instead there are endless strength and thus odds combinations, and while these collectively converge on some odds for the relevant games (although averaging the odds for the games doesn't cause the same final odds, and I don't know what kind of average would make it work; arithmetic, geometric, and harmonic all fail) I can't easily know what they converge on given how I simulate things. The best I can do is a Monte Carlo simulation that reflects those odds.|
|Apr-02-13|| ||Tiggler: <Daisuki: Tiggler, it is dependent in the sense that I can't give you one set of odds covering the two relevant games. Instead there are endless strength and thus odds combinations, and while these collectively converge on some odds for the relevant games (although averaging the odds for the games doesn't cause the same final odds, and I don't know what kind of average would make it work; arithmetic, geometric, and harmonic all fail) I can't easily know what they converge on given how I simulate things. The best I can do is a Monte Carlo simulation that reflects those odds.>|
This is a very mysterious answer. Do you not know the probabilities that are input to your monte carlo simulation for the two games?
|Apr-02-13|| ||Daisuki: Like I said, each of the four players' strengths is random relative to their ratings in each tournament. If you don't believe that it's not simple for me to come up with average odds for each game result that (importantly!) result in the average odds given at the end of the Monte Carlo simulation, then do this:|
Create four sets of two games, where the first column is a white win and is represented by RAND(), and the second column is represented by (1-respective_first_column_cell)*variable, where variable is 1/2, then 2/3, 3/4, 4/5, 5/6, 6/7, 7/8, and 8/9 as you go down cell by cell. The last column is 1-both_respective_previous_column_cells. These will be win-draw-loss odds for each game, and consider each set of two to be ordered as Carlsen, then Kramnik. Obviously for each game pair you then take KDraw*CLoss+KWin*(CDraw+CLoss) to get Kramnik's odds. Now tell me how to average the six respective game results (CWin1, CWin2, CWin3, and CWin4 being one of six sets) so that you get one game pair that results in Kramnik odds equal to the arithmetic mean of the four Kramnik results calculated earlier.
|Apr-03-13|| ||Tiggler: <Daisuki> Thanks for the explanation. I'll need to go into a long huddle and will probably flag on time before I can answer, but I appreciate it anyway.|
|Apr-03-13|| ||Daisuki: Maybe it would be better if variable was also RAND(), since I don't know ahead of time how deviant players' strengths will be relative to their ratings.|
|Apr-04-13|| ||Tiggler: <Daisuki> I don't quite understand your procedure, but I think the core of the difficulty is that you are treating each player's rating as a random variable, presumably distributed normally with a given sd. After sampling the distributions, I guess you then input the resulting ratings difference into the expected result distribution formula to sample actual results.|
There is a way to simplify this, based on a unique property of Gaussian distributions: the convolution of two (or more) Gaussians is itself Gaussian with variance equal to the sum of the separate respective variances.
Take the case where the rating of each player can have a different sd: sd1 and sd2. Then take SIGMA=SQRT(400^2 +sd1^2 +sd2^2). In a simple case in which the player ratings sd are always the same, let’s say 40 points, then SIGMA=403.980. With this value of SIGMA, we now use the following formula for expected score =(0.5 +0.5* ERF((rating1-rating2)/SIGMA)). Note that for this example the strong player always tends to underperform his rating by 1% of the rating difference, since increasing SIGMA by 1% is equivalent to decreasing (rating difference) by 1%.
Having got “expected score”, you can use whatever draw formula you like to get the win/loss/draw probability. No simulation required, and there are no approximations in this method. You get the converged solution directly.
|Apr-04-13|| ||Tiggler: It occurs to me that the standard deviation of a player's rating that should be used would depend on the number of games to which it is applied, or from which it is deduced. Thus a standard deviation of 20 points, say, for a 9-game tournament, implies a standard deviation of 60 = 20*SQRT(9) points for each game considered one-at-time.|
|Apr-04-13|| ||Daisuki: Having an expected score isn't enough, because my win-draw-loss odds depend on both rating sums and differences, as well as expected scores. Having two 2000 players face off is not considered very similar to having two 2800 players face off. So really I do need a way to average the win-draw-loss probabilities properly. Though I should tell you that there's also the problem that implementing such an average would take a fair amount of work, so I'm not expecting to do it even if I could.|
Anyway, obviously I had Carlsen as unlikely to lose and very likely to win. Ivanchuk was also considered more likely to win than Kramnik was (I'm not sure where Sonas got the opposite from; perhaps he was going based on in-tournament data evaluation, which I didn't do, or perhaps he's using a dinky advantage for white). Like I mentioned the color advantage for Carlsen relative to Kramnik was considerably more than the relative rating advantage, which was of course itself considerable. It just all stacked like that, so Kramnik got the great majority of odds from results where he won (as drawing and losing were almost the same due to Carlsen being unlikely to lose), but even if he won it was common that Carlsen would also win, eroding his chances. Generally you could simplify this into something <roughly> equal to Kramnik's win odds divided by two.
I'm not sure that Carlsen losing more would help hugely much given that he'd then have to win more. And if Carlsen lost more Kramnik would also have to lose more and thus draw less given that I'm not using player-specific expectation conversion formulas. Obviously at extremes such as drawlessness things would be very different; Kramnik's odds seem to go up to about 8.5% then, in a crude single calculation. I imagine due to color and other data differing (possibly "on the fly") other simulations were more generous to Kramnik. Given all the data I used there was really no way to give Kramnik even 10% unless I heavily altered expectations. In the end it seems that I was anecdotally more correct about Kramnik not having good chances in his game and less correct about Carlsen having a very low chance to lose (but I'm not sure how anyone could expect different; Kramnik obviously didn't).
|Apr-04-13|| ||Tiggler: <Daisuki: Having an expected score isn't enough, because my win-draw-loss odds depend on both rating sums and differences, as well as expected scores. Having two 2000 players face off is not considered very similar to having two 2800 players face off. So really I do need a way to average the win-draw-loss probabilities properly.>|
We were discussing the case of the two games in round 14, so of course the ratings and who had white and who had black were known. Can't you just plug in the expected score and the ratings into your formula to get win-loss-draw probabilities? No averaging is involved, so far as I can see. Or is there no closed form equation in your method? I already gave mine, for a given draw fraction, some pages back on the Hans Rulde games forum.
|Apr-05-13|| ||Daisuki: The ratings are effectively <not> known, because I'm using strengths to calculate expectations, and strengths vary every tournament. I already explained that the main factor is just how often Kramnik wins, and that is roughly halved to get his final tournament win percentage. It should make sense, given that Carlsen is considered unlikely to lose and that he is forced to have majority odds of beating Svidler with white, given the rating <and color> gap between them combining to on average be over 200 points (actually nearly 250). I realize Sonas was far more conservative with Carlsen and somehow ended up giving Kramnik higher win odds than Ivanchuk, but none of that is remotely possible given the data I used. I don't know what Kinghunt did to get Kramnik odds approaching those of equally random game results (i.e. approaching 1/3).|
|Apr-05-13|| ||Tiggler: <Daisuki: The ratings are effectively <not> known, because I'm using strengths>|
What are <strengths>? Apologies if you already explained it. If so, just tell me where to look, please.
|Apr-06-13|| ||Daisuki: A strength is just a rating modified at random, with a standard deviation of ~19.544 relative to the relevant March 2013 FIDE rating in my simulation's case. I use it like I would use a rating if I wasn't doing this, so I think of it like one.|
|Apr-06-13|| ||Tiggler: <Daisuki> OK, in that case one could deal with the random variations as I indicated previously. I guess these fluctations are too small to affect the win/loss/draw probabilities. You also have another feature that I know of: recent (several years) of head to head results. I don't know how you combine these with the rating information, though.|
|Apr-07-13|| ||Daisuki: Well, are my descriptions in my posts two and three back from this enough to make it clear why Kramnik had ~5% odds in my simulation? Due to ratings/strengths, colors, and also somewhat head to head results there was no way to give Kramnik high win odds or Carlsen win odds under 50%. I'm not arguing that someone not restricted to being consistent to my simulation's previous methods couldn't give substantially different odds for each game's possible results, but given the data I used there wasn't any way to get to Sonas-like Kramnik odds (my theory on this is that he's using a small white advantage like I used to; this could give Carlsen minority win odds and maybe also give Kramnik greater win odds than Ivanchuk), much less odds approaching one-third. I'm sure my win-draw-loss expectations method is flawed; what joy it would be if chess had no draws and expectations thus cleaned up immensely! ;p|
|Apr-30-13|| ||Kinghunt: Career tournament wins of Anand:
Corus 1989, 1998, 2003, 2004, 2006
Dortmund 1996, 2000, 2004
Reggio Emilia 1992
Linares 1998, 2007, 2008
Dos Hermanas 1997
Mexico City 2007
Events excluded: Only open tournaments won in his youth.
|Apr-30-13|| ||Kinghunt: Career tournament wins of Kramnik:
Dortmund 1995, 1996, 1997, 1998, 2000, 2001, 2006, 2007, 2009, 2011
Linares 2000, 2004
Dos Hermanas 1996, 1997
Moscow 2007, 2009
Wijk aan Zee 1998
Events excluded: Russian championships (no international players), Chalkidiki 1992 (insufficient strength), Hoogoven 2011 (insufficient strength and only six games), Linares 2003 (lost on tiebreak).
|Apr-30-13|| ||Kinghunt: Career tournament wins of Carlsen:
Wijk an Zee 2008, 2010, 2013
Nanjing 2009, 2010
London 2009, 2010, 2012
Bazna 2010, 2011
Bilbao 2011, 2012
Moscow 2011, 2012
Events excluded: Biel 2007 (insufficient strength)
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