< Earlier Kibitzing · PAGE 22 OF 22 ·
|Dec-13-18|| ||Tiggler: Thank you! I'll calculate the probability of 6-6 vs draw probability, when I get round to it. You already gave us somewhere the answer for 1/3. Did you use N!/(W!*D!*L!) for the above?|
|Dec-13-18|| ||john barleycorn: yes. not assuming 1/3 for each event would not allow for that formula.|
|Dec-13-18|| ||Tiggler: <john barleycorn: yes. not assuming 1/3 for each event would not allow for that formula.>|
Yes it would: the coefficients don't change; only the product Pw^W*Pl^L*Pd^D is affected by the values of the probabilities.
|Dec-13-18|| ||AylerKupp: <<Tiggler> Not scores, but sequences of game results.>|
That's correct. I used the word "scores" incorrectly. I actually calculated each of the 3^4 = 81 possible scoring sequences and I then manually deleted those that could not occur.
<The correct number is 65, in this case, because you forgot to count the complementary results. There are 23 where player A wins the match, 19 when the match is tied, and another 23 when player B wins the match.>
Yes, I got the same numbers, 23 ways that player A wins the match and 19 ways that player A ties the match for a total of 42.
But I didn't forget to count the complementary results, I deliberately ignored them. This was because I was trying to answer the question that, probably again incorrectly, I thought that <john barleycorn> was asking, namely "how many ways can <a player> win or draw a match?" If that was the question, then my answer was correct. If (as now seems most likely) that wasn't the question and he was asking how many possible ways can <either player> win or draw the match, then my answer was clearly incorrect. Hey, if you think I'm bad at probability and statistics, I'm not nearly as bad at those as I am trying to figure out what question someone is asking!
|Dec-13-18|| ||AylerKupp: <Draw probabilities (part 1 of 2)>|
<<john barleycorn> With two players being rated "equal" (in "classical chess" 2835 Elo vs. 2832 Elo being considered "equal")>
A ±3 Elo rating point differential is considered equal as far as FIDE is concerned when using their scoring probability table with only 2 digits of resolution. But how precise do you want the answer to be? That might depend on what you intend to use the results for.
For example, FIDE uses a 102-entry scoring probability table with a range of ±736 Elo rating point differential with a variable rating point range per entry. And when calculating "official" rating changes this is the table that <must> be used. But for anything else, this table is useless.
I generated an equivalent 2001-entry scoring probability table with a range of ±1000 Elo rating point differential with a fixed resolution of 1 Elo rating point per entry. This is not just for increased resolution but for convenience; when determining the scoring probability I don't need to search the table but I can calculate an index and get the applicable probability immediately. This is slightly faster but probably not noticeably so.
As far as determining the draw probabilities (if that's important to you), there are several ways:
(1) Use a games database like Internet Explorer or ChessTempo (https://chesstempo.com/game-databas...) that give you the White win, draw, and loss percentages for each opening move actually used. I like the ChessTempo database because it not only has many more games and it's pretty conscientious (unlike Internet Explorer) of recording the players; rating but it also allows you to filter the database to get the White win, draw, and loss percentages when both players are rated 2200+, 2300+, etc. up to 2700+. And, yes, the White win, draw, and loss percentages are dependent on the rating level of both players, and the higher the players' ratings the higher the draw percentage and smaller the White win and loss percentages. Which makes sense since, in order to lose a game, a player must presumably make at least one mistake (or, as the saying goes, the next to last mistake) or at least a series of inaccuracies. And the higher rated the players are the less likely they are to make mistakes or inaccuracies so it's not unreasonable that the draw percentage would increase.
I last looked at the ChessTempo data on 11-02-18 and I got the following results. The first column is their ratings, the second column is White's winning %, the third column is the draw %, the fourth column is Black's winning %, and the fifth column is the difference between White and Black winning %s:
All 38.4% 31.4% 30.2% <8.2%>
2200+ 35.0% 39.5% 25.4% <9.6%>
2300+ 33.5% 43.1% 23.4% <10.1%>
2400+ 31.9% 46.8% 21.3% <10.6%>
2500+ 29.9% 50.9% 19.1% <10.8%>
2600+ 29.4% 52.0% 18.6% <10.8%>
2700+ 28.8% 52.1% 19.1% <9.6%>
Note that even though the draw percentage increases the higher rated the players are, the ~ 10% White win/Black win differential is roughly the same regardless of the rating of the players.
|Dec-13-18|| ||AylerKupp: <Draw probabilities (part 2 of 2)>|
The ChessTempo database has a problem relevant to our needs, as do most databases, of including not just games played at Classic time controls but at all time controls (Rapid, Blitz, Correspondence) and conditions (Simultaneous exhibitions, blindfold games, casual games, etc.) without a convenient way to filter them out. As of today the database had 17,038 games of all types for players rated 2700+.
I last extracted all the games from the ChessTempo database for players rated 2700+ about a year and a half ago, and at that time the database had 14,502 games. I tried to filter out all non-Classic time control games by keying on the event name and discarding games whose event names contained the words "Blitz" (22.1%), "Rapid" (12.2%), "Blind" (4.4%), "Exhib" (0.014%), and "Simul" (0.007%), as well as those games played before (arbitrarily) 1990 (1.5%). That left 8,666 "recent" games (59.8% of all games) played at presumably Classic time controls. Not a perfect method but an improvement as far as our interests are concerned. And also note that these percentages include both tournaments and matches but the former greatly outnumber the latter.
And that resulted in a White win % = 24.7%, draw % = 61.4%, and loss % = 13.9%. A slightly higher draw % and corresponding lower win %s for White and Black than if all the games are considered but this should also be expected since, contrary to what some think, the longer the time control and the longer the players have to think, the less likely the chances for mistakes and inaccuracies. But still the White win %/Black win % differential is a similar 10.8%.
(2) If you are only interested in WCC matches then that win, draw, loss data is available from the links provided in History of the World Chess Championship which unfortunately hasn't yet been updated with the results of the recent Carlsen - Caruana World Championship Match (2018). But I looked at results of all the WCC matches (not WCC tournaments) since 1949, including the most recent one, and I got the following % for the 400 Classic time control games: White win % = 20.3%, draw % = 68.5%, and Black win % = 11.3%. Note that the WCC match draw %s (68.5%) are somewhat higher than the tournament + match draw % (61.4%) but maybe not that much higher. And the White win/Black win differential is somewhat lower (9.0% vs. 10.8%) but maybe not that much lower, particularly since the sample size is much smaller.
(3) If you know who both players are you can look at their individual results. I looked at Carlsen and Caruana's game results prior to Carlsen - Caruana World Championship Match (2018) in the ChessTempo database and I found that they had played 55 games against each other, 33 of them at Classic time controls. Carlsen had won 30.3% of the games, Caruana had won 15.2% of the games, and 54.5% of the games were drawn. Of course, most of these games were played when Caruana was much lower rated than Carlsen (an average rating differential of 64 Elo rating points) and not the 3 Elo rating points just before their recent match or their current 5 Elo rating point differential.
So, you can use probability set (1) if you're interested in draw probabilities in the general case, probability set (2) if you are interested in WCC match draw probabilities, or probability set (3) (not recommended) if you are interested in Carlsen/Caruana draw probabilities. Or, if you are really interested in the Carlsen/Caruana WCC match draw probabilities (or, for that matter, all Carlsen/Caruana WCC match probabilities) you can blend all 3 by using probability set (1) as the prior probabilities and probability sets (2) and (3) to calculate posterior probabilities using Bayes Theorem.
|Dec-13-18|| ||Tiggler: <AylerKupp> Lots of useful data, thanks. When I was looking at the question of the rapid and blitz matches that Caru and Carl were about to play, I came upon an interesting post at the respected website of 538. The particular post that I refer to is this: https://fivethirtyeight.com/feature...|
As you will see, if you read, Mr. Roeder calculated the probabilities of each match result (down to win-loss-tie of the match) at every stage during the prospective rapid, blitz and rapid phases. Elo ratings for rapid and blitz were used, and for draws the following: <We also need a measure of how likely draws are in these faster formats. I’ll use historical data. In last year’s World Rapid Championship, for example, about 30 percent of the games were draws. In last year’s World Blitz Championship (which Carlsen won), about 20 percent of the games were draws.> The author also noted, but did not use, the following: < (In real life, the two have played 23 speedier games against each other, according to Chessgames.com — Carlsen won 13, Caruana won six and four were draws.)>
The other thing to notice about this article is that the author took into account the censorship: the conditions under which the match would end without completion of the complete number of games. Though not explicitly stated, it is evident that this was done by treating the probabilities of each games sequentially, at least after the third rapid game, and states the probability that each subsequent game would actually be played. This approach may in fact be the most economical way to program the results in an algorithm. I have been considering the problem of whether the outcome probabilities could be written in closed formula, as they can for an uncensored match (one in which all games must be played). The censored case is MUCH more difficult, but that is not the point: I am not necessarily most interested in easy problems.
Furthermore, in that article nothing is said about the probability distribution of particular match scores, only of match results as win/loss/continue at each stage. The MOST probable match score was the question initially raised by <AylerKupp> and which drew my attention. This is in fact the MODE (emphasis, since <AK> previously essayed to contradict me about this) of the probability distribution. To answer that question, the probabilities of each possible result must be calculated, and, yes, the draw probability is undoubtedly determinative of the result.
|Dec-15-18|| ||john barleycorn: Guys, for me, it sounds as if we are at the stockmarket and "analysts" try to sell their "product" based on sophisticated reasoning/calculations. We all know that a share does not care about its past. The stock market is a gambling joint. There are as many "stock" experts as there are "expert" crap-shooters in the Casino. (Ask: What is an "expert" crap-shooter?)|
Therefore I'll stick to my assumption that between "equally" rated players the outcome is best predicted by considering it a gamble with each outcome likely probable. There is an argument that after a decisive game the most "probable" outcome of the next game may be a draw. However, we do not have a "proof" for that (maybe evidence based on played games under certain situations etc. pp.). The outcome of a chess game can be and demonstrably was "manipulated" almost too easily and that is the crux in calculating anything. A draw maybe a "special" result among 2 players under specific conditions. (ok, ok unless you name is Giri)
|Dec-15-18|| ||Tiggler: <john barleycorn> As you said on the <Stumpers> forum:|
< You either solve a problem or find "excuses" for not doing so.>
|Dec-15-18|| ||Tiggler: Installment #6. Probabilities of outcomes in a four game match.|
Here are the promised numerical results for rapid 4-game match between Carlsen and Caruana using their ratings at the beginning of the match, that differed by 91 points.
The formulae derived were given here:
Tiggler chessforum (kibitz #476)
Several possible draw probabilities were used. I give the results for 0.5 , 0.3333 and 0.25 .
E= 0.6200 0.6200 0.6200
Pd= 0.5000 0.3333 0.2500
3-0 0.0507 0.0932 0.1213
2.5-0.5 0.2054 0.2055 0.1838
3-1 0.1224 0.1281 0.1387
2.5-1.5 0.2651 0.1987 0.1715
2-2 0.2207 0.1974 0.1942
1.5-2.5 0.0931 0.0935 0.0884
1-3 0.0151 0.0284 0.0368
0.5-2.5 0.0254 0.0455 0.0488
0-3 0.0022 0.0097 0.0166
A wins 0.6435 0.6255 0.6153
tie 0.2207 0.1974 0.1942
B wins 0.1358 0.1771 0.1905
The above is pasted as unformatted text directly from a table, so some clarification is needed. The values are delimited by a leading space, and each is given to four decimal places. The first row is the expected one-game score from the FIDE table for rating difference of 91, from the point of view of the higher rated player. The second row is the assumed draw probability. The next block of nine rows shows the nine possible match results in a match which terminates when either player reaches a winning score. In each row the probability of that outcome is given for each of the three assumed draw probabilities. The last block of three rows contain the probabilities of a match win by the higher rated player, match tie, or match win by the 'weaker' player.
Concerning the most probable match score, inspection of the table shows that for draw probability of 50% 2.5-1.5 is most probable; for draw probability of 33.33% 2.5-0.5 is most probable; for draw probability of 25% 2-2 is the most probable outcome.
This illustrates the point that without knowing the draw probability, one cannot reliably predict the most likely outcome.
|Dec-15-18|| ||Tiggler: The actual match result of 3-0 actually vindicates two statements of <john barleycorn>: Pd = 1/3 is as good an assumption as any, and that one cannot tell when a draw is actually a draw. The third match game would have been a draw, but for Caruana's desperation attempt to win. So the result "should" have been 2.5-0.5, which is the most probable result when the draw probability is 1/3.|
|Dec-16-18|| ||john barleycorn: oops, sorry. I deleted my 2 last posts as I was on the wrong page. back to Stumpers ...|
|Dec-17-18|| ||AylerKupp: <<john barleycorn> That's why I took the Pascal/Fermat way>|
OK, I Goggled "Pascal Fermat" and after educating myself by reading some articles and watching one video (another video was over an hour long and I don't think that even I can be that verbose!) I think I understand the reasoning and the approach. But I don't understand how you got the numbers you got; e.g. "139,152 (=2*69,576) ways will end 6,5 - 5,5". Could you please elaborate?
You could then calculate the probabilities for each match score occurring <assuming that the probability of each sequence occurring is equal>. But I still see at least 2 issues:
1. Some of the match scores, like any score where the total number of points scored is over 7, will not be achieved. And similarly, any possible game result sequence where the cumulative number of points scored exceeds 6½ (except in the special case of the score being 6-5 and the player with the 6 score wins the next game) will not be achieved since the match would end at that point. And I still haven't been able to figure out how to figure that out.
2. The probabilities of win, draw, or lose for a player are not the same; not even close. So in order to calculate the probability of each game score sequence occurring you would have to calculate the probability of the result of each game in that sequence and add them all up.
Both of these seem like a lot of work.
<With two players being rated "equal" (in "classical chess" 2835 Elo vs. 2832 Elo being considered "equal")>
They are considered "equal" according to the FIDE scoring table but they're not really equal; they are only "equal" as far as FIDE is concerned when you round the probabilities to 2 significant digits. If you don't then even a 1 rating point difference gives different probabilities, although how significantly this affects the results is open to question. Using 2 significant digits for rating change calculations for a player from one rating period to another is essential if you want to calculate the "official" rating change but pretty much useless (in the sense of being fairly inaccurate) otherwise.
|Dec-17-18|| ||AylerKupp: <<john barleycorn> For the 6-6 draws after 12 games>|
OK, that tells me how you arrived at the 73,789 ways for a 6-6 score to occur. Then knowing the probabilities of wining, drawing, and losing you can calculate the probability of a the 6-6 score occurring. Which is a lot simpler and easier than calculating probabilities of each of the 73,789 cumulative game scores for a 6-6 tie!
Did you do something similar to calculate the other numbers in Tiggler chessforum (kibitz #513)? I think that this approach will not work for any match score <other> than a 6-6 tie or either a 7-5 or 6½-5½ win since the match would terminate before the 12th game was played. But it does represent a useful upper bound.
|Dec-17-18|| ||AylerKupp: <<Tiggler> Thanks for the link to a very interesting article. For a change a report on the match from someone who seems to know what he's talking about. I personally don't consider Carlsen's decision to offer a draw a bizarre one, but I also didn't know why he chose to offer it at move 30 given the complexity of the position and his time advantage relative to Caruana. I thought that his position at the time of the draw offer was better although not necessarily winning, and that he was hardly in a position to lose the game unless he made an outright blunder. And, given his time advantage, the probability of him making an outright blunder in the next 10 moves seemed to be less than Caruana's, particularly since it's usually easier to attack than to defend, and in the position after their 30th move it's Carlsen who has the initiative and would likely be doing most of the attacking.. So I thought that Carlsen's best practical chance was to play on until the time control and only after reaching it (and assuming that Caruana did not make an obvious blunder before move 40, offer the draw at move 41. But Carlsen obviously felt differently.|
I don't understand Roader's table listing the Cumulative Chance of Winning the Championship. I would have thought that given Carlsen's rating advantage over Caruana in games played at Rapid time controls and an even greater rating advantage over Caruana in games played at Blitz time controls that Carlsen's chances of winning the championship would be greater the greater the number of games played at either. And the number of games to be potentially played at Blitz time control (10) was greater than the fixed number of games to be played at Rapid time controls.
So, while he may be correct to calculate the results of individual Blitz matches as he did, <after> each match is completed (because, at least in theory and as a first order approximation), the results of any match is independent of the result of previous matches. So it would seem to me that Carlsen's <cumulative> chance of winning the championship in the Blitz section of the match would increase the greater the number of games played and at a higher percentage that he indicated. And, if each Blitz match is considered as an event independent of previous Blitz matches, why would the percentages not be the same for each match? Oh well.
And Kasparov's comment (I had seen that before, who hadn't?) about Carlsen's draw offer being "shocking" and reconsidering his evaluation of Carlsen being the favorite in rapids were silly. Carlsen made what was to <him> a pragmatic decision given his perceived superiority to Caruana at the faster time controls and his possibly being tired after the pressure of a hard match against an equally strong opponent. He likely evaluated that his chances of winning the championship during the faster time control portions of the match were greater than his winning the championship as a result of winning game 12 from the position after White's 30th move. And I can't argue with him about that conclusion.
Perhaps Carlsen's quote summed his reasoning best, “I wasn’t in a mood to find the punch.” And with a perceived advantage at the faster time controls, he didn't have to find it.
|Dec-17-18|| ||Tiggler: <AylerKupp> Fifethirtyeight.com is a website which specializes in statistical explorations of problems of topical interest, especially politics. The name refers to the number of voters in the Electoral College for US Presidential elections. |
<So, while he may be correct to calculate the results of individual Blitz matches as he did, <after> each match is completed (because, at least in theory and as a first order approximation), the results of any match is independent of the result of previous matches. So it would seem to me that Carlsen's <cumulative> chance of winning the championship in the Blitz section of the match would increase the greater the number of games played and at a higher percentage that he indicated. And, if each Blitz match is considered as an event independent of previous Blitz matches, why would the percentages not be the same for each match? Oh well.>
The calculations are sequential, and the results at the end of each stage is, as stated, cumulative. Thus the difference in the cumulative probability at each stage is just the probability of the match been won during that stage, allowing for the fact the fact that particular stage may not be played at all, having ended before that point was reached.
Regarding <johnbarleycorn>'s post in which he refers to the <Pascal/Fermat way>, this is the basis of the trinomial distribution for the result of N trials with three possible outcomes each time, given by the probabilities Pw, Pd and Pl, respectively. The number of ways a match outcome with W wins, D draws and L loses may be reached is N!/(W!D!L!) and the probability of each one of those is Pw^W*Pd^D*Pl^L . Since a given match score can be reached in multiple manners, depending on the number of draws, these probabilities must be summed over all those that result in the same score. Thus the probability of 6-6 includes the cases with 0, 2, 4, 6, 8, 10, and 12 draws.
All that applies to the case when all N games must be played, N being predetermined before the start. The actual WC matches are, in the terminology I have been using, <censored>: the match ends when either player reaches the winning score. This case is more difficult, but not so hard as I previously supposed. I'll explain in another post.
|Dec-18-18|| ||Gypsy: <AylerKupp: <....I wonder if while we have been talking about Parameter Selection processes, he has been talking Model Selection?>|
Yes, of course I was. I see no point in spending time to select the proper parameters until we settle on the correct model, or at least a model that works best for the uses that we are trying to use it for. ...>
Alright, we are in the right space then. (That is a collective 'we')
An underlying modeling assumptions of the Elo model is this:
(1) For each and every chess-player, we can associate with him/her a 1-D distribution that, on average, reflects his/her playing results against other chess-players.
(2) We equate the <mean> of this 1-D distribution with the particular player's <rating>. These means are the primary model-parameters that we need to somehow estimate from the given data for the whole rating game to have any practical meaning.
(3) The <variance, skewness, kurtosis>, (and the rest of the <cumulants>) of the distribution are the Elo model's <ancillary> parameters.
(4) In the basic model, devised by Elo and used by FIDE, all the ancillary parameters are uniformly fixed: the variances are fixed by setting each standard deviation uniformly to 200; and the rest of the cumulants are fixed uniformly to 0 by the gaussian assumption.
The model implied by the auxiliary parameter-fixing described in (4) is minimal in the sense that only the primary variables, means/ratings, are computed from the data. A different fixing of the ancillary parameters may produce a different minimal model, but, again, only the means/ratings are computed.
I recall <Frogbert> reporting some time ago that his number crunching probably means that the distribution for most players is somewhat skewed and also leptokurtic (heavy-tailed): Most players earn more points from playing down than from playing up; and upsets are more common than they should be under gaussian model.
There also is a possibility to use non-minimal models, where some of the ancillary parameters are estimated directly from the data, along the principal variables. Unfortunately, there tends to be a heavy penalty in the amount of additional data needed for robust estimation of these extra parameters.
With extra parameters to play with, models are prone to start learning the data itself rather than abstract the essential information from it.
Hope this makes some sense.
|Dec-18-18|| ||john barleycorn: <AylerKupp: ... But I don't understand how you got the numbers you got; e.g. "139,152 (=2*69,576) ways will end 6,5 - 5,5". Could you please elaborate?>|
because under "equally" strong players the result should be symmetric, meaning 6,6 - 5,5 is as likely as 5,5 -6,5.
<You could then calculate the probabilities for each match score occurring <assuming that the probability of each sequence occurring is equal>. But I still see at least 2 issues:
1. Some of the match scores, like any score where the total number of points scored is over 7, will not be achieved. And similarly, any possible game result sequence where the cumulative number of points scored exceeds 6½ (except in the special case of the score being 6-5 and the player with the 6 score wins the next game) will not be achieved since the match would end at that point. And I still haven't been able to figure out how to figure that out.>
The Pascal/Fermat way as I called it, is a way to simulate the result when a match is terminated before all games are played. let's say one player (in a 12 game match ) wins the first 7. then the match stopps and he is declared the winner. Is it fair to split the price money 60-40 in that case? Assume the match was played on until 12 games were completed. any result like 7-5, 9,5 - 2,5 etc could be possible. What does 7-0 then mean and as I asked before how would your sample space look like?
<2. The probabilities of win, draw, or lose for a player are not the same; not even close. So in order to calculate the probability of each game score sequence occurring you would have to calculate the probability of the result of each game in that sequence and add them all up.>
Kindly provide the probabilities of win, draw or lose between "equally rated" players.
<They are considered "equal" according to the FIDE scoring table but they're not really equal; they are only "equal" as far as FIDE is concerned when you round the probabilities to 2 significant digits. If you don't then even a 1 rating point difference gives different ...>
I knew that when I cracked the joke that you would tell us that 3 points in Elo difference would materialize in 50,237 games for the higher rated player. But sorry, there is one thing I do not understand. You said that the higher Elo rating and thus the FIDE ranking will materialize and just before you said <They are considered "equal" according to the FIDE scoring table but they're not really equal>.
Where is the difference between a 2835 rated player, one rated 2832? In the opening, the middle game or endgame? In their age, height or shoe size?
|Dec-18-18|| ||AylerKupp: <<Tiggler> As you said on the <Stumpers> forum: You either solve a problem or find "excuses" for not doing so.>|
I think that Yoda said it best: "Do. Or do not. There is no try." https://www.youtube.com/watch?v=BQ4....
|Dec-18-18|| ||AylerKupp: <<Tiggler> Fifethirtyeight.com is a website which specializes in statistical explorations of problems of topical interest, especially politics.>|
Yes, I'm familiar with FiveThirtyEight. I even have (and, more importantly, have read :-) ) Nate Silver's book, "The signal and the noise".
<The calculations are sequential, and the results at the end of each stage is, as stated, cumulative. Thus the difference in the cumulative probability at each stage is just the probability of the match been won during that stage, allowing for the fact the fact that particular stage may not be played at all, having ended before that point was reached.>
But, since the predictions would have been made prior to the start of the match, or at least prior to the tart of the Blitz section, wouldn't it be more appropriate to calculate the probabilities of the match ending after 2, 4, 6, 8, and 10 Blitz games being played? After all, the title of the table is "Cumulative Chance of Winning Championship".
|Dec-18-18|| ||AylerKupp: <Probabilities of winning, drawing, or losing> (part 1 of 2)|
<<john barleycorn> Kindly provide the probabilities of win, draw or lose between "equally rated" players.>
I can't, since I haven't convinced myself of a way to calculate the p(D) based on the information provided by either FIDE or Dr. Elo. I have tried, as I indicated to Tiggler in Carlsen - Caruana World Championship Match (2018) (kibitz #2822), ways to calculate it but I haven't found any to my satisfaction. I think I do have an approach but I have to check it out further.
The best I can do until then is regurgitate the p(W) + p(D)/2 (a.k.a. Scoring probability) for 2 "equally" rated players. The first column is the actual rating differential and the second column is the probability:
If you use my extended precision tables with one entry per rating differential then you have the following probabilities for the p(W) + p(D)/2:
Expanded precision tables:
The numbers are displayed after rounding to 6 decimal places since that's the precision required to display the numbers and still ensure that the probabilities in increments of 1 Elo rating point are unique.
But the following information may be useful based on the ChessTempo database on May 5, 2017 for games between players each rated 2700+ and at least it shows how different the win, draw, and loss percentages are overall for all games.
<Rapid Games: 1,776>
White wins 548 <30.9%>
Draws 818 <46.1%>
White loses 410 <23.1%>
<Blitz Games: 3,205>
White wins 1,215 <37.9%>
Draws 1,106 <31.7%>
White loses 974 <30.4%>
|Dec-18-18|| ||AylerKupp: <Probabilities of winning, drawing, or losing> (part 2 of 2)|
So, as expected, the drawing percentage for games at Rapid time controls is less than the drawing percentage for games at Classic time controls, and the drawing percentage for games at Blitz time controls is less than the drawing percentage for games at either Classic or Rapid time controls.
But these percentages are based almost exclusively on <tournament> games and not <match> games. And, as I pointed out in Tiggler chessforum (kibitz #525), the probabilities for White wins, draws, and losses in WCC matches (400 games) are different than for tournaments. For the 400 WCC games at Classic time controls the breakdown is:
White wins 81 <20.3%>
Draws 274 <68.5%>
White loses 45 <11.3%>
The corresponding breakdown for 8.879 mostly tournament games at Classic time controls is:
White wins 2.329 <26.2%>
Draws 5,153 <58.0%>
White loses 1,397 <15.7%>
There have only been 15 games played at Rapid time controls, including this match at the WCC level and NO games at Blitz time controls. For the 15 games played at Rapid time controls:
White wins 8 <53.3%>
Draws 6 <40.0%>
White loses 1 <6.7%>
If this match's games are not included (as they wouldn't be if you were attempting to predict the most probable score of this match, the equivalent results for the 12 games played at WCC matches prior to this one at Rapid time controls are:
White wins 5 <41.7%>
Draws 6 <50.0%>
White loses 1 <8.3%>
Of course, for Blitz these are very few number of games so the statistics are probably not very reliable in the general sense.
If you do need separate win, draw, and loss probabilities for 2 "equally" rated players both rated 2700+ (or for that matter 2800+ in this case) I would suggest for games at Classical time controls and based on the typical 1-4-1 Normal distribution gross approximation something like if you don't want to consider White's advantage:
White wins <16.7%>
White loses <16.7%>
But that would be just a guess.
|Dec-18-18|| ||john barleycorn: <AylerKupp: <Probabilities of winning, drawing, or losing> (part 1 of 2)|
<<john barleycorn> Kindly provide the probabilities of win, draw or lose between "equally rated" players.>
I can't, since I haven't convinced myself ...>
thanks, and you are not alone here. the learning I take from this is that the "weighting" of outcomes between "equally" rated players are guesswork (more or less educated). The Laplace formula is as good as any other in this case. Meaning to say, we do not know.
|Dec-18-18|| ||Tiggler: Many posts here suddenly, and I ought to reply to them all, I suppose. I was about to post my thoughts on the <censored> trinomial problem, as I had promised, when further talk of the Pascal/Fermat way pulled me up short. I had thought that referred just to the trinomial extension of the Pascal triangle, which arises in consideration of the binomial distribution. But now I see it refers to incomplete games. But not, so far as I can see, to games terminated because one player has actually already won,|
So here are my thought on that case. The winning game, if there is one, in a sequence of games in which the maximum number of games is fixed, is the last one actually played. It must be won by the match-winning player, or it may be drawn. If the possible winning scores are enumerated, then the score prior to the play of the last game must have been either one point or half a point less, for the winning player, because he cannot have lost the game that decided the outcome. Now, the score before the last game was played was reached without any influence of censorship. That means its probability can be calculated from the trinomial distribution governing the number of game then completed without regard to censorship. Only in the last game can the censorship have taken effect.
The rest is just manipulation: enumerate the possible winning scores, calculate the probability of the scores that must have preceded the last game using the trinomial distribution, and then multiply these by the probabilities of win or draw to reach the desired score. The sum of those probabilities, if there is more than one, yields the desired probability for each final score. Finally, it is wise to check by summing the probabilities for all possible end results, to ensure that the total is one.
|Dec-31-18|| ||Tiggler: A new years blessing. In as much as there are not only a few who believe in a divine being, and not wishing to offend such entity, if existent, and whereas it may be all such superstition as rational thought might suggest, nevertheless, and notwithstanding: God bless you all in the new year.|
< Earlier Kibitzing · PAGE 22 OF 22 ·
Advertise on Chessgames.com