The value of White in terms of Elo pts - A statistical study

[Norm Pollock]

The database I used for the study (slightly updated) is now online. It is quite large (44M, so a broadband connection is recommended.) This database (normbk03.pgn) has 199,375 games of master level (2300+ elo) players. Only long time control human v human games. Removed include duplicates, rapid, blitz, blindfold, simultaneous, and Internet games. Also removed are variations, comments, nags and fens. All games have results and both elo values. The games are "clean" and "unique". They were played from 1991 to the present.

To download, go to crafty-chess.com (run by Peter Skinner),
downloads, Pollock, then pick normbk03.zip.

Direct link:

http://www.crafty-chess.com/down/Pollock/normbk03.zip

[Sven Schüle]

For comparison of 2 players you would still take the full elo (the weighted average of the white and black elos) as you do now.

The practical uses of a full/white/black elo system are:
(1) You can tell if a player excels with a particular color, or on the other hand, if the color does not matter.
(2) You can better predict the outcome of a single game.

It would be impossible to have a "true" White-side World Champion because you cannot have a match where both players are playing white. But you could say that one player is the best in the world with white, and another is best in the world with black, and a third is the best overall.

Hi Norm,

how should such white/black elo should be calculated from games?

Consider an event where a (human) player A with an established rating plays 9 opponents, with established ratings, too. 5 games with white, 4 with black. Say A has 2400 "full elo", opponents in A's "white" games have 2300 "full elo" on average, and those in "black" games have 2350. On average, all 9 opponents have 2322. A scores 3,5/5 with white and 2,0/4 with black. Expected was about 3,2/5 with white and about 2,3/4 with black, so A's total score of 5,5/9 roughly matches the expected score.

Now how do you get A's "white/black elo" after that event? I do not understand this up to now, could you explain it, please? Should there be any influence from the involved players' previous "white/black elo" ratings?

Perhaps one should even start with a simpler example: two players B and C having established ratings Bfull, Bwhite, Bblack, Cfull, Cwhite, Cblack are playing a match of N games, how to calculate their new full/white/black ratings after the match?

Sorry for insisting on the details

Sven

[Norm Pollock]

For comparison of 2 players you would still take the full elo (the weighted average of the white and black elos) as you do now.

The practical uses of a full/white/black elo system are:
(1) You can tell if a player excels with a particular color, or on the other hand, if the color does not matter.
(2) You can better predict the outcome of a single game.

It would be impossible to have a "true" White-side World Champion because you cannot have a match where both players are playing white. But you could say that one player is the best in the world with white, and another is best in the world with black, and a third is the best overall.

Hi Norm,

how should such white/black elo should be calculated from games?

Consider an event where a (human) player A with an established rating plays 9 opponents, with established ratings, too. 5 games with white, 4 with black. Say A has 2400 "full elo", opponents in A's "white" games have 2300 "full elo" on average, and those in "black" games have 2350. On average, all 9 opponents have 2322. A scores 3,5/5 with white and 2,0/4 with black. Expected was about 3,2/5 with white and about 2,3/4 with black, so A's total score of 5,5/9 roughly matches the expected score.

Now how do you get A's "white/black elo" after that event? I do not understand this up to now, could you explain it, please? Should there be any influence from the involved players' previous "white/black elo" ratings?

Perhaps one should even start with a simpler example: two players B and C having established ratings Bfull, Bwhite, Bblack, Cfull, Cwhite, Cblack are playing a match of N games, how to calculate their new full/white/black ratings after the match?

Sorry for insisting on the details

Sven

Hi Sven,

Suppose player A has a full elo of 2400 and he is playing a 10 game match with player B who has a full elo of 2300. Suppose neither player has ever before been assigned a white or black elo.

Because each has never been assigned a white or black elo, we would use a 16 point adjustment (or some other arbitrary adjustment). Therefore A now has a white elo of 2416 and a black elo of 2384. B now has a white elo of 2316 and a black elo of 2284.

In game 1, A's 2416 white elo is playing B's 2284 black elo. The result of game 1 would affect A's white elo and B's black elo for game 3, and so forth for each odd-numbered game. Full elos would be computed as usual.

In game 2, the result of game 1 would not matter for A's black elo and B's white elo. A's 2384 black elo would be playing B's 2316 white elo. The result of game 2 would affect A's black elo and B's white elo for game 4, and so forth for each even-numbered game. Full elos would be computed as usual.

Suppose A scored 3.5-1.5 in the odd-numbered games where he played white, and suppose A scored 2.5-2.5 in the even-numbered games where he played black. His full elo change would be based on all 10 games. His white elo change would only be based on the odd-numbered games. His black elo change would only be based on the even-numbered games.

Let's suppose the new elo for A calculates to 2405 full elo, 2440 white elo and 2370 black elo. Since A played an equal number of games with white as with black (5 each), the average of the white and black elos should equal the full elo (except for round-off errors). Otherwise the full elo would be the weighted average of the white and black elos based on how many games played as white and how many games played as black.

-Norm