Value of center formations?

[Henk]

I wonder what would be value of center formations like:

white [c2] d4 e4 f2
black [c7] d6 e5 f7

white c2 e4 f2
black c7 d6 f2

white c4 d4 f2
black c7 e7 f7

etc.

Maybe just play some hundred games with these positions and collect results.
Or would it be thousands of games. If so then they make no difference.

https://www.chessable.com/blog/best-chess-formations/

[dangi12012]

You could load the open lichess database -
https://database.lichess.org/

Load it into this with --ply 999 and limit to above 2k elo
https://github.com/nguyenpham/oobs

Then you can query it with PQL directly. (not documented in the readme - but here: https://github.com/nguyenpham/ocgdb)
If memory serves it takes around 11 seconds for a query like that written in PQL per GB of data (which is many tens of thousands of games)

What you want is the average winrate when these patterns occur for a player.

Excercise for the reader: Find all center formations and create a scatter plot of winrate vs formation. I am sure that there are some negative correlated center formations as well (IE very bad structure).

[Chessnut1071]

You could load the open lichess database -
https://database.lichess.org/

Load it into this with --ply 999 and limit to above 2k elo
https://github.com/nguyenpham/oobs

Then you can query it with PQL directly. (not documented in the readme - but here: https://github.com/nguyenpham/ocgdb)
If memory serves it takes around 11 seconds for a query like that written in PQL per GB of data (which is many tens of thousands of games)

What you want is the average winrate when these patterns occur for a player.

Excercise for the reader: Find all center formations and create a scatter plot of winrate vs formation. I am sure that there are some negative correlated center formations as well (IE very bad structure).

rather than a scatter plot, you might try a designed experiment using a 2-way ANOVA table with interactions-columns[Openongs] & rows[defenses]. Without the blocking factors the data would be too noisy.