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No matter whether I test it online on lichess or locally: Stockfish shows different variations and evaluations depending on the number of variations displayed.

Where does this behaviour come from, and how to find out SF's true objective evaluation for a position?

These images are from a locally run Stockfish 15, where it analyzed the position after 1. e4. It was interesting to watch how the different openings rise and fall in Stockfish's evaluation. For example, Stockfish considered the Ruy Lopez as the best for White against 1. ..e5, but didn't find anything against it after looking deeply enough, so it switched to the Italian, which it had thought not to be promising for an edge all the time up to depth 44. This is similar to what we see in top-level chess nowadays, the Italian game is back after a long break!

Seven variations after 1. e4 , local SF 15

However when rerunning the experiment with only 3 variations, Stockfish still tries to combat the Berlin defense. It even thinks that the Caro-Kann is the best reply against e4 and only changed mind at depth 46, where it considered 1. ..e5 the best and 1. ..c5 slightly better than the Caro-Kann.

Three variations after 1. e4, local SF 15

I have noticed the same on lichess frequently. (Side note: this isn't a question about openings; it happens in other positions as well.)

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  • Please use the search function. This question has been asked and answered before.
    – Brian Towers ♦
    Sep 8, 2022 at 13:14
  • Are you sure? I could only find two questions, that are related, but not quite the same: 22300 and 25278.
    – Hauptideal
    Sep 8, 2022 at 13:41
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    I would guess that the answer at chess.stackexchange.com/questions/25278/… would also apply here. Having to search the other moves would certainly change what's in the hashtables.
    – D M
    Sep 9, 2022 at 2:19
  • @DM that's what I assume, too. But as this is a FAQ site that also helps people who google stuff it makes sense to put an answer to this question, so that it can be found when searching for an answer online.
    – Hauptideal
    Sep 9, 2022 at 13:40
  • Ask ChatGPT something naunced twice in a row.
    – Starship - On Strike
    Apr 5 at 19:18

1 Answer 1

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This is largely due to the mathematics behind the game in my opinion, you let it run from 1. e4, this is an inconceivable amount of possible games/positions, so in short, the engine is calculating an extreme amount of variations and lines. Compare this to if you did this with an endgame or even a late middlegame, I suspect that no matter how many lines you ask it to calculate (within reason) it will come up with the same best move more or less every time, and will begin to rank other moves in relation to the best.

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  • Any sources? Anything more than in my opinion?
    – Starship - On Strike
    Apr 25 at 17:00
  • Im not sure how youd even go about googling this question to find sources, but its mathematically intuitive due to the sheer volume of possible games from the first couple moves, Claude Shannon could be a person to look into. [en.wikipedia.org/wiki/Shannon_number]
    – Crogmcrob
    May 1 at 5:32
  • They way you would go about googling it would be too.... type in a related search term and look at the results. In my opinion answers generally arent a great fit here. If you some actual calculations/sources then the answer is valid. Otherwise, it has no validity as it is backed by nothing but "i think" and "in my opinion" @Crogmcrob
    – Starship - On Strike
    May 1 at 9:47
  • Shannons number is estimated on the assumption that the average position has 40 legal moves, so the formula for the amount of possible games after n plies (each move by a player is a ply) would roughly be on the order of 40^n, which shoots off to infinity very quickly, 40^8 for example is 6.5536 x 10^12. This takes computing time to calculate and is subject to massive change at deeper calculation
    – Crogmcrob
    May 2 at 2:10

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