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After decades away from tournament play, I decided to play in an open tourney my child entered. Trying to gauge how my game might have changed during my time away, I found an online rating estimator at Chess Maniac. I was surprised to find that it estimated me 500 points stronger than my (USCF) rating - which actually went down about 60 points at the tournament. So, how accurate is this estimator?

Edit: I realize the Chess Maniac page in question is just an estimator. I'm looking for answers indicating that it tends to overrate or underrate players and by how much. If it doesn't have a tendency one way or another, I'd like to know that.

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Absolutely inaccurate. ELO rating is determined by playing a whole game of chess, not by solving a few positions without any pressure.

Please play some online games for better estimate of your playing strength.

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  • Thanks for the answer. That said, I'm aware of how ELO ratings are calculated; I was hoping to get more specific answers (e.g. "It usually overrates a player by 250 points"). As for playing online, I've played a few dozen games on chess.com; it has generally been a very frustrating experience (to wit: chess.stackexchange.com/questions/22236/… as well as other problems). – GreenMatt Oct 15 '18 at 17:54
  • @GreenMatt I am afraid that you won't get the answer you want. Chess playing strength (or ELO rating if you wish) is determined by lots of factors such as knowledge of tactics, endgames, openings, how easy you tire, blunder rate, etc. It is very possible to have two players of similar rating that are very different in style, e.g. one a wild attacking player that wins but also loses many games due to blunders; and the other a quiet player who draws most of his games, but blunders little. It is completely unrealistic trying to estimate playing strength based on ten puzzles. – user1583209 Oct 15 '18 at 19:19
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As I wrote in a comment above, I also believe that it is unrealistic for many reasons.

You could also think about it from a technical point of view. How would you program the evaluation function of that website (No idea whether that code is public or not.)?

The best you could do I believe is that you have each of the ten positions analyzed by an engine. Then you match the entered moves to the engine evaluation.

  1. In the simplest case you just accept the best possible move and assign something like 200 ELO points to each correct answer (so as to cover roughly 2000 ELO point range from beginner level to World Champion). So you have a huge error just from the method used (not including all the other factors I mentioned in the comment).

  2. If you want to do it a bit more fancy you could also accept second, third, etc. best moves and give ELO points for them, perhaps weighted (with an arbitrary weight). As you can see, all of this seems a bit random.


In summary a few points not related to the technical difficulties mentioned above:

  • The ELO system is made to measure the relative strength of players, not the absolute strength (which might not even exist) as this site is trying to achieve. So by design, the ELO rating only tells you how strong you are relative to all other ELO rated players, not how strong you are overall.

  • Players might have the same rating but completely different strength and weaknesses. How do you expect to measure all the different factors with just 10 puzzles?

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  • 1
    It is possible to come up with a reasonable, if imperfect test, by getting say 1000 players with a variety of ELO ratings, have them take the test, and then build a statistical model. You could estimate the error bars, answering the question. Of course, that is a lot of work, so I doubt that anything like that underlies the Chess Maniac test. If anyone did all that work, it would be worthy of publication! – itub Oct 15 '18 at 20:42
  • @itub: While I can't say how Chess Maniac came up with their "model" (if there actually is anything resembling a statistical model), you can see how the test works, since it is in JavaScript that you can download. That part works on the latter 1/2 of your idea: Answering A on position 1 yields one rating value; B another value and so on. Then the stated rating for the test taker is the average of the values over all the positions at the end. – GreenMatt Apr 2 at 18:09

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