Measuring methods often operate on different scales. For instance, if you have $8 you have twice as much money as someone who has $4. But an earthquake that is an 8 on the richter scale is 10 times as strong as an earthquake that is a 7, which means an 8 earthquake is 10,000 times as strong as a 4 earthquake.
Now, can we assume that (in theory) a chess player that is twice as good as another chess player will score twice the amount of points in games between the two? That is, the twice-as-strong player will score 66.667% in games between them, whereas the weaker player will score 33.333%, or half as much as the stronger player.
Using this site's win expectancy formula (and assuming I have done the math correctly), a 66.667% score is supposed to happen when a player is approximately 120.5 points higher than their opponent. Which means (in theory) that a player 120.5 points above you is twice as good as you are. If you played 3 games you would expect to win one and lose two, or to lose one and draw two.
Now, the scale doubles every (approximately) 120.5 points. This means that a player that is 241 points above you is 4 times as good (in 5 games you would expect to win one and lose 4), a player 361.5 points above you is 8 times as good, a player 482 points above you is 16 times as good, etc.
This means that the top player on the site is a staggering 1, 412 times as good as the "average" 1200 player, meaning that if he played a 1200 player 1413 times he would expect to win 1412 of them and lose once, or win 1411 of them and draw twice - a score of 99.93%.
Comments?
Originally posted by Jasen777Yes, you've done the math correctly.
Measuring methods often operate on different scales. For instance, if you have $8 you have twice as much money as someone who has $4. But an earthquake that is an 8 on the richter scale is 10 times as strong as an earthquake that is a 7, which means an 8 earthquake is 10,000 times as strong as a 4 earthquake.
Now, can we assume that (in theory) a chess pla ...[text shortened]... of them and lose once, or win 1411 of them and draw twice - a score of 99.93%.
Comments?
Isn't math wonderful?
Originally posted by Jasen777That's right but in the end the thing known as the human factor will probably slightly change that statistical results.
Measuring methods often operate on different scales. For instance, if you have $8 you have twice as much money as someone who has $4. But an earthquake that is an 8 on the richter scale is 10 times as strong as an earthquake that is a 7, which means an 8 earthquake is 10,000 times as strong as a 4 earthquake.
Now, can we assume that (in theory) a chess ...[text shortened]... hem and lose once, or win 1411 of them and draw twice - a score of 99.93%.
Comments?
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Originally posted by ivan2908For me, it also depends a lot on playing styles. Some people who have a mathematical rating below mine play in a way that I find hard to beat - and vice-versa.
That's right but in the end the thing known as the human factor will probably slightly change that statistical results.
Interesting reading but, isn't the idea that a player twice as good as you will beat you twice and lose to you once (on average in 3 games) an assumption?
I have always thought that generally speaking, the better player will continue to win - of course, I'm not saying I am right. I'm just interested to know how a player that is "twice as good as you" equates to losing 2 and winning 1 on average.
It is an assumption, I asked if we could assume it. I think it's rather a semantics issue though. If a baseball team wins 2 out of 3 games in a series, were they twice as good in those three games? Depends what you mean by "twice as good."
The doubling (or halving depending on which way you are going) of expected results every 120.5 points is still there in any case.
Originally posted by Jasen777"Twice" as good is subjective of course, as how "good" you are is not something that can be measured concretely. Still, I think that if A, with "twice" the chess skill of B, plays B, A will win every time or almost every time.
Measuring methods often operate on different scales. For instance, if you have $8 you have twice as much money as someone who has $4. But an earthquake that is an 8 on the richter scale is 10 times as strong as an earthquake that is a 7, which means an 8 earthquake is 10,000 times as strong as a 4 earthquake.
Now, can we assume that (in theory) a chess ...[text shortened]... hem and lose once, or win 1411 of them and draw twice - a score of 99.93%.
Comments?
If A beats B 2 of every 3 games, and B wins the other 33%, then A is only very slightly better than B.
Originally posted by Jasen777Sorry, I didn't read the opening post completely and missed that you had noted it as an assumption. I would be inclined to agree with sh76 though 🙂
It is an assumption, I asked if we could assume it. I think it's rather a semantics issue though. If a baseball team wins 2 out of 3 games in a series, were they twice as good in those three games? Depends what you mean by "twice as good."
The doubling (or halving depending on which way you are going) of expected results every 120.5 points is still there in any case.
Originally posted by HabeascorpThe 1,000 + games is mostly theoretical, I don't think anyone would expect such an event to happen. A better way to look at it is that a true 1200 level player has less than a one in a thousand chance to beat a 2460 player.
are we also assuming that after playing a gm a 1000 odd times the player is not improving at all? In that case he should give up chess.