Then, as all the rest of us said, it hugely depends on what kind of sub-200-game beginners you're thinking about.Jocke wrote:Exactly!I think having that ranking is a definition of being good at go. That's my motivation.
Which ranking does always beat beginners?
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billywoods
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Re: Which ranking does always beat beginners?
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Bill Spight
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Re: Which ranking does always beat beginners?
Then don't ask.Boidhre wrote:It may be the OP wants to avoid the situation of sitting down across from someone, getting thrashed, and asking how long they've been playing and getting "3 months" back as an answer after they themselves have been playing two years.Dusk Eagle wrote:I disagree that being able to always beat beginners at any game, not just Go, defines being good at the game. Good players can always beat players who can always beat players who can always beat players who can always beat players who can always beat beginners at a game.
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Visualize whirled peas.
Everything with love. Stay safe.
- Bantari
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Re: Which ranking does always beat beginners?
Very interesting and unusual question. Interesting.
Still, if I had to answer this, I see two aspects.
1. What rank can one reach within 200 games, and
2. What rank will always beat that rank.
Taking into account that ranks indicate only statistical likelihood of beating a person, there might never be a 100% (or 0% chance.) Which brings up a third question:
3. What winning likelihood would you accept? 10%? 1%? 0.1%? 0.01%? Or what?
Question #1 is crucial here, and I have no clue how to answer. Not sure if anybody does - it just depends on too many factors.
In more practical terms, as a very inexact approximation, I'd say this:
1. Lets assume average person can reach up to 10k if they are really trying hard, and
2. Lets assume 1d has enough winning chance against 10k to satisfy you.
So just get to 1d and you're golden. Heh...
Still, if I had to answer this, I see two aspects.
1. What rank can one reach within 200 games, and
2. What rank will always beat that rank.
Taking into account that ranks indicate only statistical likelihood of beating a person, there might never be a 100% (or 0% chance.) Which brings up a third question:
3. What winning likelihood would you accept? 10%? 1%? 0.1%? 0.01%? Or what?
Question #1 is crucial here, and I have no clue how to answer. Not sure if anybody does - it just depends on too many factors.
In more practical terms, as a very inexact approximation, I'd say this:
1. Lets assume average person can reach up to 10k if they are really trying hard, and
2. Lets assume 1d has enough winning chance against 10k to satisfy you.
So just get to 1d and you're golden. Heh...
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WARNING: This post might contain Opinions!!
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WARNING: This post might contain Opinions!!
Re: Which ranking does always beat beginners?
I would say 0.1 %, one in a thousand.Bantari wrote:Very interesting and unusual question. Interesting.
Still, if I had to answer this, I see two aspects.
1. What rank can one reach within 200 games, and
2. What rank will always beat that rank.
Taking into account that ranks indicate only statistical likelihood of beating a person, there might never be a 100% (or 0% chance.) Which brings up a third question:
3. What winning likelihood would you accept? 10%? 1%? 0.1%? 0.01%? Or what?
Question #1 is crucial here, and I have no clue how to answer. Not sure if anybody does - it just depends on too many factors.
In more practical terms, as a very inexact approximation, I'd say this:
1. Lets assume average person can reach up to 10k if they are really trying hard, and
2. Lets assume 1d has enough winning chance against 10k to satisfy you.
So just get to 1d and you're golden. Heh...
I think this was a very interesting answer.
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Claint
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Re: Which ranking does always beat beginners?
Since you have defined the expected win percentage, which is 0.001, I can answer the problem from, say my side, using ELO rating system and EGF ranks.
I am pretty sure I haven't played 200 19x19 games yet, but I might be close or over if you count the smaller board games. My rating is 5-6kyu ish KGS. Let's say, I am the guy and I am 6kyu.
According to ELO formula the expected win percentage is magnified 10 times with each rating difference of 400. So for 0.001 chances, you need a rating difference of 1200.
So with EGF ranks: 6kyu = 1500 ELO rating.
That means we need a guy with 2700 rating. Which again with EGF ratings is equal to 7 dan amateur or 1d professional.
Note: This is a quick calculation and is probably wrong, since according to reference, EGF modifies the normal ELO formula somewhat. But 0.001 is a mighty difference.
Ref:
http://en.wikipedia.org/wiki/Elo_rating_system
http://en.wikipedia.org/wiki/Go_ranks_and_ratings
I am pretty sure I haven't played 200 19x19 games yet, but I might be close or over if you count the smaller board games. My rating is 5-6kyu ish KGS. Let's say, I am the guy and I am 6kyu.
According to ELO formula the expected win percentage is magnified 10 times with each rating difference of 400. So for 0.001 chances, you need a rating difference of 1200.
So with EGF ranks: 6kyu = 1500 ELO rating.
That means we need a guy with 2700 rating. Which again with EGF ratings is equal to 7 dan amateur or 1d professional.
Note: This is a quick calculation and is probably wrong, since according to reference, EGF modifies the normal ELO formula somewhat. But 0.001 is a mighty difference.
Ref:
http://en.wikipedia.org/wiki/Elo_rating_system
http://en.wikipedia.org/wiki/Go_ranks_and_ratings
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Boidhre
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Re: Which ranking does always beat beginners?
I don't think you can assume that winning percentage in go increases by the same factor for each rank once you go past 2 or 3 stones difference.Claint wrote:Since you have defined the expected win percentage, which is 0.001, I can answer the problem from, say my side, using ELO rating system and EGF ranks.
I am pretty sure I haven't played 200 19x19 games yet, but I might be close or over if you count the smaller board games. My rating is 5-6kyu ish KGS. Let's say, I am the guy and I am 6kyu.
According to ELO formula the expected win percentage is magnified 10 times with each rating difference of 400. So for 0.001 chances, you need a rating difference of 1200.
So with EGF ranks: 6kyu = 1500 ELO rating.
That means we need a guy with 2700 rating. Which again with EGF ratings is equal to 7 dan amateur or 1d professional.
Note: This is a quick calculation and is probably wrong, since according to reference, EGF modifies the normal ELO formula somewhat. But 0.001 is a mighty difference.
Ref:
http://en.wikipedia.org/wiki/Elo_rating_system
http://en.wikipedia.org/wiki/Go_ranks_and_ratings
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illluck
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Re: Which ranking does always beat beginners?
Agreed, I doubt I'd lose a game in a thousand against KGS 6k.Boidhre wrote:I don't think you can assume that winning percentage in go increases by the same factor for each rank once you go past 2 or 3 stones difference.Claint wrote:Since you have defined the expected win percentage, which is 0.001, I can answer the problem from, say my side, using ELO rating system and EGF ranks.
I am pretty sure I haven't played 200 19x19 games yet, but I might be close or over if you count the smaller board games. My rating is 5-6kyu ish KGS. Let's say, I am the guy and I am 6kyu.
According to ELO formula the expected win percentage is magnified 10 times with each rating difference of 400. So for 0.001 chances, you need a rating difference of 1200.
So with EGF ranks: 6kyu = 1500 ELO rating.
That means we need a guy with 2700 rating. Which again with EGF ratings is equal to 7 dan amateur or 1d professional.
Note: This is a quick calculation and is probably wrong, since according to reference, EGF modifies the normal ELO formula somewhat. But 0.001 is a mighty difference.
Ref:
http://en.wikipedia.org/wiki/Elo_rating_system
http://en.wikipedia.org/wiki/Go_ranks_and_ratings
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Re: Which ranking does always beat beginners?
You also need to play thousand games to lose one in a thousand.
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illluck
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Re: Which ranking does always beat beginners?
We are talking about ptobability here - even if my winning chance is 99.999% I could still lose the first game.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
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Bill Spight
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Re: Which ranking does always beat beginners?
Actually, if you play 693 games at those odds, the probability is 0.500 that you will lose one.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
Edit: Corrected, as lightvector pointed out.
Last edited by Bill Spight on Sun Jun 02, 2013 1:48 pm, edited 2 times in total.
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
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lightvector
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Re: Which ranking does always beat beginners?
Do you mean 690? (Actually, 692 or 693):Bill Spight wrote:Actually, if you play 69 games at those odds, the probability is 0.500 that you will lose one.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
(999/1000)^692 ~= 0.50040
(999/1000)^693 ~= 0.49990
An interesting source of statistics about winning chances between mismatched players in even games can be found here:Claint wrote:My rating is 5-6kyu ish KGS. Let's say, I am the guy and I am 6kyu.
According to ELO formula the expected win percentage is magnified 10 times with each rating difference of 400. So for 0.001 chances, you need a rating difference of 1200.
So with EGF ranks: 6kyu = 1500 ELO rating.
That means we need a guy with 2700 rating. Which again with EGF ratings is equal to 7 dan amateur or 1d professional.
Note: This is a quick calculation and is probably wrong, since according to reference, EGF modifies the normal ELO formula somewhat. But 0.001 is a mighty difference.
http://gemma.ujf.cas.cz/~cieply/GO/statev.html
The most basic versions of the Elo model assume that if we have players A, B, C and the odds of player A beating B are 1:x (that is, a probability of 1/(x+1)), and the odds of player B beating C are 1:y, then the odds of player A beating C are 1:xy.
If we assume that this is true, then roughly eyeballing the stats on the G+4 column and multiplying up the odds for a 6k beating a 2k, and a 2k beating a 3d, and a 3d beating someone around 6d or 7d, it does actually look like you need to go up to about 6d or 7d before you reach an odds ratio of 1:999. Indeed, 0.001 is a mighty difference.
However, it's common wisdom that the chance of a weaker player beating someone much stronger falls off much faster than this sort of model would suggest. The stats on that site seem to bear this out. In almost every case, if you multiply up the appropriate odds ratios for the stats in the G+1 column and/or G+2 columns to get a prediction for the G+4 column, you obtain an odds ratio that's smaller than the actual one. For example, multiplying up the ratios for 6k beating 5k, 5k beating 4k, 4k beating 3k, 3k beating 2k, you get a odds ratio of 1:2.69, but the empirical ratio for 6k beating 2k is 1:3.67.
Whereas this effect is clear in the G+4 column, if you do the same thing for the G+3 column, you only barely see this effect. If you do it for the G+2 column, you basically don't see it at all and the simplistic model matches up pretty well relative to the noise in the stats.
Based on this trend and some intuition, it seems likely that the effect will continue to increase as the strength difference between the players increases, although there's no data to pin down how fast it will increase. But playing with these stats in a spreadsheet with various extrapolations for how fast, it looks like anywhere from about 2d to 5d would be a plausible guess for how strong you need to be to have only a 0.1% chance of losing against a 6k.
Last edited by lightvector on Sun Jun 02, 2013 1:47 pm, edited 1 time in total.
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Bill Spight
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Re: Which ranking does always beat beginners?
Thanks.lightvector wrote:Do you mean 690? (Actually, 692 or 693):Bill Spight wrote:Actually, if you play 69 games at those odds, the probability is 0.500 that you will lose one.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
(999/1000)^692 ~= 0.50040
(999/1000)^693 ~= 0.49990
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
At some point, doesn't thinking have to go on?
— Winona Adkins
Visualize whirled peas.
Everything with love. Stay safe.
- Shinkenjoe
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Re: Which ranking does always beat beginners?
OK. After winning 693 ganes with this porbability you have to star fearing.Bill Spight wrote:Actually, if you play 693 games at those odds, the probability is 0.500 that you will lose one.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
Edit: Corrected, as lightvector pointed out.
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illluck
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Re: Which ranking does always beat beginners?
Sorry, still wrong XDShinkenjoe wrote:OK. After winning 693 ganes with this porbability you have to star fearing.Bill Spight wrote:Actually, if you play 693 games at those odds, the probability is 0.500 that you will lose one.Shinkenjoe wrote:You also need to play thousand games to lose one in a thousand.
Edit: Corrected, as lightvector pointed out.
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Re: Which ranking does always beat beginners?
You have to be very afraid - that your opponent has gained seven or eight stones while you've been training him...illluck wrote:Sorry, still wrong XDShinkenjoe wrote:OK. After winning 693 ganes with this porbability you have to star fearing.Bill Spight wrote: Actually, if you play 693 games at those odds, the probability is 0.500 that you will lose one.
Edit: Corrected, as lightvector pointed out.