Chess will never be solved, here's why

@4640
No, not at all.
> 99% of ICCF draws are perfect games. In none is the 50-moves invoked.
You can safely ignore the 50-moves rule.
The solution to chess is the same as with the 50-moves rule.

Likewise the stalemate rule plays no role either.
As AlphaZero has shown, even if stalemate is made a win, chess still remains a draw.

The 50-moves rule and the stalemate rule change the optimal value of some positions,
but such positions are not reached in a perfect game.

I proclaim that your proclamation is false.

Beat that!

Has chess been solved? No
Can chess be solved? Yes, it takes 5 years on cloud engines.
Will chess be solved? Maybe, it depends on somebody paying 5 million $ for the cloud engines and the human assistants during 5 years.

Have humans walked on Mars? No
Can humans walk on Mars? Yes
Will humans walk on Mars? Maybe, it depends on somebody paying billions of $ to build and launch a spacecraft.

Hi, can you read my name backwards and change your diapers?

Reported.  Enjoy your vacation.

@4642
"I proclaim that your proclamation is false."
++ I prove, you proclaim.

You have yet to prove anything.

@4646
"You have yet to prove anything."
++ I have proven many things so far.
Some do not read or do not understand.

Sveshnikov proclaimed:
"Give me five years, good assistants and the latest computers
- I will bring all openings to technical endgames and "close" chess."
I have proven him right.
In 5 years there are 157788000 seconds. The latest computers calculate a billion positions per second. So 3 such computers - or 3000 desktops - can calculate all 10^17 (hundred million billion) relevant chess positions in 5 years to weakly solve chess.

@Elroch proclaimed the 50-moves rules plays some role in weakly solving chess.
I have proven him wrong.
Using statistics and probability - which he should understand - and looking at a sufficiently large tournament with sufficient level and a sufficient number of independent players e.g. the 30th ICCF World Championship finals I have proven that several 1000 of such games are perfect games with no error. In none of those perfect games and in none of the imperfect games either was the 50-moves rule invoked. Thus the 50-move rule plays no role in weakly solving chess.
The weak solution of chess without the 50-moves rule is the same as with the 50-moves rule.

No, and no.

Chess is not closed out, and there is no proof of any perfect games whatsoever.

@4648
"No, and no."
++ Because you proclaim so?

"Chess is not closed out"
++ Chess is not yet weakly solved, but can be weakly solved in 5 years.

"there is no proof of any perfect games whatsoever"
I have provided proof. If you do not understand it, then that is OK.

You haven't provided proof.  You have provided faulty claims of perfect games based on imperfect evaluations.  These claims are no better than the ARB Systems guy or that Tasmanian Devil crackpot.

"I have provided proof. If you do not understand it, then that is OK"

The proof was probability calculations based on statistics, but I've shown you how they are inaccurate because there are unknown factors there (hence we dont even know how inaccurate the calculations truely are) and this is why you cant call it proof. Sure, chess is likely a draw and possibly there are games in which no mistakes happen, but it is not mathematically proven.

Oh guys,

we receive daily and several times a day feeding messages from this blog.
It’s up to the one who wants to be right and has the biggest head. You’re seriously starting to inflate us with your desires to always be right. There is not one to put in front, except for the few who have understood that it was better to withdraw from these fruitless discussions.
In a good way, hello

@4650
"You have provided faulty claims of perfect games based on imperfect evaluations,"
++ No, I have not based my proof on any evaluations, just on statistics and probability.
I first provided a simplified proof with high school math only.
@Elroch then proposed a Poisson distribution of errors / game.
So I repeated the calculation using the Poisson distribution, which is college math.

I try to explain again, hoping you read and understand.
The 30th ICCF World Championship finals had 136 games = 127 draws + 9 decisive games.
An error (?) is a move that worsens the game state from draw to loss, or from win to draw.
A blunder or double error (??) is a move that worsens the game state from win to loss.

First assume chess being a white or black win.
Try to fit a Poisson distribution so there is 127/136 probability of an odd number of errors.
It is impossible.
Thus chess is a draw.

Now assume chess a draw.
Try to fit a Poisson distribution so there is 9/136 probability of an odd number of errors.
It is possible with a mean value of 0.071 errors / game.
This leads to 127 perfect games with 0 errors,
99.7% certain, 0.3% chance of 1 game with 2 errors (?) and (?) that cancel out.
and 9 games with exactly 1 error (?),
99.91% certain, 0.09% chance of 1 game with 3 errors: either ?,?,?, or ? and ??
So that leads to 127 perfect games from the 30th ICCF World Championship Final.

Likewise we harvest perfect games from the other ICCF WC Finals.

In none of the ICCF WC finals was the 50-moves rule invoked.
So the 50-moves rule was not invoked in perfect games, a subset of the weak solution of chess.

The same reasoning applied to e.g. the Zürich 1953 Candidates' Tournament also gives 74 perfect games without telling which games these are.

@tygxc, you have a problem distinguishing between "unlikely" and "impossible".  For reference, these are not the same thing.  You don't even know what a proof is.

Even a known-to-be-optimal probability model + empirical data can never provide certainty.

[I won't confuse things by drawing on the fact that the model concerned is an imperfect simplification].

Your posts are a raging blunderfest. For example, post #4647 contains 7 falsehoods.

...and it goes on to say who’s right.
One thing is certain:
- failures will be solved, or will not be solved, before you have finished your cockfighting!

Poisson distribution doesn't work in this context because when an error occurs in a chess game, this error will change the position and the continuation of the game, affecting probability of future errors. 

Say chess is a win. Theres a certain probability, likely very high that an error will occur because at the start there are very few lines that force a win. Once an error occurs, the probability for another error to be made is now lower because there are more lines available to force a draw.

@4654

"you have a problem distinguishing between "unlikely" and "impossible"."
++ No. It is impossible to fit a Poisson distribution with 127/136 probability of an odd number of errors. Try it yourself if you do not believe that.

"A known perfect probability model + empirical data can never provide certainty."
++ I even indicated the certainty: 99.7% certain that the 30th ICCF WC Finals had 127 perfect games with 0 error, 99.91% certain that it had 9 games with 1 error.

"the model concerned is an imperfect simplification"
++ Please then come up with a model that you think is more perfect and show what results that model yields. The result will be more or less the same.
I first came up with a simpler model with high school math only. It lead to 126 perfect games, 9 games with 1 error and 1 game with 2 errors that cancel out.
Then you proposed Poisson for errors / game. So I did that and the results are about the same.

@4656

"when an error occurs in a chess game, this error will change the position and the continuation of the game, affecting probability of future errors."
++ That may well be right, but the Poisson calculation shows 127 games with 0 error and 9 with 1 error. There is only 0.3% chance of a game with 2 errors, which might affect each other. So the possible affection of errors plays no role. Then there is only 0 or 1 error, then errors affecting each other does not happen.

"Say chess is a win."
++ It is impossible to fit a Poisson distribution with 127/136 chance of an odd number of errors.

"at the start there are very few lines that force a win"
++ OK let us assume there is only 1 line that forces a win.
What would that look like? 1 e4? 1 d4? 1 c4? 1 Nf3? 1 a4? 1 g4? Let us assume 1 e4 were a forced win, like Rauzer said in the previous century. We have ICCF WC Finals games starting with 1 e4 and ending in draws. Some of these go 1...e5. Some go 1...c5 as well, but let us disregard that.

What would the winning line look like? 2 Nf3? 2 Nc3? 2 d4? 2 Bc4? 2 f4? 2 a3? 2 g4?
Let us assume 2 Nf3. We have ICCF WC Finals games going 2 Nf3 and ending in draws.
Some of these go 2...Nc6, some go 2...Nf6 as well, but let us disregard that.

What would the winning line look like? 3 Bb5? 3 Bc4? 3 d4? 3 Nc3? 3 c4? 3 a3? 3 g3?
Let us assume 3 Bb5. We have many ICCF WC Finals games going 3 Bb5 and ending in draws. Some of these go 3...Nf6, some go 3...a6 as well, but let us disregard that.

You see, each time the move of the one supposedly winning line meets like 2 lines drawing in ICCF WC Finals games, taking years with 2 ICCF (grand)masters and engines.

"Once an error occurs, the probability for another error to be made is now lower because there are more lines available to force a draw."
++ If chess were a white win, then as soon as white makes an error, we have a drawn position. The probability of another error does not matter then. The examples above show that if white makes no error, then there are not 1, but 2 lines available to force a draw.

"It is impossible to fit a Poisson distribution with 127/136 chance of an odd number of errors"

Because poisson distribution does not apply to this scenario. Errors would need to be independent events that have no interference with each other. It couldnt fit poisson distribution, because it expects each event to have the same, independent probability. With chess it could go for example like this (ofc im pulling random numbers out of somewhere):

Probability for white to make error #1: 99,9%

Probability for black to make error #1: 5%

Probability for white to make error #2: 2%

Probability for black to make error #2: 9%

And each error made will determine probability for next error for both sides depeding on the position and quality of error. You can't attempt to use poisson distribution for this chaos. This is why you cannot come up with proof based on the fact that hypothesis "chess is a forced win for white" doesnt fit poisson distribution with the stats you use.

And edit.

"What would the winning line look like?"

Ofc I dont know what it would look like. I dont even think it exists, all im saying these models cant prove it doesnt exist.