Chess will never be solved, here's why

@8838

"wheres your data on the number of errors"
++ Another case: Tata Steel Masters 2023
91 games: 60 draws + 31 decisive games

Assuming Chess a white or black win:
60/91 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
No lambda satisfies this equation.
Thus Chess cannot be a white or black win.
Thus Chess is a draw.

Assuming Chess a draw:
31/91 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
lambda = 0.57 errors / game average satisfies this.

Games with 0 errors: 51
Games with 1 error: 29
Games with 2 errors: 9
Games with 3 errors: 2
Games with 4 or more errors: 0

Of course it can be known, @Optimissed. Huge numbers of people "know" all sorts of things that are either uncertain or untrue. But rigour is something else.

AI chess computers beat any human player. AI knows better than us, it is solved. 

tru

THATS NOT DATA

"Assuming Chess a white or black win:
60/91 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
No lambda satisfies this equation.
Thus Chess cannot be a white or black win.
Thus Chess is a draw."

this is objectively false.  even if the assumed poisson distribution were true, this data in no way proves that chess is a draw.

its a probability distribution.

not an absolute.

jesus christ you probably failed middle school math

tygxc the tata steel isnt error data.

you cant be this blind.

@8864

"jesus christ you probably failed middle school math"
++ I do not know about Jesus Christ, or about you, but I studied more math than any here...

@8863

"THATS NOT DATA"
++ Of course that is data.
The data are: 91 games = 60 draws + 31 decisive games.

The explanation is:
Chess is a draw.
Games with 0 errors: 51
Games with 1 error: 29
Games with 2 errors: 9
Games with 3 errors: 2
Games with 4 or more errors: 0

Now come up with an alternative explanation:
Chess is ... a draw / a white win / a black win.
Games with 0 errors: ...
Games with 1 error: ...
Games with 2 errors: ...
Games with 3 errors: ...
Games with 4 or more errors: ...

I dont need to "come up with an alternative explanation"

All I need to do is show how yours isnt sufficient proof.

I learnt this in 5th grade math.

You claim a proof but all you can give is an explanation.  typical.

There was multiple games at tata steel with 4 or more errors lmao

@8867

"I dont need to come up with an alternative explanation" ++ You cannot.

@8868
"There was multiple games at tata steel with 4 or more errors" ++ No.

Abdusattorov_Nodirbek vs Van_Foreest_Jorden 

Erigaisi_Arjun vs Carlsen_Magnus

had 5+ errors

and thats just from the last 2 rounds.

So your "data"

is OBJECTIVELY WRONG.

k

@8870

I tend to agree that Erigaisi - Carlsen had 5 errors:
17 a3?, 17...g5?, 26 Qd3?, 26...Nf7?, 42 Qf1?
I will look at Abdusattorov - van Foreest.

Studying math does not mean somebody understands anything about the data.  If you plug garbage numbers into an equation, you get garbage out.

Your "errors premise" has always been wrong.  If you instead assume chess to be a win, then what are the number of errors per game?  If you were to assume that chess had a single forced winning line for white, how many errors per game for the average human or engine?  One error for white, the first that deviates from the single line.  Black cannot even *make* an error in that case, by your logic.

This comes down to the same faulty thinking every time.  You rely on the notion that chess is already a draw, and you pretend that we can determine where the draws are lost.  You believe that engines are good enough to evaluate all errors (minus 1 in 100,000, which you somehow think humans will then catch) that change the theoretical game state.  They are not.  This error chart you have posited here is a joke.  For every single game you can add pairs of errors that trade the game state back and forth (using your own narrowed definition of "error"), so each formula should say:

Blah blah games 1 error + 2N errors

...or somesuch, as you found out when claiming 0 games with 4 or more errors and were promptly shown a game with at least 5 errors.  

Let me just say that if your explanations fall down in <5 minutes, then the chances of your 3 GMs and some cloud engines making it 5 years without error are a joke.  There will be trillions upon trillions of mistakes...and that is being kind.  The "mistake" pile will be larger than the number of games ever played by humanity...

Seems unlikely.

In any case you claim rests uncomfortably with you not understanding what solving a combinatorial game means. Presumably nothing related to that was in your studies, you haven't gained understanding of the subject since and, unlike most mathematicians, you don't immediately understand the difference between a deduced result and uncertain statistical inference.

Waiting for 8 posts after this to post, and 21 after that.

I posted earlier (#8819)

...

You don't need to guess what the distribution of blunders would be using SF15 by looking at games where the theoretical result of the starting position is unknown, the player is not SF15 and there's no chance of deciding which moves are blunders.

We have actually weakly solved competition rules chess (amended to make the game soluble) for positions with 7 or fewer men on the board, no castling rights and no positions since the previous ply count 0 position that are considered the same for the purposes of the triple repetition rule. 

So you only need to look at a series of games in some of those positions using SF15 itself. Then there's no guesswork.

To make it simple I suggest this series which I've already gone to the trouble of producing for you. Twelve games, twelve draws.

@cobra91 has already checked these games with the Syzygy tablebase and produced a table of blunders.

Now be a good lad and tell us what your Poisson distribution is for those games using only the results without any info from Syzygy and we can see how closely it matches - you've been avoiding it for months. While you're about it, you claim to be able to say what the theoretical result of the initial position is from that distribution, so tell us what that says as well.

For the record, I predict you will continue to do neither because you're interested only in posting crap.

My prediction has so far turned out to be spot on.

So we'll have to do it for you (as usual).

Applying your reasoning to the games shown.

12 games: 12 draws + 0 decisive games

Assuming the initial position a white or black win:
12/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
No lambda satisfies this equation.
Thus the initial position cannot be a white or black win.
Thus the initial position is a draw.

Assuming the initial position a draw:
0/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
lambda = 0 errors / game average satisfies this.

Games with 0 errors: 12

But Syzygy tells us (with the aid of @cobra91):

The position is a Black win.

Games with 0 errors: 0

Games with 1 error: 4
Games with 2 errors: 0
Games with 3 errors: 3
Games with 4 or more errors: 5

(error = @tygxc error = half point blunder or full point blunder counted twice.)

So what went wrong?

Answer: your reasoning.

You have used the hypothesis that blunders per game occur with probabilities that form a Poisson distributution. You gave no reasons why they should and the hypothesis is intuitively unlikely. Obviously they don't.

So you can now check those results and stop posting that argument.

For the record, I predict you will continue to do neither because you're interested only in posting crap.