Chess will never be solved, here's why

@8941

"what a node is" ++ A node is a position plus history and evaluation.
A position is a diagram plus side to move, castling rights and en passant flag.
A diagram is the location of men on the board.

@8942

"all four need to be true for a Poisson distribution"
++ All four apply when the tournament is sufficiently large and sufficiently strong.

   why

DID  I ASK 

Is that guess based just on the fact that you can't think of a Poisson process that might explain it, you can't think of any answer to @MEGACHE3SE's post #8940 , it can't fit the game results in situations where those can be known and it's led you to predict blunder rates for SF that are thousands of times too low in situations where those can be identified, or do you have even more evidence for it? 

hi!

ferret

Tygxc is it not true that multiple errors can occur on a single move?

Tygxc you do realize that the fact that you assume a probability distribution as part of a proof in the first place automatically invalidates the proof?

Idk bro I used to win math competitions exclusively dedicated to proofs when I was younger

@8948

"assume a probability distribution as part of a proof in the first place automatically invalidates the proof?"
++ No, you see this the wrong way.
A theory is good when it can explain observed facts.
A theory is bad when an observed fact contradicts it.
Observed fact: a strong chess tournament has 136 games = 121 draws + 15 decisive games.

Assuming a Poisson distribution leads to:
Chess is a draw
120 games with 0 errors
15 games with 1 error
1 game with 2 errors that undo each other
0 games with 3 or more errors

Now try to come up with any 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 errors: ...

@8947

"Tygxc is it not true that multiple errors can occur on a single move?"
++ Yes: 2 errors (??) can occur in a single move: a blunder (??) that turns a win into a loss.
However, if the observed tournament is sufficiently strong as above @8951, then no games with 3 errors occur and thus no cases of a double error (??).

@8928

You can read here:

'A node, in turn, is a chess position with its evaluation and history, i.e. castling rights, repetition of moves, move turn, etc.'

'Higher speeds are possible to reach only with cloud servers: from 100,000 to 1 million kN/s and even higher if needed.'

@8944

"fit the game results in situations" ++ that are irrelevant to weakly solving Chess.
Of all 10^44 legal positions only the drawn positions are relevant to weakly solving Chess.
Of these only a small subset is needed to weakly solving Chess.
Both 1 e4 e5 and 1 e4 c5 are probably draws, but weakly solving Chess only needs one.

“However, if the observed tournament is sufficiently strong as above @8951”

but that tourney isn’t necessarily sufficiently strong. 
Even if it was it still doesn’t follow Poisson.  The fact that it would even be POSSIBLE for a double error automatically disqualifies it.

Tygxc we aren’t talking about ‘suitable explanations’ we are talking about proofs.  Do you know how to read or understand basic logic?  

you also make an EXTREMELY BASIC ERROR IN LOGIC.

JUST BECAUSE THERE ARE A LOT OF DRAWS AND THE POISSON DISTRIBUTION PREDICTS A BUNCH OF DRAWS DOESNT MEAN THAT THE POISSON DISTRIBUTION IS CORRECT.

You also are ignoring every other way chess doesn’t follow Poisson distribution assumptions.

the fact that you didn’t even bother validating those requirements for the distribution before using it indicates that you really have no idea what you are talking about.

@8957

"that tourney isn’t necessarily sufficiently strong."
++ The ICCF WC Finals is sufficiently strong. As calculated there are at most 2 errors / game,
so there never occurs a double error and the errors are not coupled.
The Tata Steel Masters is not sufficiently strong, as at least one blunder (??) or double error occured changing a win to a loss.
Even then Poisson is a good approximation and the calculation is almost correct.

"the fact that it would even be POSSIBLE for a double error automatically disqualifies it"
++ No, something that could occur but does not occur does not disqualify.

Even if it does occur in rare cases like in Tata Steel Masters, it is still a good approximation.
Maybe for Tata Steel Masters the real distribution of errors per game differs slightly from the one calculated by the Poisson distribution, but the conclusions stay the same.

@8958

"we aren’t talking about ‘suitable explanations’ we are talking about proofs"
++ Newtonian mechanics and gravitation was proven by its ability to explain observed trajectories of planets. Einsteinian relativity was proven by its ability to explain observed differences of Mercurius, Moons of Jupiter, Neptunus and deflection of starlight by the Sun during a solar exclipse.

"Do you know how to read or understand basic logic?" ++ Yes, more than you.

"JUST BECAUSE THERE ARE A LOT OF DRAWS AND THE POISSON DISTRIBUTION PREDICTS A BUNCH OF DRAWS DOESNT MEAN THAT THE POISSON DISTRIBUTION IS CORRECT."
++ You misunderstand the argument. The only plausible way to explain the number of draws and decisive games is to reach the conclusion that chess is a draw, that > 99% of ICCF WC draws are perfect games with optimal play from both sides and < 1% are games with 2 errors that undo each other, that the few decisive games contain exactly 1 error.

Try yourself to come up with an alternative and plausible 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 errors: ...
Games with 5 errors: ...

"doesn’t follow Poisson distribution assumptions"
++ How do you know it does not follow? What then do you think it follows?

"validating those requirements for the distribution before using"
++ That is done in many sciences. I gave an example: a voltage V accelerates an electron with mass m and charge e, what speed v does it reach? Solution: Assume v << c speed of light. Newtonian mechanics apply. Conservation of energy: Ve = mv²/2.  Thus v = Sqrt (2Ve / m). Now check. If v << c, then the result is correct, else switch to relativistic mechanics:
Ve = mc²/Sqrt(1 - v²/c²)

"you really have no idea what you are talking about" ++ I do, you do not.

“The ICCF WC Finals is sufficiently strong. As calculated there are at most 2 errors / game, ”

That hasn’t been calculated yet lmao

"we are talking about proofs" ++ Me too. Newton's mechanics was proven by its ablility to explain observed motions of planets. Einsteinian mechanics was proven by its ability to explain observed differences in the motions of Mercurius, the Moons of Juipter, and deflection of starlight by the Sun during a solar eclipse.

"Do you know how to read or understand basic logic?" ++ I do, you do not.

"BECAUSE THERE ARE A LOT OF DRAWS AND THE POISSON DISTRIBUTION PREDICTS A BUNCH OF DRAWS DOESNT MEAN THAT THE POISSON DISTRIBUTION IS CORRECT"
++ You do not understand the argument. Observed facts are 121 draws in 136 games.
The only plausible explanation is: chess is a draw, 120 games with 0 errors, 15 games with 1 error, 1 game with 2 errors, 0 games with 3 or more errors.

"chess doesn’t follow Poisson distribution assumptions"
++ How do you know? What distribution do you think the errors / game follow?

"validating those requirements for the distribution before using"
++ That is done in many sciences. Example: A voltage V accelerates an electron with charge e and mass m, what speed v does it reach? Solution: assume v << c speed of light. Newtonian mechanics apply: conservation of energy: Ve = mv²/2. Thus v = Sqrt (2Ve/m). Now check. If v <<c then the result is valid, else switch to relativity: Ve = mc²/Sqrt(1 - v²/c²).

"you really have no idea what you are talking about" ++ I do , you do not.