Chess will never be solved, here's why

Hypthesis 6. An error is not only what change a draw in a loss or a win in a draw, as you stated before. A more severe error can change a win in a loss. So if, let's say, Black makes a mistake and the game change from drawn to lost, but White makes a blunder, the game can turn into a win for Black, thus you can have a decisive game with an even number of errors. The same reasoning applies to drawn games, which may contain an odd number of errors.

#2378
"Petrosian drew more than Kasparov, he played better than Kasparov."
++ That is not what I said. The draw rate in a top tournament is indicative of the level of play. I took the most ICCF world championship at 5 days per move, engines allowed as an example of high level play.
The lifetime record of a player is indicative of the level of opposition he faced:
how much they were above their peers.
Petrosian had a 55% draw rate, Kasparov 47%, Morphy 8%.

"An error is not only what change a draw in a loss or a win in a draw, as you stated before."
++ error (?) = move that changes the game state from draw to loss, or from won back to draw
blunder (??) = double error = move that changes the game state from won to lost

I deleted the first part before you posted. Actually, Kasparov's percentage of losses is lower than Petrosian's (not considering exhibitions, blitz games, etc.) and yes, the number of errors in a game depends on both players, so I will take my time to explain better what I mean.

Following your reasoning, that would imply that the probability to make a blunder is lower than the probability to make an error, but that's another hypothesis.

#2379
"the probability to make a blunder is lower than the probability to make an error"
++ Yes, that is correct.
In ICCF they do make errors once in a while, but blunders are very, very rare.

We cannot know how many errors and blunders (in that precise meaning we gave) they made, until chess is strongly solved. If a game ends in a draw, it could be because White makes an error and Black makes an error; or because White makes an error, Black makes a blunder, White another blunder and Black another error, for example. Nor we can treat "errors" measured as difference between the expected (by the avaluation function) values of positions before and after a move, like errors measured as difference between the game-theoretic values of positions before and after a move.

#2381
"We cannot know how many errors and blunders (in that precise meaning you gave) they made, until chess is strongly solved."
++ Yes, we can.  It is enough for chess to be ultra-weakly solved, i.e. the game-theoretic value of the initial position has been determined, i.e. for chess = a draw. We do not know exactly how many, but we know if it is odd or even. We do not know which moves, but we know their parity.

"ultra-weakly solved means that the game-theoretic value of the initial position has been determined, weakly solved means that for the initial position a strategy has been determined to achieve the game-theoretic value against any opposition, and strongly solved is being used for a game for which such a strategy has been determined for all legal positions." - van den Herik

By the way "a strategy has been determined" means that one strategy is enough.
So once 1 e4 e5 is proven a draw, there is no need to investigate if 1 e4 c5 draws as well or not.

"Yes, we can": W-H-Y? Even supposing that the game value is a draw? Even if a blunder counts as two errors, how do you know that the probability to make a blunder is much less than the probability to make an error? Read again my previous post, I have added some things before I could read yours. The server does not notify me of new posts, don't know why. I have to refresh the page every time. And sorry, I am making a lot of errors; have edited this post at least 10 times. I apologize.

No it doesn't as I already pointed out.

The strategy for White

Play 1.Qa7+

is not a weak solution for this position, but it does achieve the game-theoretic value against any opposition. 

#2383
"how do you know that the probability to make a blunder is much less than the probability to make an error?"
++ Because a blunder is a double error. Example: 1 e4 (draw) e5 (draw) 2 Ba6? (lost)  Nxa6 (won) 3 Nf3 (lost) Qh4?? (lost). A blunder (??) does in one move what two errors do in two moves. That is why the probability of 1 blunder = probability of 2 (consecutive) errors. It is just the same whether you throw away the win (?) on move 7 and then lose (?) on move 8, or blunder (??) on move 7.

"I am making a lot of errors; have edited this post at least 10 times. I apologize."
++ No problem, I sometimes edit for clarity too. I always quote the phrase I reply to.

#2384
You should write a letter to van den Herik.

No, you should stop talking nonsense. Are you incapable of thinking for yourself?

There's no point in offering to weakly solve chess for us if you don't even know what "weakly solve" means. 

#2387
I am perfectly capable of thinking for myself and of seeing that you talk nonsense.
I adopt a generally accepted definition by one of the leading authorities in the world.
You then come up with a nonsense refutation, as usual.

It's not a generally accepted definition. I think you are possibly the only person posting on the the thread that would accept it.

But then you also accepted the Wikipaedia definition with which it is incompatible.

Like the White queen, you find it easy to believe six impossible things before breakfast.

#2389

"It's not a generally accepted definition." ++ Sorry, but it is in the world of game theory.

"I think you are possibly the only person posting on the the thread that would accept it."
++ If that were true, then I would be the only sensible person on the thread.

"But then you also accepted the Wikipaedia definition with which it is incompatible."
++ I was inclined towards accepting the Wikipedia definition, but I got criticised for that, as people did not accept Wikipedia as a reliable source. Anyway there already is enough confusion about many things, so we should at least agree to accept the same definitions as given by reputable sources. If we start to assign different meanings to the same word, then the confusion can only grow. You cannot refute a definition. You can only accept it.
https://reader.elsevier.com/reader/sd/pii/S0004370201001527?token=5EA63AC2A4C94EC9333C82E5E39A3A25210F8FC184F9458F8141645F002E759F7CEABB7B09D64FA30FBEB25ABD8CB6E2&originRegion=eu-west-1&originCreation=20220317135832 

To me this is hypothesis 7. The effect of a blunder is that of two consecutive errors; that does not mean it has the same probability to occur as two consecutive errors. Besides, you are assuming that the probability to make an error is little, so the probability to make two consecutive errors can be neglected. But you infer that the error rate per move is low, by the fact that many games are drawn, as if they contained only a few errors, while you have admitted that:

So you assume that the number of errors per game is low, from that deduce the error rate per move is low, from that a much lower probability of two consecutive errors, and from that you assume that the probability to make a blunder is as low as two consecutive errors, because it has the same effect. If the error rate per move, instead, was not so low, the engine would not really know what it's doing (more on that later), and even if the last assumption was correct, the probability of two consecutive errors or a blunder could not be neglected.

#2391

"The effect of a blunder is that of two consecutive errors; that does not mean it has the same probability to occur as two consecutive errors."
++ Can you agree that the probalitity of an error on move 8 and an error on move 12 is the same as the probability of an error on move 8 and an error on move 9? If not, then would you think it is higher or lower? If yes, then why would it not be the same as a double error on move 8?

"you are assuming that the probability to make an error is little, so the probability to make two consecutive errors can be neglected."
++ I constatate from ICCF results that the vast majority of games end in draws (by agreement, repetition, or table base), and thus many games have an even number of errors and very few have an odd number of errors. The only logical explanation for this fact is that there are more games with 0 errors, less with 1 error, less with 2 errors, less with 3 errors... From that I conclude that the probability of an error in the ICCF World Championship is little indeed. I do not assume, I conclude from the data. You can even replicate that with a previous ICCF World Championship, or with the Yekaterinburg Candidates' Tournament.

"So you assume that the number of errors per game is low, from that deduce the error per move is low, from that a much lower probability of two consecutive errors, and from that you assume that the probability to make a blunder is as low as two consecutive errors, because it has the same effect."
++ No, I count much more draws than decisive games in the ICCF World Championship. From that I deduce there are more games with an even than an odd number of errors. From that I deduce that the number of errors per game must be low.

"the probability of two consecutive errors or a blunder could not be neglected"
++ I do not neglect it. Maybe some of the draws in the ICCF World Championship contain an error (?) by white and an error (?) by black. Maybe some contain an error (?) by white, a blunder (??) by black, and an error (?) by white. However, the probability of 4 errors must be lower than the probability of 2 errors and lower than the probability of 0 errors to explain the data.
I cannot tell for sure that all the drawn games contain no errors and all the decisive games contain 1 error. There may be a few drawn games with 2 or 4 errors and there may be decisive games with 3 or 5 errors.
If you were to look at the tournament table of the championship of your local club, then you would notice far less draws, indicative of a far higher error rate.

It is a perfect move.

But the point is, a strategy for one player to lose is just not what is meant by a weak solution (whatever Prof van den Herik and @tygxc say).

Sorry, @MARattigan, I have posted that before realizing you added another post with a definition of "weak solution"; I have deleted it to better study your previous post. I don't want to deal with both you and @tygxc at the same time, so to reduce my error rate .

#2392
"Perhaps we could agree to assign that meaning."
++ No, your personal alternative definition is not generally accepted.
I could propose my own personal alternative definition, which you would not accept either.
"To weakly solve chess is to prove that black can draw against all reasonable white moves."
Let us just accept the generally accepted definition by van den Herik, leading authority in that field.
Otherwise let us accept a definition in Wikipedia,
or from any reputable encyclopedia e.g. Britannica.
If we cannot agree on the meaning of a word,
then we do not start an argument about it, we look it up in a dictionary, e.g. Webster.