Chess will never be solved, here's why

A slight advantage according to...?

As I said you cannot calculate probability of error like this, because all errors are different. If few lines exist that force a win, it would be extremely unlikely to avoid an error that leads to deviating from one of these lines.

After the first error there could be plenty of lines to force a draw, hence probability for a second error occurring to deviate from one of these lines is significantly less likely.

You dont have the data necessary to calculate the probabilities.

So, imperfect engines evaluating other imperfect engines gives what kind of results...?

@4417

"all errors are different"
++ No, an error (?) is a move that changes the game state from draw to loss or from won back to draw. A blunder (??) or double error is a move that changes the game state from won to loss.

"After the first error there could be plenty of lines to force a draw" ++ No, starting with a draw any error leads to a lost position. In a lost position no line can force a draw.

"You dont have the data necessary to calculate the probabilities."
++ 136 games, 17 ICCF (grand)masters with engines, 50 days / 10 moves is enough and reliable data to calculate.

++No it's not.

A draw was agreed, but after 51...Rf2 it's a win for Black.

(I know that for a fact, it's just I can't play the win against Stockfish.)

You didnt quite catch my point. You can't calculate probability for an x amount of errors occuring during a game, because the first error will have a different probability from the next one. To avoid error #1 you might have 3 possible lines you could follow to force a win against perfect play for example. To avoid error #2 you might have 20 possible lines you could follow to force a draw.

Your calculations and logic is over simplified.

@4421
"Your calculations and logic is over simplified."
++ I use simple, high school math.
However, try to come up with another plausible error distribution under any of the 3 hypotheses: chess is a draw / white win / black win that explains:
127 draws, 6 white wins, 3 black wins
Number of games with 0 errors = ...
Number of games with 1 error = ...
Number of games with 2 errors = ...
Number of games with 3 errors = ...
You can use whatever simple or complicated logic and calculations you want.
Just give the numbers.
I predict you cannot find any plausible distribution under hypotheses white wins or black wins.
I also predict you cannot find any plausible distribution under hypothesis draw that differs substantially from the 126 with 0 error, 9 with 1 error, 1 with 2 errors I found.

++So it is, now you mention it.

I should have said 46...Rf2.

@4420
No, even if the draw by 3-fold repetition were not claimed, 51...Rf2 is a draw just the same and not a win for black. So it is no fact and you do not know.
Whether you can win that against Stockfish or not does not matter.
A win is a win, a draw is a draw, regardless of whether you can play that against Stockfish or not.

++Sorry, I know 46...Rf2 a win for Black.

You have to prove me wrong.

@4424
No, 46...Rf2 is a draw just the same. 47 bxa6 bxa6 48 Rb3 Bc5 49 Rb8 Rf1+ 50 Kc2 etc.

Actually that would be entirely unsurprising given the paucity of double error games. The statistically surprising thing would be that black is benefiting much more from the errors than white.

And this is where your understanding comes to an end. For some reason you are unable to take the step that acknowledges that an outcome which is unlikely remains logically possible.

It's like if you toss a coin 20 times and get 20 heads, to be certain the coin is biased is a blunder. Rather it is appropriate to think it is likely the coin is biased and to observe that if it were not biased, the result would still occur 1 in a million times.

[It would be hazardous to try to reason from this retrospectively, as we could do likewise for many "unlikely" outcomes. Eg all tails, or alternate heads and tails. All outcomes are actually equally unlikely! This is why it is important to specify the outcomes of interest before experiments begin.]

Anyhow, you are in an analogous position, saying the coin is definitely biased. You can be probably right while being definitely wrong to be certain.

The sound way to infer from results is to start with an a priori belief as to what the value of the game of chess is and then modify that belief using Bayesian inference as you gather data. Bayesian probability was elegantly described as "the logic of science" by Jaynes, and never reaches boolean certainty, but rather very strong levels of belief.

Do you even understand that the most you can do is merely to show that this outcome would be very unlikely? Is that really too difficult for you?

@4431
"show that this outcome would be very unlikely"
++ But come up with any other explanation but 126 with 0 error, 9 with 1 error, 1 with 2 errors.
You can use whatever you like, Poisson distribution or geometrical distribution, whatever, but show figures, not words.

@4428
"You need to consider all the others"
++ No, the onus of proof is on the one who claims a win with equal material.
The onus of proof in on the one who claims a draw with unequal material.

@4429
"The sound way to infer from results is to start with an a priori belief as to what the value of the game of chess is and then modify that belief using Bayesian inference as you gather data."
++ I know more about statistics and probability than coin flips. If you say my simple high school math way is not 'sound', I am willing to accept that. But then please come up with a 'sound' or 'rigourous' error distribution of number of games with 0, 1, 2, ... errors to explain 127 draws, 6 white wins and 3 black wins.