Chess will never be solved, here's why

++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.

++No, the onus of proof is normally on the one who claims irrespective, but we've apparently decided to ignore that convention and instead place the onus of disproof on everybody else..

@4434
"the onus of proof is normally on the one who claims irrespective"
++ You claimed 'I know that for a fact', so you have to prove it. Material is even, so a win claim is unlikely to start with. There is no positional advantage. A draw it is.

When I say 1 e4 e5 2 Ba6 is a black win, that is because being a bishop up is known to be winning. I do not have to prove more, but I did and showed a forced checkmate in 82.

@4436
So the 860 rated player 'knows' and does not have to prove.
I at least showed the forced checkmate in 82 of 1 e4 e5 2 Ba6, though it is obvious.

Since you insist.

I'll use exactly the same method as your proof so you can't fault it.

So 46...Rf2 is a forced checkmate in 10 moves.
But what if white plays differently?
Then white loses in a different way.

@4437
47 Rxe5+? is a losing mistake. Then white is down a full rook without any compensation and thus lost without even needing further proof. 47 bxa6 draws as proven @4426.

Now you are welcome to show where in my proof that 1 e4 e5 2 Ba6 is a forced checkmate in 82 for black you see any possible improvement for white.