Chess will never be solved, here's why

@4491
"That is the crux of the question: is there any way for white (or black) to force a win from the starting position, or will the result be a draw if both sides play perfectly on every move?"
++ No, that is the crux of the question of another thread with even more posts.
So that thread is about ultra-weakly solving chess.
The answer is of course that chess is a draw with best play from both sides.

The crux of the question here is if chess will or can be solved and how long that takes.
As chess is a finite game it can be solved.
Strongly solving chess i.e. a 32-men table base needs 10^44 legal positions, too much for now.
Weakly solving chess i.e. demonstrating how black can draw against all reasonable white moves can be done with 10^17 sensible, reachable, relevant positions, doable in 5 years.
However hiring 3 (ICCF) (grand)masters and renting 3 cloud engines for 5 years costs $3 million.

The amount of wins, losses and draws over x amount of games isn't possibly enough data to determine whether a win can be forced or whether chess is a draw. This is a waste of time.

@4494
"The amount of wins, losses and draws over x amount of games isn't possibly enough data to determine whether a win can be forced or whether chess is a draw."
++ 17 ICCF (grand)masters with engines, 50 days / 10 moves, 2 years / game is quantitatively and qualitatively enough data to determine whether chess is a draw, a win, or a loss.
For more data, you can use 1360 games of the last 10 ICCF World Championship Finals.

Asume chess were a win for white or black. Then to explain 127 draws in 136 games there would need to be 127/136 probability for a game with an odd number of errors. Now try to fit a Poisson distribution to achieve that. It turns out impossible. So chess is a draw.

Thus assume chess is a draw. Then to explain 9 decisive games in 136 games there need to be 9/136 probability for a game with an odd number of errors. Now fit a Poisson distribution to achieve that. It is possible with average 0.070984 errors per game. That means there is 99.75% probability that the 127 drawn games are perfect games with no errors and 99.91% probability that the 9 decisive games contain exactly 1 error.

This is kind of like concluding it's impossible to win a lottery after X number of lotteries go by without a winner.  Doesn't work.

@4496
"it's impossible to win a lottery after X number of lotteries go by without a winner"
++ You do not understand the statistical argument and thus come up with a nonsense comparison.

Go ahead and explain how it's nonsense.  Also, your quote/snippet is out of context once you remove "This is kind of like concluding...".

@4498
"Go ahead and explain how it's nonsense."
++ The comparison is nonsense.
Of course after X number of lotteries go by without a winner says nothing about whether winning the lottery is possible or not.
The argument goes that there is a certain probability of an error in a game, and that this probability can be inferred by observing the outcome of a sufficiently large number 136 of games of sufficient quality. It is impossible to fit 'chess is a win' with the observed data.

If chess is a forced win for white, it would require an errorless game. An errorless game has not happened in your sample of 136 games. 

No, clearly the argument doesn't "go" ...you'll have to explain it in your own words.  Can you?

You should probably start off with the obvious...why you think that ICCF games can be proven to be of sufficient quality...and what exactly "sufficient quality" is supposed to represent, in this case.  Because it's completely circular as you have explained it on many occasions...if you cannot explain why ICCF games have reached some "quality" threshold that is absolute and not relative to current engine playing strength, and if you cannot do so without using engine-derived and/or engine-dependent statistics to prove that engines have reached this threshold (which is like asking toddlers to tell you if they can run fast, when their only frame of reference is a playground full of toddlers running around), then you have nothing.

@4501
"If chess is a forced win for white, it would require an errorless game.
An errorless game has not happened in your sample of 136 games."
++ There are 3 possibilities: chess is a draw, a white win, or a black win.
Observed: 133 games = 127 draws + 6 white wins + 3 black wins.

Assuming chess a white or black win no Poisson distribution of the errors / game fits.
Thus chess is no white or black win.

Assuming chess is a draw, a Poisson distribution with mean value 0.070984 error / game fits.
The 127 draws are 99.74868% sure to be errorless.
The 9 decisive games are 99.9148% sure to contain exactly 1 error.
We can even pinpoint the 1 error: it is usually the last move.

@4502

"why you think that ICCF games can be proven to be of sufficient quality"
++ A World Championship Finals, 17 ICCF (grand)masters with engines, 50 days / 10 moves.

"why ICCF games have reached some "quality" threshold that is absolute"
++ There is no such threshold.
Take the 1953 Zürich Candidates' Tournament:
210 games = 118 draws + 49 white wins + 43 black wins
Assume chess a draw.
Fit a Poisson distribution so the probability of an odd number of errors is (49 + 43) / 210.
Result: mean value = 1.044 error / game.
Games with 0 errors: 74
Games with 1 error:  77
Games with 2 errors: 40
Games with 3 errors: 14
Games with 4 errors:  4
Games with 5 errors:  1

Now assume chess is a white or black win.
Fit a Poisson distribution so the probability of an odd number of errors is 118 / 210. 
Result: impossible fit

Conclusion: chess is a draw with best play from both sides.

But there's no reason to suppose that it's a Poisson distribution.

If there is a forced win, nobody's figured out how to do it yet - probably because it involves deep calculations which are beyond the ability of existing players.So it wouldn't be surprising if everybody made the same 'mistake' of missing the win.

@4505

"But there's no reason to suppose that it's a Poisson distribution."
++ There is reason, a Poisson distribution is derived from the Binomial distribution and applies to many similar stochastic processes.

"If there is a forced win, nobody's figured out how to do it yet" ++ OK

"probably because it involves deep calculations" ++ speculative

"So it wouldn't be surprising if everybody made the same 'mistake' of missing the win."
++ No, this would be extremely surprising. How would such a win look like? 1 a4? Even that has been played several times by grandmasters. It runs against all human and engine logic.

Or as it was explained to me, as far as we know there has never been a game with no errors. 

Good morning,
Thank you for your answers to all. It is true that e.e4 and 1.d4 are unanimously cited as the first reference shot. But I also agree with 1.c4 which can be quite embarrassing for blacks.
I read that in machine-to-machine matches, to avoid draws, humans impose the first moves to force machines to play from various positions and thus sometimes unbalanced. And even like that there are about 3/4 of draws against 1/4 of blank wins. Which means that the advantage of the first shot of the whites can hardly be missed by the blacks and gives at best a draw for the blacks.
Happy end of days

@4507
" there has never been a game with no errors"
++ They lied to you.
99.7% of ICCF WC draws are games with no errors.
Even Zürich 1953 had 74 games with no errors.

@4508
"the advantage of the first shot of the whites can hardly be missed by the blacks and gives at best a draw for the blacks."
++ The advantage of 1 tempo is not enough to win,
so chess is a draw with best play from both sides.
You cannot queen a tempo.

Yes, of course, but this one-shot advantage is decisive for machine-to-machine matches, which is why blacks do best with a draw. No?