Chess will never be solved, here's why

“Figure 2. AlphaZero self-play game outcomes under different time controls. As moves are determined in a deterministic fashion given the same conditions, diversity was enforced by sampling the first 20 plies in each game proportional to their MCTS visit counts.”

@8821
++ Yes, the openings were indeed enforced for diversity. It makes sense.
If they had 10,000 games with say the Berlin Defense of the Ruy Lopez, then somebody could object that the Sicilian or the Queen's Gambit would yield entirely different results.

If it doesn’t matter then why did you object so vehemently?  
Again, where does it show the amount of errors per game?

@8823

The errors per game follow from the number of decisive games.
10,000 games at 1 s/move: 11.8% decisive games.
Thus the probability of an odd number of errors must be 11.8%.
Assume a Poisson distribution of the errors / game.
That is a plausible assumption also used for radioactive decay, phone calls etc.
0.118 = Poisson (1; lambda; 0) + Poisson (3; lambda, 0) + Poisson (5; lambda; 0) + ...
Solution: lambda = 0.134 errors / game average
Decisive games with 1 error: 1176
Decisive games with 3 errors: 4
Decisive games with 5 or more errors: 0

1,000 games at 1 min/move: 2.1% decisive games.
0.021 = Poisson (1; lambda; 0) + Poisson (3; lambda, 0) + Poisson (5; lambda; 0) + ...
Solution: lambda = 0.021 errors / game average
Decisive games with 1 error: 21
Decisive games with 3 or more errors: 0

You can't reliably count errors per game without having solved chess first.

“Thus the probability of an odd number of errors must be 11.8%. “

It could be even too

“Assume a Poisson distribution of the errors / game. 
That is a plausible assumption…”

it actually isn’t a plausible assumption.  It assumes rare events that aren’t interconnected. Errors are rare and connected

Deja vu .

A Poisson distribution explicitly assumes for it to be very likely to be zero errors, which is simply false

@8826

"It could be even too"
++ No, as chess is a draw every decisive game must contain an odd number of errors.

"It assumes rare events that aren’t interconnected. Errors are rare and connected"
++ If they are rare they are not connected. At 1 min/move all 21 decisive games contain 1 error. So there is no 2nd or 3rd error that can be connected.

@8828
"A Poisson distribution explicitly assumes for it to be very likely to be zero errors"
++ That is the case here. 
'Across all variations the percentage of drawn games increases with longer thinking times.
This seems to suggest that the starting position might be theoretically drawn in these chess variants, like in Classical chess'

@8825

"You can't reliably count errors per game without having solved chess first."
++ Yes, we can: using statistics.
Ultra-weakly solving chess does not need weakly solving first.
Weakly solving Chess does not need strongly solving first.
A weak solution includes an ultra-weak solution.
A Strong solution includes all weak solutions.

Tf no you can’t every single move in each game could be an error for all we know

Optimized you are continue to strawman.  Please do some reading comprehension or something

@8883

"every single move in each game could be an error for all we know"
++ No, then there would be much more decisive games.

@8836

"E4 e5 Bc4 Bc5 Qh5 Nf6 Qxh7 is a black win with an odd number of errors"
++ Yes: any win has an odd number of errors.
In this case 3...Nf6? is an error and 4 Qxh7?? is a blunder or double error.
"someone can blunder a win to a loss on a single error"
++ That is a blunder or double error (??).

"Why is something unconnected if it is rare?"
++ If there is only 1 error in a game, then there is no 2nd error to be connected to.

"you have no evidence of the number of errors" ++ Yes, statistical evidence.

"THERE IS A LOW AMOUNT OF ERRORS"
++ Low amount of decisive games -> low amount of errors.

"they weren’t saying that chess was proven to be a draw"
++ Read the paper. Even if stalemate is made a win or some other rules changes intended to increase game decisiveness, Chess still stays a draw.

That is a blunder or double error (??).
THEN OBVIOUSLY THE EVENTS ARE CONNECTED

You do realize that statistical evidence means nothing in terms of proof?

"you have no evidence of the number of errors" ++ Yes, statistical evidence.

but you don’t have statistical evidence.  
You baselessly assumed a distribution, and then used the results of the distribution to justify your use of the distribution. 

@8838

"THEN OBVIOUSLY THE EVENTS ARE CONNECTED"
++ At 1 min/move there is at most 1 error/game, no blunder (??). So there is no connection.
Besides, you can also fit a binomial distribution, of which the Poisson distribution is a tail-end approximation. The result is just the same.

@8839

"statistical evidence means nothing in terms of proof"
++ If you do not know statistics, then that is your problem. Of course statistics are meaningful in terms of proving. Quantum mechanics and thermodynamics are entirely founded on statistics.

@8840

"you don’t have statistical evidence"
++ I have. I have access to results of many games and can derive conclusions using statistics.

"You baselessly assumed a distribution" ++ Not baselessly, but on solid base.

"used the results of the distribution to justify your use of the distribution"
++ That is done in many sciences.
Example: an electron with charge e and mass m is accelerated by a voltage V.
What speed v does it reach?
Solution:
Assume v << c speed of light. Then Newtonian mechanics apply. Thus eV = mv²/2 thus v = sqrt (2Ve/m) . Now check: if v<<c then the assumption and the result are valid, else apply relativity.

That is a plausible assumption in the case of radioactive decay, 'phone calls etc. because in those cases a Poisson process is plausible, which would necessarily result in a Poisson distribution.

You don't postulate a Poisson process.

A Poisson process firstly needs a measurable space (time intervals in the cases you quote) where the probability of events occurring is proportional to the measure (as likely to get twice the number of 'phone calls or radioactive decays in double the time interval etc.).

You don't describe any measurable space, but from what you say a game of any length would constitute a constant measure in some space if you have a Poisson process in mind. That involves the assumption that the expected number of blunders in a game lasting 20 moves is the same as the expected number of blunders in a game lasting 200 moves. That is NOT plausible.

Of course a Poisson distribution could apply without a Poisson process being responsible. Do you have some other argument for it?

You don't need to guess what the distribution of blunders would be using SF15 by looking at games where the theoretical result of the starting position is unknown, the player is not SF15 and there's no chance of deciding which moves are blunders.

We have actually weakly solved competition rules chess (amended to make the game soluble) for positions with 7 or fewer men on the board, no castling rights and no positions since the previous ply count 0 position that are considered the same for the purposes of the triple repetition rule. 

So you only need to look at a series of games in some of those positions using SF15 itself. Then there's no guesswork.

To make it simple I suggest this series which I've already gone to the trouble of producing for you. Twelve games, twelve draws.

@cobra91 has already checked these games with the Syzygy tablebase and produced a table of blunders.

Now be a good lad and tell us what your Poisson distribution is for those games using only the results without any info from Syzygy and we can see how closely it matches - you've been avoiding it for months. While you're about it, you claim to be able to say what the theoretical result of the initial position is from that distribution, so tell us what that says as well.

For the record, I predict you will continue to do neither because you're interested only in posting crap.