Chess will never be solved, here's why

Waiting for 8 posts after this to post, and 21 after that.

I posted earlier (#8819)

...

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.

My prediction has so far turned out to be spot on.

So we'll have to do it for you (as usual).

Applying your reasoning to the games shown.

12 games: 12 draws + 0 decisive games

Assuming the initial position a white or black win:
12/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
No lambda satisfies this equation.
Thus the initial position cannot be a white or black win.
Thus the initial position is a draw.

Assuming the initial position a draw:
0/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
lambda = 0 errors / game average satisfies this.

Games with 0 errors: 12

But Syzygy tells us (with the aid of @cobra91):

The position is a Black win.

Games with 0 errors: 0

Games with 1 error: 4
Games with 2 errors: 0
Games with 3 errors: 3
Games with 4 or more errors: 5

(error = @tygxc error = half point blunder or full point blunder counted twice.)

So what went wrong?

Answer: your reasoning.

You have used the hypothesis that blunders per game occur with probabilities that form a Poisson distributution. You gave no reasons why they should and the hypothesis is intuitively unlikely. Obviously they don't.

So you can now check those results and stop posting that argument.

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

Is the "26" bit what you scored?

ey yo tygxc you know whats funny is that i can PROVE that chess is a win with your logic.

assume a poisson distribution of errors.

use the dataset of my games.

there are more decisive games than draws, so chess must be a win.

hi

This has been pointed out to him a dozen times .  It's ignored.

@8875

"If you instead assume chess to be a win, then what are the number of errors per game?"
++ Then no consistent numbers of errors per game can be found.

Take this tournament, which is larger and stronger that Tata Steel Masters 2023.
https://www.iccf.com/event?id=79897 
136 games = 121 draws + 15 decisive games

Assume Chess to be a win for either white or black.
Then there must be 121 / 136 probability for an odd number of errors.
121 / 136 = Poisson (1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
There exists no lambda that satisfies this equation.
Thus Chess can be no white or black win.
Thus Chess is a draw.

Now verify: assume Chess a draw.
Then there must be 15 / 136 probability for an odd number of errors
15 / 136 = Poisson (1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
lambda = 0.124 errors / game average satisfies this

Thus:
Games with 0 errors: 120
Games with 1 error: 15
Games with 2 errors: 1
Games with 3 or more errors: 0

@8883
"use the dataset of my games"
++ That is neither a sufficiently large nor a sufficiently strong tournament.
As a 1462 player you probably err every move and so do your opponents.
That is why you get more decisive games than draws.

@8882

"You can not use Stockfish to determine if a chess game had zero errors with perfect play."
++ I do not use Stockfish,
I use statistics on a sufficiently large and sufficiently strong tournament.

"Stockfish does not play perfect chess" ++ Indeed Stockfish does not play perfect chess.
At 17 s / move on a 10^9 nodes/s cloud engine it errs once every 100,000 positions.

"Stockfish is still improving"
++ Yes, with new versions, better hardware, more time per move, it makes fewer errors.

I posted earlier (#8880)

I posted earlier (#8819)

...

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.

My prediction has so far turned out to be spot on.

So we'll have to do it for you (as usual).

Applying your reasoning to the games shown.

12 games: 12 draws + 0 decisive games

Assuming the initial position a white or black win:
12/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
No lambda satisfies this equation.
Thus the initial position cannot be a white or black win.
Thus the initial position is a draw.

Assuming the initial position a draw:
0/12 = Poisson(1; lambda; 0) + Poisson (3; lambda; 0) + Poisson (5; lambda; 0) + ...
lambda = 0 errors / game average satisfies this.

Games with 0 errors: 12

But Syzygy tells us (with the aid of @cobra91):

The position is a Black win.

Games with 0 errors: 0

Games with 1 error: 4
Games with 2 errors: 0
Games with 3 errors: 3
Games with 4 or more errors: 5

(error = @tygxc error = half point blunder or full point blunder counted twice.)

So what went wrong?

Answer: your reasoning.

You have used the hypothesis that blunders per game occur with probabilities that form a Poisson distribution. You gave no reasons why they should and the hypothesis is intuitively unlikely. Obviously they don't.

So you can now check those results and stop posting that argument.

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

Again my prediction was spot on. Maybe it's talking to @Optimissed too much, but I'm beginning to think I've got precognition. Nostradamus was an amateur.

Let me reiterate. The above proves

YOUR ASSUMPTION THAT THE NUMBER OF HALF POINT BLUNDERS PER SF15 v SF15 GAME WILL FOLLOW A POISSON DISTRIBUTION IS WRONG. CAN YOU STOP POSTING IT PLEASE?

(I predict the answer will be "++No. I'm only interested in posting crap.")

@8889

You post crap.

Your position cannot result from the initial position by optimal play from both sides
and is thus not relevant to weakly solving Chess.

This is no sufficiently large and sufficiently strong tournament starting from the initial position, so that is not what my reasoning about distribution or errors applies to.

N

@8894

"And how would you know this to be a fact." ++ Because it is a won position.
Weakly solving Chess only hops from drawn position to drawn position. 

"It is a legal chess position that can be calculated like any other positions."
++ Strongly solving Chess needs all 10^44 legal positions.
Weakly solving Chess only needs the 10^18 relevant positions, all draws.

@8896

"Nor if your weakly solving involves hopping from position to position with SF15 it doesn't."
++ It does. At 17 s/move on a 10^9 nodes/s it hops from draw to draw as becomes clear once it reaches the 7-men endgame table base draw or a prior 3-fold repetition. If it would occasionally err, as it is expected to do once in 10^5 positions, then that will show by not reaching the 7-men endgame table base draw and can be easily repaired by retracting that move.
If you want to emulate 17 s/move on a 10^9 nodes/s cloud engine using a 10^6 nodes/s desktop, then you need 4.7 h/move.
If you want to reach a win in N moves, then you need time to calculate at 2N depth.
Hopping from draw to draw is easier as it does not require looking ahead to 2N.

"it is a legal chess position that can be calculated like any other position."
++ Yes, it is a legal chess position. Yes, it can be calculated. No, it is not relevant to weakly solving Chess. What you present is a set of move sequences from a won position. No, it is not a sufficiently large, sufficiently strong tournament, where the Poisson distribution applies.

wat

@8923

"extrapolation that assumes an exponential relationship between blunder rates and think time doesn't work" ++ I fail to see what is wrong. I have 3 data points: decisive games at 1 s/move, decisive games at 1 min/move and 0 decisive games at unlimited time.

"Here is @cobra91's complete table." ++ Of the 4 cases only the drawn KRPP vs. KRP is relevant.

There clearly are anomalies in your data. The error rate should descend monotoneously and it does not. How come it jumps up and down erraticly? There is something wrong with your data.
What are your sequences for this KRPP vs. KRP position? Please repeat or link.

"Er, which move exactly?" ++ The principle is easy. White tries to win, black tries to draw. If all white tries fail, then Chess is weakly solved. Whenever a draw or a white lossis reached, a white move needs retracting. Whenever a white win is reached, a black move needs retracting. According to my calculations the latter case is exceptional as I calculate being right 99,999 out of 100,000 positions. You doubt that result, OK. I doubt your data. Let us look into your data.

i"f N>1000 ply" ++ Chess is not 1000 ply deep. If there are w non-transposing choices per move, then all white moves and all black moves d moves deep lead to w^(2d) positions. There can be no more positions than there are legal positions, so w^(2d) < 10^44.