Chess will never be solved, here's why

Splendid.

Do provide your reasoning, since you have inadvertently omitted it so far.

[Also, thanks for posting about the ICCF rules. On seeing it I realised I had forgotten to post after I read them myself, although of course my post about hope in hopeless positions was black humour].

Coming from you, shouldn't that read "don't make personal attacks, you idiot, dimwit x2, dullard psycho?"

I'm being nice and only using the the most direct namecalling/insults you've done in the past...3 days on one thread.

I can supply the hardware.

@4398
"Do provide your reasoning, since you have inadvertently omitted it so far."
++ I provided my reasoning before.
30th ICCF World championship 136 games = 127 draws, 6 white wins, 3 black wins.
https://iccf.com/event?id=66745 
Now there are 3 hypotheses: D) chess is a draw, W) chess is a white win, L) chess is a black win.
Define an error (?) as a move that changes the game state from a draw to a loss or from a win back to a draw. Define a blunder or double error (??) as a move that changes the game state from a win to a loss.
Under D) it means 127 games with an even number of errors and 9 games with an odd number of errors.
Under W) it means 6 games with an even number of errors, 127 games with an odd number of errors and 3 games with an even number of errors at least 2
Under L) it means 3 games with an even number of errors, 127 games with an odd number of errors and 6 games with an even number of errors at least 2.
The only consistent explanation is that D) is true and the distribution of errors is:
126 draws with 0 error, 9 decisive games with 1 error, 1 draw with 2 errors.
Please feel free to come up with an alternative error distribution even under a different hypothesis.

 

good point

Explain this further. I'm missing the part where you point out why for example this theoretical outcome couldn't be the case, and only D) is consistent.

W) it means 6 games with an even number of errors, 127 games with an odd number of errors and 3 games with an even number of errors at least 2

 

Here is an interesting stat. Early in TCEC, Houdini and Rybka 4 were within about 30 points of 3100 (on opposite sides) and had 23 drawn games out of 40 (5 games lost by Houdini and 12 lost by Rybka -  a 58.75% score). Surely lots of perfect games?

The current strongest engine is almost 400 points stronger, enough to get  92% score, showing those weak engines from the past would blunder in almost every game.

I wonder how bad today's engines are? Who would have said Houdini was that lousy at the time?

If perfect games exist, why isn't every game perfect?

Is it because of bad luck or is it because the player who played the perfect game is so bad they forgot how to do it?

me just wondering what will fools arguing on the thread getting by posting lakhs of passages to someone who will not listen?

@4399
"If perfect games exist, why isn't every game perfect?"
++ Because of human errors.
Even in correspondence humans sometimes forget one move and err as a consequence.

@4398

"Here is an interesting stat. Early in TCEC, Houdini and Rybka 4 were within about 30 points of 3100 (on opposite sides) and had 23 drawn games out of 40 (5 games lost by Houdini and 12 lost by Rybka -  a 58.75% score). Surely lots of perfect games?"

++ Nowadays in TCEC they impose slightly unbalanced openings to avoid all draws. I do not know if that was the case then.
On to your question: D = 17 / 40
E = Sqrt (1 + (1/(2*D))^2) - 1/(2*D) = 0.3675
Perfect games: 17,
games with 1 error: 15,
games with 2 errors: 5,
games with 3 errors: 2,
games with 4 errors: 1 

 

@4405
"And computer errors. (Only tablebases are perfect)"
++ Yes, that is right, but with more time computers approach perfection.

@4398

"Explain this further. I'm missing the part where you point out why for example this theoretical outcome couldn't be the case, and only D) is consistent.
W) it means 6 games with an even number of errors, 127 games with an odd number of errors and 3 games with an even number of errors at least 2"
++ Let us assume W): chess is a white win and try to explain the observed 127 draws, 6 white wins, 3 black wins. At first sight you could think:
6 games with 0 error,
127 games with 1 error,
3 games with 2 errors.
However, that is not plausible.
How could there be so many games with 2 errors and so few with 0 or 3 errors?

I found a Perfect game of chess on YouTube here are the moves thank me later.

1. e4 c6 2. d4 d5 3. Nd2 dxe4 4. Nxe4 Bf5 5. Ng3 Bg6 6. Nf3 Nd7 7. h4 h6 8. Bd3 Bxd3 9. Qxd3 e6 10. Bf4 Qa5+ 11. Bd2 Qc7 12. O-O-O Ngf6 13. Nf1 c5 14. Qe2 cxd4 15. Nxd4 Bc5 16. Nb3 Bb6 17. Rh3 Qc6 18. Rg3 Qe4 19. Re1 Qxe2 20. Rxe2 Bc7 21. Rc3 Rc8 22. g3 O-O 23. Nc5 Nxc5 24. Rxc5 a6 25. a3 Nd7 26. Rc3 Rfd8 27. a4 Bd6 28. Rxc8 Rxc8 29. Kd1 Rc6 30. a5 Ne5 31. c3 f5 32. Kc2 Kf7 33. b4 Nf3 34. Kb3 e5 35. Be3 f4 36. Nd2 Nxd2+ 37. Bxd2 Bb8 38. Re4 g5 39. h5 fxg3 40. fxg3 Rf6 41. g4 Ke6 42. c4 Rf3+ 43. Kc2 Bd6 44. b5 Rf2 45. Kd1 Rf3 46. Re3 Rf8 47. Re4 Rf2 48. Re2 Rf3 49. Re3 Rf8 50. Re1 Rf3 51. Re3

It’s a forced draw in 51.

Well, that's it then .  A forced draw in 51, chess is solved...

So, how is this a "forced" draw?

@4408

This may well be a perfect game indeed.

Not forced in the way that the moves have to be played but forced as far these moves do not get played the other engine has a slight advantage over the other

Not all errors are alike. If there are only few lines that force win for white, there are always errors in every game played because the margin of error is very small. With this hypothesis games with 0 errors would never happen.

 

However, once diviated from the lines that lead to a forced win, there are plenty of lines that lead to a draw with accurate play. This would explain why there are so many draws, as its easier to find ways too draw than it is to win, even if a forced win is available.

 

If you're trying to use probability here, you are thinking too simple.

 

 

@4412
If the probability of 1 error is E = 127 / 136,
then the probability of 2 errors would be E² = 0.87
hence the expected number of games with 2 errors would be 119 and not 3 as observed.
Under the hypothesis that chess were a white win it is not possible to find a consistent error distribution explaining the observed 127 draws, 6 white wins, 3 black wins.