Chess will never be solved, here's why

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 

More even than crores of them.

 

@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.

A slight advantage according to...?

Stockfisch 15 and LZ0

LC0

As I said you cannot calculate probability of error like this, because all errors are different. If few lines exist that force a win, it would be extremely unlikely to avoid an error that leads to deviating from one of these lines.

After the first error there could be plenty of lines to force a draw, hence probability for a second error occurring to deviate from one of these lines is significantly less likely.

You dont have the data necessary to calculate the probabilities.

 

 

So, imperfect engines evaluating other imperfect engines gives what kind of results...?

 

@4417

"all errors are different"
++ No, an error (?) is a move that changes the game state from draw to loss or from won back to draw. A blunder (??) or double error is a move that changes the game state from won to loss.

"After the first error there could be plenty of lines to force a draw" ++ No, starting with a draw any error leads to a lost position. In a lost position no line can force a draw.

"You dont have the data necessary to calculate the probabilities."
++ 136 games, 17 ICCF (grand)masters with engines, 50 days / 10 moves is enough and reliable data to calculate.