Chess will never be solved, here's why

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.

++No it's not.

A draw was agreed, but after 51...Rf2 it's a win for Black.

(I know that for a fact, it's just I can't play the win against Stockfish.)

You didnt quite catch my point. You can't calculate probability for an x amount of errors occuring during a game, because the first error will have a different probability from the next one. To avoid error #1 you might have 3 possible lines you could follow to force a win against perfect play for example. To avoid error #2 you might have 20 possible lines you could follow to force a draw.

Your calculations and logic is over simplified.

 

@4421
"Your calculations and logic is over simplified."
++ I use simple, high school math.
However, try to come up with another plausible error distribution under any of the 3 hypotheses: chess is a draw / white win / black win that explains:
127 draws, 6 white wins, 3 black wins
Number of games with 0 errors = ...
Number of games with 1 error = ...
Number of games with 2 errors = ...
Number of games with 3 errors = ...
You can use whatever simple or complicated logic and calculations you want.
Just give the numbers.
I predict you cannot find any plausible distribution under hypotheses white wins or black wins.
I also predict you cannot find any plausible distribution under hypothesis draw that differs substantially from the 126 with 0 error, 9 with 1 error, 1 with 2 errors I found.

++So it is, now you mention it.

I should have said 46...Rf2.

@4420
No, even if the draw by 3-fold repetition were not claimed, 51...Rf2 is a draw just the same and not a win for black. So it is no fact and you do not know.
Whether you can win that against Stockfish or not does not matter.
A win is a win, a draw is a draw, regardless of whether you can play that against Stockfish or not.

++Sorry, I know 46...Rf2 a win for Black.

You have to prove me wrong.

@4424
No, 46...Rf2 is a draw just the same. 47 bxa6 bxa6 48 Rb3 Bc5 49 Rb8 Rf1+ 50 Kc2 etc.