Chess will never be solved, here's why

@6037

"10^120 = Shannon's number (number of games of chess)"
++ No, the number of Chess games lies between 10^29241 and 10^34082 because of transpositions.
https://wismuth.com/chess/longest-game.html 
That is why the number of positions counts: 10^44 legal of which 10^17 relevant.

Whatever the number of games of chess, it's still a BIG number, comparable to 52 factorial.

@6041
But the number of legal positions 10^44 and the number of relevant positions 10^17 are smaller.

We can be sure the number of positions needed for a valid weak solution (as defined in the peer-reviewed literature on the subject) is much greater than 10^17.

The number of positions needed to fail to weakly solve chess (according to the correct definition) happens to be exactly 10^17.

@6043
What is your number and what is your calculation?
I gave my number 10^17 relevant positions to weakly solve Chess and two calculations to arrive at it.

And still a BIG number compared with 10^120 whether the 50 move and 3-fold repetition rules are included or not. (Infinite if not.)

Rather amusing point about the video is that it goes wrong almost immediately. First it says that if you have a room of 23 people, there is about 50% chance of two people having the same birthday. This is correct. Then it says, "so if you walk into a room with 23 people in it, the probability of someone having the same birthday as you is 50%".

No!

Specifying you as one of the two people leaves only 23 possible pairs that could match rather than the (23 * 22 / 2) pairs that the correct fact refers to. The chance of one of the 23 people having the same birthday as you is about 1 - (364/365)^23 ~= 6% (ignoring leap years).

@6047
"You are not at all good at taking on board important points like that."
++ What points?
I am sure 10^17 is a good estimate for the number of positions relevant to weakly solving chess.
I am also sure 1 e4 e5 2 Ba6? loses for white.
I am also sure 1 a4 cannot be better than 1 e4 or 1 d4.
I am also sure 1 Nh3 cannot be better than 1 Nf3.

Ah, but @tygxc could reduce the number of sensible relevant birthdays to 3 bringing your probability of a match up to 94,134,790,219 / 94,143,178,827 (=50% for all practical purposes).

Lol, I had the exact same thought/reaction.  That's a Tygxc-level blunder in thinking right there.

And this is the problem with such stories...they lose accuracy in the translation, because somebody lacking in expertise incorrectly equates two things that are not the same at all.

Just for accuracy, I don't believe that was tygxc's video. I'm pretty sure he'd know better than that.

I didn't say it was.  I said it shows the kind of thinking that characterizes Tygxc's mistaken assumptions and extrapolations .  He takes a value, doesn't realize the parameters of what that value actually represents, then uses that value in some equation where he conflates it with some other value that actually conforms to a different set of criteria.

TL;DR version:  he likes to do apple math with lots of oranges.

I figured that's  what you meant....but my comment was intended to keep others from misinterpreting. 

I also like to head off misinterpretations at the pass, so I understand the impulse.

@6031
"my comment was intended to keep others from misinterpreting"
++ Others misinterprete all the time.
They doubt what is right and are sure about the false.
I present facts & figures and back them up.

So why don't you back up your claim of being able to determine the game theoretic result of a position and the number of errors in games played from the position by applying your calculations to the games here and back up your claim of 1 error in 10^20 moves at 17 secs. per move on your 10^9 nps machine (which curiously didn't change when you dropped  the time from 60 hours to 17 seconds).

No reason to restrict yourself to just draws or just KRPP vs. KRP positions. After the first single error you're bound to have both wins and draws in the game, so that's a pathetic excuse.

And contrary to what you keep posting as you try to wriggle out of the exercise, one of the positions is a drawn KRPP vs. KRP position.

@6036

"1 error in 10^20 moves at 17 secs. per move on your 10^9 nps machine
(which curiously didn't change when you dropped  the time from 60 hours to 17 seconds)."
++ 17 s on the 10^9 NPS engine corresponds to 60 h on the engine of the paper.

"No reason to restrict yourself to just draws" ++ Yes: only drawn positions are relevant to weakly solving chess: hopping from the initial drawn position to other drawn positions until a 7-men endgame table base draw or a prior 3-fold repetition.

"or just KRPP vs. KRP positions"
++ Rook endings occur most. Rook endings can be draws despite one pawn down.
A rook ending is a major way for black to achieve the game-theoretic value of the draw.

"one of the positions is a drawn KRPP vs. KRP position anyway."
++ No, see top right in your image: White is winning DTZ 101.
It is a draw only by the 50-moves rule. Black can draw from the initial position without the 50-moves rule, so this position cannot be reached from optimal play by both sides.

Why don't you stop wriggling and post your calculations for the games here? Then we can stop all this pointless discussion about your proposal to solve chess in five years.