Chess will never be solved, here's why

@tygxc, you have a problem distinguishing between "unlikely" and "impossible".  For reference, these are not the same thing.  You don't even know what a proof is.

Even a known-to-be-optimal probability model + empirical data can never provide certainty.

[I won't confuse things by drawing on the fact that the model concerned is an imperfect simplification].

Your posts are a raging blunderfest. For example, post #4647 contains 7 falsehoods.

...and it goes on to say who’s right.
One thing is certain:
- failures will be solved, or will not be solved, before you have finished your cockfighting!

Poisson distribution doesn't work in this context because when an error occurs in a chess game, this error will change the position and the continuation of the game, affecting probability of future errors. 

Say chess is a win. Theres a certain probability, likely very high that an error will occur because at the start there are very few lines that force a win. Once an error occurs, the probability for another error to be made is now lower because there are more lines available to force a draw.

@4654

"you have a problem distinguishing between "unlikely" and "impossible"."
++ No. It is impossible to fit a Poisson distribution with 127/136 probability of an odd number of errors. Try it yourself if you do not believe that.

"A known perfect probability model + empirical data can never provide certainty."
++ I even indicated the certainty: 99.7% certain that the 30th ICCF WC Finals had 127 perfect games with 0 error, 99.91% certain that it had 9 games with 1 error.

"the model concerned is an imperfect simplification"
++ Please then come up with a model that you think is more perfect and show what results that model yields. The result will be more or less the same.
I first came up with a simpler model with high school math only. It lead to 126 perfect games, 9 games with 1 error and 1 game with 2 errors that cancel out.
Then you proposed Poisson for errors / game. So I did that and the results are about the same.

@4656

"when an error occurs in a chess game, this error will change the position and the continuation of the game, affecting probability of future errors."
++ That may well be right, but the Poisson calculation shows 127 games with 0 error and 9 with 1 error. There is only 0.3% chance of a game with 2 errors, which might affect each other. So the possible affection of errors plays no role. Then there is only 0 or 1 error, then errors affecting each other does not happen.

"Say chess is a win."
++ It is impossible to fit a Poisson distribution with 127/136 chance of an odd number of errors.

"at the start there are very few lines that force a win"
++ OK let us assume there is only 1 line that forces a win.
What would that look like? 1 e4? 1 d4? 1 c4? 1 Nf3? 1 a4? 1 g4? Let us assume 1 e4 were a forced win, like Rauzer said in the previous century. We have ICCF WC Finals games starting with 1 e4 and ending in draws. Some of these go 1...e5. Some go 1...c5 as well, but let us disregard that.

What would the winning line look like? 2 Nf3? 2 Nc3? 2 d4? 2 Bc4? 2 f4? 2 a3? 2 g4?
Let us assume 2 Nf3. We have ICCF WC Finals games going 2 Nf3 and ending in draws.
Some of these go 2...Nc6, some go 2...Nf6 as well, but let us disregard that.

What would the winning line look like? 3 Bb5? 3 Bc4? 3 d4? 3 Nc3? 3 c4? 3 a3? 3 g3?
Let us assume 3 Bb5. We have many ICCF WC Finals games going 3 Bb5 and ending in draws. Some of these go 3...Nf6, some go 3...a6 as well, but let us disregard that.

You see, each time the move of the one supposedly winning line meets like 2 lines drawing in ICCF WC Finals games, taking years with 2 ICCF (grand)masters and engines.

"Once an error occurs, the probability for another error to be made is now lower because there are more lines available to force a draw."
++ If chess were a white win, then as soon as white makes an error, we have a drawn position. The probability of another error does not matter then. The examples above show that if white makes no error, then there are not 1, but 2 lines available to force a draw.

"It is impossible to fit a Poisson distribution with 127/136 chance of an odd number of errors"

Because poisson distribution does not apply to this scenario. Errors would need to be independent events that have no interference with each other. It couldnt fit poisson distribution, because it expects each event to have the same, independent probability. With chess it could go for example like this (ofc im pulling random numbers out of somewhere):

Probability for white to make error #1: 99,9%

Probability for black to make error #1: 5%

Probability for white to make error #2: 2%

Probability for black to make error #2: 9%

And each error made will determine probability for next error for both sides depeding on the position and quality of error. You can't attempt to use poisson distribution for this chaos. This is why you cannot come up with proof based on the fact that hypothesis "chess is a forced win for white" doesnt fit poisson distribution with the stats you use.

And edit.

"What would the winning line look like?"

Ofc I dont know what it would look like. I dont even think it exists, all im saying these models cant prove it doesnt exist.

@4659

"poisson distribution does not apply to this scenario."
++ It is a plausible distribution for many similar stochastic processes.
@Elroch proposed it in criticism to my simpler high school math model.
You have to prove it does not apply or at least make plausible it does not apply and why.

"Errors would need to be independent events that have no interference with each other."
++ I showed the 136 games have 127 games with 0 error and 9 with 1 error and 0 with 2 or more errors. When there only is 0 or 1 error, then dependence of 2 or more errors plays no role.

"it expects each event to have the same, independent probability" ++ That does not matter: calculation shows only 0/1 error so dependence of 2 or more errors plays no role.

"With chess it could go for example like this:
Probability for white to make error #1: 99,9%
Probability for black to make error #1: 5%
Probability for white to make error #2: 2%
Probability for black to make error #2: 9%"
++ That makes no sense. Why would the same ICCF (grand)master have 99.9% probability to err when playing white and only 5% when playing black?
On the contrary: white wins more than black because white has a lower probability to err than black: white has a wider choice of good moves than black.

What do you mean? Do you mean 1 a4 is a forced win for white and nobody plays that and they all err playing junk like 1 e4 or 1 d4?

"each error made will determine probability for next error for both sides"
++ As said: 136 games, 127 with 0 error, 9 with 1 error, < 0.4 with 2 errors.
A change in probability for the next error plays no role as there is no next error.

... look at these two moderated (doris) they don’t even see what is happening around them anymore because they are so proud and certain to be right!
.. and they continue tirelessly on the same discussion to boast, both to be right!
In addition, tygxc is obliged to deceive himself in 4654 and 4656 to accentuate his words and his pride. Anything!

..  moderated (insulting) )Doris), I tell you!!!

A little humility would do you three good, if we count the dupplification of tygxc!! ha ha ha!

Change the subject "les Moderated (doris)", we’ve had enough

Try to be nice in your posts (Doris/Mod)

"It is a plausible distribution for many similar stochastic processes"

Give an example of a process that is similar to chess, where the previous event affects the probability of the next one, that we can predict with poisson distribution.

I already explained why poisson distribution doesnt apply here. Because each event (error) will affect the probability of the next event, and we cannot predict how.

"That makes no sense. Why would the same ICCF (grand)master have 99.9% probability to err when playing white and only 5% when playing black?" 

If chess was a forced win, imagine there is only one line to follow against perfect play to force a win. It would be very unlikely to avoid an error. Black could have an easier task, more lines available to maintain current evaluation, thus lower probability for an error. As I said, those numbers are just an example, dont bother analysing them. After any given error there could be a 0-99,9% chance for another error, depending on how many available lines would maintain current evaluation in the position.

"What do you mean? Do you mean 1 a4 is a forced win for white and nobody plays that and they all err playing junk like 1 e4 or 1 d4?"

As I said, I'm not gonna bother speculating lines. It could be a long series of only moves for white to force a win. All I'm saying is your calculations dont prove that chess is a draw, even if it most likely is.

I'm new to this discussion, am I not allowed to participate?

This is a rip-off of my forum.

And it’s stuck.

On chess.com, we value respect and kindness. Please do not swear, post offensive messages, and apply verbal abuse to your forum contributions. We thank you for helping out on chess.com and being a part of the community, but please do not be rude to anyone.

Thanks.

Je suis poli. Je suis francophone te je traduis mes messages. Désolé si la traduction de "coqs" (Cooks) a été mal traduite. Je ne comprends d'ailleurs pas pourquoi ça été mal traduit ?!

Toujours est-il

I am polite. I am francophone and I translate my messages. Sorry if the translation of "coqs" (Cooks) has been incorrectly translated. I don’t understand why it was poorly translated?!

Still, it must be the longest blog published on chess with its 234 pages!

@4663

"Give an example of a process that is similar to chess"
++ Number of phone calls,
number of radioactive decays,
number of soldiers killed by horses,
number of stars per unit of space,
number of patients arriving,
number of meteorite strikes,
number of photons on a detector,
number of river floods...

"Because each event (error) will affect the probability of the next event"
++ When there is only 0 or 1 event, there is no next event to be affected.

"If chess was a forced win, imagine there is only one line to follow against perfect play to force a win. It would be very unlikely to avoid an error. Black could have an easier task, more lines available to maintain current evaluation, thus lower probability for an error."
++ But some black responses would leave more than only moves. 
Let us assume 1 e4 e5 2 Nf3 Nc6 3 Bb5 were a white win and were the only way to win for white. After 1 e4 b5 or 1 e4 f5 white then has more than 1 way to win. 
Likewise after 1 e4 e5 2 Nf3 a6 or 2...g5 etc. white then has more than 1 way to win. 
So black would also have a series of only moves to force white to play one only move to win.
If chess were a win, a white error would not be more probable than a black error.

"After any given error there could be a 0-99,9% chance for another error"
++ For the ICCF WC Finals there is only 0 or 1 error, there is no other error.
For the 1953 Zürich Candidates there are 59 games with 2 to 5 errors, so the calculation might have < 25% error due to interdependence.

"I'm not gonna bother speculating lines." ++ Fair enough, but the popular lines 1 e4 and 1 d4 are heavily played and have many draws and lines like 1 a4 would defy logic if winning.

"When there is only 0 or 1 event, there is no next event to be affected"

There is no mathematical way to prove these games only had 0-1 errors.

"Number of phone calls, 

number of radioactive decays, 

number of soldiers killed by horses, 

number of stars per unit of space, 

number of patients arriving, 

number of meteorite strikes, 

number of photons on a detector, 

number of river floods..."

And what do these events have in common? One event of phone call will not affect the probability of the next one. In chess all events affect the next one, therefore we can't predict them with poissons distribution. We can't predict errors or calculate their probability, or make any distributions with this system.

I forecast that @tygxc will fail to learn from your observation.

To be precise it is a forum, not a blog, since it is open to contributions rather than being restricted to one or a few authors.

It is quite a long way from being the longest forum: I believe this one is not the longest forum either despite its 2092 pages!

@4670

"There is no mathematical way to prove these games only had 0-1 errors."
++ Yes there is. 30th ICCF WC Finals. Assume a Poisson distribution. Fit a Poisson distribution. Result: 127 games with 0 error, 9 games with 1 error, < 0.4 game with >1 error.

"In chess all events affect the next one"
++ There is no next one in ICCF WC finals. No error =  draw, 1 error = loss.

"we can't predict them with poissons distribution"
++ Let us assume for whatever reason Poisson were not applicable.
Please then come up with an alternative, plausible distribution of errors by whatever means.

With Poisson: the 30th ICCF WC finals: 127 draws with 0 error, 9 decisive games with 1 error (?).
Zürich 1953 Candidates: 74 draws with 0 error, 77 decisive games with 1 error (?), 40 draws with 2 errors (?) that undo each other, 14 decisive games with 3 errors, either 3 lone errors (?), or an error (?) and a blunder (??),  4 draws with 4 errors, either 4 lone errors (?), or an error (?), a blunder (??) and an error (?), 1 decisive game with 5 errors: either 5 lone errors (?), or 3 errors (?) and a blunder (??), or 1 error (?) and 2 blunders (??).

30th ICCF WC Finals:
0 errors: ... games,
1 error:   ... games,
2 errors: ... games,
3 errors: ... games.

Zürich 1953 Candidates:
0 errors: ... games,
1 error:   ... games,
2 errors: ... games,
3 errors: ... games,
4 errors: ... games,
5 errors: ... games,
6 errors: ... games,
7 errors: ... games.