Chess will never be solved, here's why

Random events do not happen for a "reason". They happen with a non-zero probability. What you are doing is thinking "if errors are common enough for there to be a game with 2 errors, it is certain there are games with 1 ".

This is not actually any sort of reasoning, it is a mistaken conclusion.

Correct reasoning would involve estimating the probability that there is a game with 2 errors without there being a game with 1 error, and show that it has low probability. You can be sure that with any reasonable assumptions the probability would be non-zero.

[In addition you are making an unwarranted assumption of independent errors. In truth if there is a crucial line that one player is likely to miss, it is probably more likely the other one will too, especially if the players are very similar in their analysis procedure (latest Stockfish). This means both players may pass it by, making a double error. Every chess player has seen such examples even in human games where the players are not so similar].

No.

I can guarantee neither Ghostess nor any one of the many other people who read and contributed to the thread you mean (probably the biological evolution one) ever saw such a fallacious post. That's because it did not exist - it is your mistake.

For balance, neither have you made any posts claiming you were paraplegic or have leukemia.

@tygzc

How do you know a more powerfull machine would not find a win ?

Presumably a joke, but given your proven level of mathematical skill one can never tell.

What would you calculate to be the odds of an odd number of errors (half points blundered on @tygxc's basis) in all 106 games?

Given that the initial position is either a theoretical win or a draw, the two two figures should add up to 1. Do they?

What are the odds of there being an odd number of errors in the 13 games posted here, given that the initial position should be a White win and they all end in draws? (It's a test of your stupendously high IQ.)

It is interesting how you mix comically obvious lies with genuine blunders.

Yes. You failed.

He will be equally unhappy to hear I exposed his lying about Ghostess, by telling her about it.

Regarding the probability of all games having an even number of errors versus all games having an odd number of errors, it is perfectly reasonable to assume that the probabilities are independent and identically distributed for each of the games.

In that case the model is very simple indeed. In each independent game, there is a probability of p of an even number of errors and (1-p) of an odd number of errors.

Given the assumptions, this means that the probability of an even number of errors in all 106 games is p^106, and the probability of an odd number of errors in all 106 games is (1-p)^106.

With @Optimissed stated answer for the all even case, we can deduce p=0.6318. From his stated answer for the all odd case we can deduce (1-p) = 0.6318. These two are not consistent.

DOES NOT COMPUTE. DOES NOT COMPUTE. TRANSISTOR OVERLOAD. PHUT!

It's @Optimissed's lucky day!

I see the number of games was 106, not 160 as in my calculation - DUH! - and @Optimissed is assuming the probability of an even number of errors is 0.5, so the same for an odd number of errors. That at least makes the probabilities consistent - credit where credit is due.

The problem is that we have no reason for believing the probability of an even number of errors is 0.5. This could be assumed, but why would we assume this?

Bayesian analysis can't solve chess, but it is the right way to deal with such questions. What should be done is to first choose a prior distribution for the probability of having an even number of errors. We also need a prior distribution for the true value of chess (three probabilities for whether white has a forced win, black has a forced win or there is a forced draw).

Then we can do the Bayesian inference step by using the evidence we have (a set of 106 draws). The problem is that there are two distinct types of errors when one side is winning - either a move that gives away a draw or a move that turns a win into a loss. This messes up the whole idea of even and odd numbers of errors.

One idea is to consider a full point error as two simultaneous half point errors, and count it as two such. This allows us to deal with parity.

Well I'd agree with the last sentence of #11263 at any rate.

Given that all the games started from the standard starting position and all finished in draws the probability of there being an even number of errors in each game (counting a half point blunder as an error and a full point blunder as two, as @tygxc does) is just the probability that the starting position is a draw.

Curiously @Optimissed assigns a value of 1/2¹⁰⁶ to that proposition and finishes up in the unenviable situation of both claiming to know that the starting position is a draw and asserting that the probability of it being true is about 0.00000000000000000000000000000000123... .

According to @Optimissed's idea of how these things should be calculated, the more games the ICCF people draw, the less likely it is that the starting position is in fact a draw.

Ok

How could it be "copying your post" when firstly it has no text that you wrote in it and secondly it is a description of how to infer that the probability of an even number of errors is very different to 0.5?

Admittedly, you don't seem very concerned at consistency between your posts.

@8643

"Random events do not happen for a reason" ++ The outcome of 106 ICCF WC Finals games is no random event, there are underlying logical reasons.
There is a small bit of randomness by the human factor.

"They happen with a non-zero probability." ++ Yes. There is a nonzero probability of a decisive game in the ongoing 30 games. You estimated it at 10^-8. I estimate it at more.

"if errors are common enough for there to be a game with 2 errors, it is certain there are games with 1" ++ or 0, or 3, or 4 or 5. If there is no underlying non random coupling between errors, then games with 2 errors also require games with 1 or 3 errors. If there is an underlying strong coupling between errors, like in autoplay of 1 entity, then indeed games of 1 or 3 errors can be absent. However, if then there are games with 2 errors, then there should also be games of 0 or 4 errors.

"In truth if there is a crucial line that one player is likely to miss, it is probably more likely the other one will too, especially if the players are very similar in their analysis procedure"
++ Players are not similar and use different engines, at different depths and widths, with different tunings and with different times per move.

"Every chess player has seen such examples even in human games where the players are not so similar" ++ In over the board play there is psychology. The clock is ticking. I make an incorrect sacrifice. You cannot calculate it through and decide to decline my incorrect sacrifice. Correspondence is different. You have 5 days/move average. You can use all computers you want.

@8645

"How do you know a more powerfull machine would not find a win ?"
++ Because the present machines at 5 days/move found the draw.
They are at zero error now. The more powerful machine cannot get below 0 error.
The more powerful machine can do the same in 5 h/move, or 5 min/move, or 5 s/move.

100 pages or so 'disappeared' from another forum and many from yet another.
I believe @Optimissed has been muted by chess.com.
His account shows as online right now.
But his posts seem to have disappeared from at least three forums.

@8652

"Bayesian analysis can't solve chess, but it is the right way to deal with such questions."
++ If you want to discuss that, then let us take an easier example to reason on.
Say I toss a loaded coin 106 times and it lands heads 106 times.
If we now toss the loaded coin 30 more times, what is the probability it lands tails at least once?

How do either of you know that a more powerful machine will not do worse? Have you tried a contest?

Below from a set of attempted SF v SF mates in KNNvKP at time per move between 1 sec and 35 mins. (BR blunders under FIDE basic rules, CR bunders under FIDE competition rules.) In both cases the blunder rate is higher with 35 mins. think time per move than 1 sec. and overall the blunder rate shows a tendency to increase as the think time increases (which is equivalent to increasing the CPU speed).

Blunders identified from the Syzygy tablebase.

Without trying it out you can't say if the same will happen from the starting position. And not necessarily even then. If the results are equal it says nothing, because you can't in general spot the blunders.