Chess will never be solved, here's why

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!

Incorrect.

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.

In the absence of any meaningful analysis to observe the numbers of errors occurring in games that end in draw, since to perform it, the object of this thread would need to be accomplished, we fall back on the observation that the numbers of odd and even numbers of errors per game are equal. In actual fact there will be a weighting towards lower numbers of errors, assuming that errors are actually rare events. Therefore, there will be a weighting towards odd numbers of errors where an odd number is lower than an even number, except at the very beginning of the series. It depends on the modality, below which there is likely to be a similar weighting towards an even number which is higher than an odd number. Since we don't know the results of this, the only realistic approach is to treat the odds and evens regarding numbers of errors as equally likely and to accept the calculation I gave, with some reservations.

You must also know very well that your pretended illustration was random noise. If you were brighter than you are, you would know that I am dumber than I appear to look and I won't even be able to understand your argument, which therefore you shouldn't have given. Or you would know that I'm a lot brighter than you pretend to think I am and can therefore see clearly when you're attempting to con other people.

Yet not when it's to your advantage to give the argument you are now criticising, since you gave it at least three times to criticise ty. I am far more justified giving that argument in this circumstance than you were to give it obviously incorrectly.

Surrounding yourself with people like RATMAR can't help you much either.

Maybe you're hoping to appear clever in contrast with him.

Deliberately misinterpreting the simple calculation makes you appear really, really dumb. That is going to hold for all the non-dumb people commenting on and reading this thread.

I think you are so vain that you think even the non-dumb people will believe you regarding any subject, since they wouldn't dare to criticise you. In my opinion, it's about time they did. Your complete lack of mental focus hardly instils a focussed response in others. Perhaps that is also what you hope to achieve.

How come you're copying my post and pretending to point out what I'd pointed out over an hour before?

Oh, I forgot. It's because you're vain, narcissistic and dishonest. Silly me.

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.

Some gm's assisted by engines got a lot of draws?
So what?
I believe Elroch already refuted tygxc's claims about that a while back.
Martin is finding further ways to refute?
If these long series of draws continue - will tygxc's claims be strengthened?

Not necessarily.
Its a fact - that the stronger the two opponents - the higher the percentage of draws between them - provided that they're evenly enough matched.
And its obvious why - fewer mistakes - plus the mistakes are smaller.
A factor of similiarly programmed engines is also obviously relevant.
I believe as the gap in strength between the computers and the GM's continues to increase with the computers becoming ever-stronger compared to the GM's then we would see these massive series of draws continue - with the human programming of the computers also a factor.
The GM using a computer gets his mistakes corrected by the computer before he the GM makes his move. So does his opponent. There's going to be a lot more draws.
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tygxc wants to 'draw inferences' from the results?
Can his inferences be debunked?
Well perhaps not to him.
But try: five years from now - another big match is held (regardless of other matches happening up to then)
then try hypothetically - one side uses today's engines - the other side uses the top of the line engines of that future time ...
do the draws still happen?
Is a point reached where the weaker engine still can draw even if a top of the line engine is many thousands of times stronger?
Obviously - if chess had been 'solved' or 'solved enough' then that point would probably be reached.
But that isn't the case on either count.
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Plus tygxc has already conceded that chess can't be solved.
tygxc might do a lot 'better' on his 106 games angle than on 'taking the square root' though .....
whatever his angle on 'alternatives to solving' happens to be or how invalid it is.

But how do the rest of us, without the big red telephone, know?