Throwing out draws to calculate Elo

[Dann Corbit]

It is not trivial to do measurements at the 0.1% accuracy level. There are many issues that could potentially have an effect of that magnitude.

Are you quoting me? I believe I have said that very same thing in this thread.

[hgm]

Another interesting factor is that the Elocalc output shows that the engines have the same Elo within the confidence interval.

Not really: it is at the very edge of the confidence interval. So one could just as well say the Elo calculation proves with confidence that one engine is stronger than the other.

Of course this is to be expected, as the confidence interval quoted by the Elo calculation is the 95% interval (i.e. 2 sigma) for two-sided error bars, i.e. 97.5% for one-sided excess. Which is exactly the LOS.

In other words, you are only within the eror bars because they were quite generous (95%) error bars. And the more generous the error bars are, the less it means to stay within them. Accepting error bars this large is tantamount to saying "I don't care is the result suggests with 97.5% confidence that A is stronger than B; I will still consider them the same in that case". But that should then apply to the LOS as well, or you are comparing apples and oranges. A LOS of 97.5% means the result is not conclusive enough for you, if you accept such large error bars on the calculated Elo.

Are you quoting me? I believe I have said that very same thing in this thread.

No, this is all well known to engine testers. See earlier discussions with Bob Hyatt some 10 years ago.

[Dann Corbit]

It is true that my result is rather unexpected (especially after only 14,000 games), but if I had let the games run all the way to 100,000 as I had originally intended, the LOS value would have been very close to either 0 or 1 as Ajedrecista proved in his analysis.
Edit: Last night, when I went to sleep, it was not as lopsided. I was surprised this morning when I took a look at it.

[Ajedrecista]

Hello Dann:

I think this analysis is pretty good. Like I said, long matches destroy LOS.
Now, here is the burning question:
In a short, N game match, do you trust LOS since there is randomness in the outcome of chess games?
IOW, is there any reason to believe that the accuracy of wins, losses and draws of a ten game contest is higher for a LOS calculation than for an Elo calculation?

Thank you. I would say something that has been said before: LOS purpose should be for compare engines that we know that are different, not exact copies: Ctrl+C, Ctrl+V and rename the last one. You are watching dispersion, which is explained with the concept of standard deviation. If you flip a balanced coin twice, the most probable outcome will be one head and one tail (whatever the order of them), but it is not the only possible result! More probable does not imply uniqueness. The same when the number of coin tosses raise, already explained in my statement c).

Answering your questions: you can not conclude anything from short matches ('short' should be defined). Elo measure is fine but it is not an standalone measure: it comes with error bars, which are proportional to 1/sqrt(n) at first approximation. I mean, it is not the same to measure +40 Elo (some info missing from my POV) that with a given confidence level (for example 95%) you measure +40 ± 100 Elo or +40 ± 4 Elo. LOS is related with these error bars using scores instead of Elo:

Code: Select all

Using a trinomial distribution with:

W = wins/games
D = draws/games
L = loses/games

W + D + L = 1

Score = µ = W + D/2

Sample standard deviation = s = sqrt{[µ*(1-µ) - D/4]/(n - 1)}
D = 1 (100%) would bring s = 0: there is not dispersion in the sample. The same with µ = 0 (0%) and µ = 1 (100%), which represents L = 1 (100%) and W = 1 (100%), respectively.

Lower bound and upper bound of scores are µ - z*s and µ + z*s with a given confidence level represented by z (for example, z ~ 1.96 for 95% confidence level in a two-sided test).

If µ > 0.5:
Looking for superiority: lower bound = µ - z0*s > 0.5 (score > 50% --> Elo > 0).
z0 < (µ - 0.5)/s
LOS = Prob.(z < z0) = Prob.[z < (µ-0.5)/s] > 50% (one-sided test).

If µ < 0.5, one should search µ + z0*s > 0.5; z0 < -(µ - 0.5)/s; LOS = Prob.(z < z0) = Prob.[z < -(µ - 0.5)/s] < 50%.

Finally, I get +3.91177 Elo in your +3085 =7932 -2928 match instead of your +2.49297. LOS value looks fine. Although I want to advise: if you are measuring Elo with a fixed game match, never stop it before the match is finished or you will introduce a bias. Disgracefully, I can not prove it due to my limited statistics skills, but someone smarter can.

Regarding 'the engines have the same Elo within the confidence interval', well... if you have a measure of +x ± x Elo with 95% confidence (rounding of first and second x should be taken with care), then LOS = 0.5*(1 + 0.95) = 0.975 = 97.5%.

------------

@HGM: I get a sample standard deviation of 0.2780% from those 13945 games, which is in line with your 0.277%. Otherwise, I do not see anything wrong with z ~ 2.02 in a single trial... other thing is if it would have been |z| > 40. The wrong thing is stopping the match before it finishes: first design a size, then run it until the end. Otherwise you bias the result stopping when things are temporarily good or bad, just a sort of cherry picking.

Regards from Spain.

Ajedrecista.

[hgm]

It is true that my result is rather unexpected (especially after only 14,000 games), but if I had let the games run all the way to 100,000 as I had originally intended, the LOS value would have been very close to either 0 or 1 as Ajedrecista proved in his analysis.
Edit: Last night, when I went to sleep, it was not as lopsided. I was surprised this morning when I took a look at it.

No, why do you think that? For equal engines you expect the LOS to randomly fluctuate in the range 0.16-0.84 (i.e. between 50% +/- one standard deviation), only occasionally moving out of that interval, no matter how many games you do.

[Dann Corbit]

It is true that my result is rather unexpected (especially after only 14,000 games), but if I had let the games run all the way to 100,000 as I had originally intended, the LOS value would have been very close to either 0 or 1 as Ajedrecista proved in his analysis.
Edit: Last night, when I went to sleep, it was not as lopsided. I was surprised this morning when I took a look at it.

No, why do you think that? For equal engines you expect the LOS to randomly fluctuate in the range 0.16-0.84 (i.e. between 50% +/- one standard deviation), only occasionally moving out of that interval, no matter how many games you do.

Because of this:
b) Since LOS can be related with z-score, which can be proportional to sqrt(n), which tends to infinity, z-score tends to ±infinity and LOS tends to 0 or 1.

This is the conclusion of Ajedrecista, backed by careful analysis.
Now, I can see why you do not believe me, since I apparently fail to grasp simple mathematical concepts like trichotomy (as multiple attempts to teach me seem to have not had the desired effect). But I know math well enough to follow someone else's work.

Incidentally, Ajedrecista is right that stopping the games short introduced a bias. It wasn't evil intent on my part, I simply wanted the cores to analyze 404 candidate mate positions for a test set that I am building.

[Dann Corbit]

It is true that my result is rather unexpected (especially after only 14,000 games), but if I had let the games run all the way to 100,000 as I had originally intended, the LOS value would have been very close to either 0 or 1 as Ajedrecista proved in his analysis.
Edit: Last night, when I went to sleep, it was not as lopsided. I was surprised this morning when I took a look at it.

No, why do you think that? For equal engines you expect the LOS to randomly fluctuate in the range 0.16-0.84 (i.e. between 50% +/- one standard deviation), only occasionally moving out of that interval, no matter how many games you do.

Because of this:
b) Since LOS can be related with z-score, which can be proportional to sqrt(n), which tends to infinity, z-score tends to ±infinity and LOS tends to 0 or 1.

This is the conclusion of Ajedrecista, backed by careful analysis.
Now, I can see why you do not believe me, since I apparently fail to grasp simple mathematical concepts like trichotomy (as multiple attempts to teach me seem to have not had the desired effect). But I know math well enough to follow someone else's work.

Incidentally, Ajedrecista is right that stopping the games short introduced a bias. It wasn't evil intent on my part, I simply wanted the cores to analyze 404 candidate mate positions for a test set that I am building.

And in case you think I am not being truthful about the 404 mate positions, here they are:

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[Ajedrecista]

Hello:

It is true that my result is rather unexpected (especially after only 14,000 games), but if I had let the games run all the way to 100,000 as I had originally intended, the LOS value would have been very close to either 0 or 1 as Ajedrecista proved in his analysis.
Edit: Last night, when I went to sleep, it was not as lopsided. I was surprised this morning when I took a look at it.

No, why do you think that? For equal engines you expect the LOS to randomly fluctuate in the range 0.16-0.84 (i.e. between 50% +/- one standard deviation), only occasionally moving out of that interval, no matter how many games you do.

I think there has been a bit of misunderstanding: I was talking about the tails of the distribution, where |z| value should be high with 10^100 games in the most extreme cases. Since z = (2*score - 1)*sqrt(n) according with my previous analysis (corrections are always welcome!), it does not imply that sqrt(n) dominates z because 2*score - 1 is an infinitesimal of similar order most of the time and the multiplication of them is not ±infinity always. I was trying to say that LOS could be very, very close to 0% or 100% between equal engines in long matches, but not always!

Regards from Spain.

Ajedrecista.

[hgm]

Because of this:
b) Since LOS can be related with z-score, which can be proportional to sqrt(n), which tends to infinity, z-score tends to ±infinity and LOS tends to 0 or 1.

Sorry, that is bullshit. z-scores will remain on the order of 1 during the entire test. By definition. They are the ratio of the actual deviation and the standard deviation. And the standard deviation is defined as a typical magnitude of the actual deviation. They will fluctuate, for sure, and sometimes stray out of the interval [-1, 1] (namely 32% of the time), and even more rarely out of [-2, 2] (5% of the time), but they will always return to within [-1, 1] when you continue the match.

[Alayan]

Either the math predict that the LoS of a given run converges on average towards 50% (i.e., despite the expected divergence in absolute wins and losses, the LoS goes towards 50% faster, and an increasingly small proportion of random runs would show a LoS beyond 50 +- delta %) or LoS is broken.

Doing 100K runs of 100K coinflips is deceptive because of course you'd expect about 1K runs to have a LoS >= 99%.

But if you just take a single run and increase the number N of coinflips, either LoS converges to 50% as N goes towards infinity (assuming the coinflip is perfectly fair) or it's broken.

I'm not convinced at all LoS has such a diverging behavior as claimed earlier in this thread, however.