Interesting position in the 5k–4k game Tournament Game: Through the Years: Long Correspondence (59567) R:1 (vblandy vs Jazz)
That game is progressing at about a move a week.
Interesting position in the 5k–4k game Tournament Game: Through the Years: Long Correspondence (59567) R:1 (vblandy vs Jazz)
That game is progressing at about a move a week.
The point is that we are not talking about a quantized model dealing with discrete (let alone integer) values, but rather the capacitance equations treat these values as real numbers. My original point is just to say that applying a model dealing with real numbers to something that is integers (i.e. number of games), one has to convert back to integers somehow (such as via rounding), since a fractional number of games finished is ill-defined.
While charge is fundamentally quantized (by the charge of an electron or proton), the other key quantities that the capacitance equations work with, such as voltage, current, and capacitance, are not.
However, applying any sort of simple model like this to human behavior is bound to be only a weak approximation. It is still interesting, if imprecise, to attempt to fit models to the aggregate statistical behavior of a population and speculate about the accuracy and implications.
You’re right, S_Alexander. What you’re talking about is more like radioactive decay. To put it slightly differently, each of N initial undecayed atoms has a certain probability, P, of decaying during the next second so the expected total number decaying during that time is NP, after which the number of atoms left will be N - NP = N(1-P). After T seconds there will therefore be N(1-P)^T atoms left.
So, is there a constant probability of any given game finishing in the next second?
Is the data for this tournament available somewhere in a machine-readable form?
Search for “API” in this topic
Thanks @Lys, just caught up with all your previous posts - amazing work!
Is your code online anywhere?
Can you answer the other question (easily): what does a graph of
p(date) = games_completed_on(date) / total_games_incomplete(date)
against date look like?
If this tournament is like radioactive decay then p(date) should not depend on date.
I’d probably want to filter out disqualifications. It would be interesting to compare wins, resignations and timeouts.
Thanks.
I didn’t write a single line of code.
I’m using an ETL tool (extract, transform, load) called FME to transfer informations from OGS API to a database.
If you don’t have that software, my files are pretty useless.
I think I can do that, but I don’t have daily data.
I sample data almost weekly.
We can extract details from each single game page but it’s time consuming and I didn’t yet.
Disqualified players are still playing their games.
This information too is only in game’s page.
It doesn’t matter what the period is or even if it’s constant, what matters is the proportional rate of change:
games_ended_this_period / games_not_yet_ended / length_of_period
If the period is measured in days, that gives you the probability of a game ending in any given day.
@denishowe: I have two charts for you.
Since the sampling interval isn’t fixed but it varied a bit in the past, I’m calculating the daily average of completed games for each period (which is exactly games_ended_this_period / length_of_period). It’s the orange line in the bottom right of my dashboard.
Here I put that in comparison with games still in progress (which is games_not_yet_ended):
Do they look similar?
Here is p(date) with reference lines for maximum, minimum and average values.
These numbers are pretty small!
The line seems to be more stable in later months, while it was changing a lot until december 2020.
This could be caused by the period lenght: my samples were more frequent then, so recent data are more flattened because of the average over longer intervals.
What do you think? Are we radioactive?
Wow, brilliant work, thanks very much for these.
Yes, we are definitely radioactive!
If I’m reading the graphs correctly, there’s a 1% (±0.5%) chance of a given game finishing on a given day and a half-life of three months. I.e. the number of unfinished games halves every three months. So, by
the first round should have one game left 39 months after July 2020, which is October 2023.
Vsotvep: So, with about 300 moves for a complete games, each round will take approximately 5 to 6 years. 12 rounds, says we’ll be done in 60 to 80 years.
Looks like that was an overestimate then.
Is this graph from May 5 ? May 4th, or April 5th. (Maybe should use ISO date format YYYY-MM-DD ?)
Looks like for May 4 based on prior graph
The opening is still in progress in Tournament Game: Through the Years: Long Correspondence (59567) R:1 (Jaap09 vs papageno_GO)
Yes.
If I had published this dashboard online, you’d see your standard, based on your local settings and the browser you use.
But that’s a snapshot from my pc, so it shows Italian standard (day/month/year).
I think that’s the best choice and I use it a lot in filenames. But I don’t in my dashboards just because (in the correct environment) they are responsive to user local settings.
That remains to be seen…
This is just as likely to be an underestimate I think
Thanks for the clarification. Glad it wasn’t a cut/paste error, just a localization ambiguity.
fair point
Perhaps, but some games only about 1/10th of the way through, so it could still take 5 to 6 years for the round!
1468 players out of 2233 have no games in progress. That’s about 66% or 2/3.
About 1/3 of groups have no games in progress.
I googled for radioactive decay and found this formula:
N(t)=N(0)*e^(-λt)
which is just a simple exponential witth negative exponent.
The software I’m using has automatic trend lines, includind exponential, so I tried that and here is the result:
As you can see, the trend line starts below my first sample point (day 15), then crosses two times actual curve and ends below the last point (day 313). Also the N(0) value is 9535, which is quite different from our actual games count: 10038.
The automatic trend line has a starting point too low and isn’t curved enough.
I wasn’t satisfied with that, so I tried to draw my own trend line, starting from N(0)=10038, and here it is:
It seems to me that exp curve still isn’t curved enough.