Rus
United States Mountain View California

Edit: I was assuming tournament2 format was the same as tournament1 format, hence the analysis of the top 4. See a later post for the analysis of top 16, which is more relevant to tournament2.
Having accumulated 15 prestige points after the 4th round of the 2nd octgn tournament (US division), I began wondering if there is still any chance of me making it to the final 4. This resulted in a more general statistical analsysis that anybody can use.
I ran 10K independent trials of Swiss rounds 4 and 5 of the US Octgn tournament, starting from current round 3 standings and scores. I ignored any round 4 games that have already happened and assumed that every match will have random outcomes of 06, 60, 24, 42, 33 with probabilities 0.28, 0.28, 0.18, 0.18, 0.08, respectively (emperically measured from existing games).
Below are a couple graphs from that simulation.
From the above graph you can tell that the current 1st player (AndrewRogue) has an 86% chance of staying in the top4, whereas the next 3 players only have a 2540% chance of staying in the top4. Furthermore, you stand some reasonable chance of making it to the final 4 even if you're currently ranked as low as 15, but you need to win both of the next two matches. If you're 32 or lower, your chances are 0.
From the above graph, you can tell that in order for you to make it to the top4, your prestige better be 22 or higher. Sadly, having tied my round 4 match, the best I can do is 21 points, and even then my chances of final4 are only 0.2%.
There are of course some technical deficiencies with my simulations and lots of factors I am not accounting for (e.g. I am not using the OCTGN tiebreaker pairing, just whatever the MATLAB sorting routine does, which I think explains the oscillations in graph1). However, I believe these results hold to a good approximation.
I can actually make some more detailed predictions about the final scores of the top4 finishers and other thing (in a later post).

Elad David Amir
United States New York New York

Re: can you make it to the final 4 in the OCTGN tournament? (statistical analysis)
Brilliant! Post an analysis showing how a bunch of people have a small chance of reaching final four, causing aforementioned people to drop from the tournament, thus increasing the OP's chance!! Such creative cunning puts even the finest Jinteki CEOs to shame!



Re: can you make it to the final 4 in the OCTGN tournament? (statistical analysis)
What I'm really curious about is what score you are likely to need to make the top 16 and continue the tournament after the swiss rounds.
Edit:
rbelikov wrote: Edit: just checked the tournament rules, looks like tournament 2 is not the same as tournament 1! So, we have 6 Swiss rounds and then the top 16 of US and Europe go to 5 singleelimination rounds? I'll have to redo my analysis... Greatness. I can't wait.

Rus
United States Mountain View California

Re: can you make it to the final 4 in the OCTGN tournament? (statistical analysis)
IirionClaus wrote: Brilliant! Post an analysis showing how a bunch of people have a small chance of reaching final four, causing aforementioned people to drop from the tournament, thus increasing the OP's chance!! Such creative cunning puts even the finest Jinteki CEOs to shame!
You uncovered my secret plan! I am undone!
But seriously, I haven't even considered that my post might have this effect. I sincerely hope nobody drops! (And I was under the impression that signing up for the tournament obligated you to stay with it.)
(PS if somebody drops because their chance of top 4 is negligible, then this will only improve my chances negligibly.)
Yan wrote: What I'm really curious about is what score you are likely to need to make the top 16 and continue the tournament after the swiss rounds. I was under the impression only the top 4 players continue after the Swiss rounds (hence my analysis), similar to the 1st octgn tournament. Is this not the case?
Edit: just checked the tournament rules, looks like tournament 2 is not the same as tournament 1! So, we have 6 Swiss rounds and then the top 16 of US and Europe go to 5 singleelimination rounds? I'll have to redo my analysis...

Geoff Hollis
Canada Edmonton Alberta

Re: can you make it to the final 4 in the OCTGN tournament? (statistical analysis)
1. What about making the top 16, rather than the top 4? The top 16 prestige earners in both divisions make the cut to single elimination, after all.
2.
rbelikov wrote: I ignored any round 4 games that have already happened and assumed that every match will have random outcomes of 06, 60, 24, 42, 33 with probabilities 0.28, 0.28, 0.18, 0.18, 0.08, respectively (empirically measured from existing games).
It is interesting to wonder how much this assumption might affect the simulation's validity. In practice, there's at least two important things going on in a swissstyle tournament. First, each successive round results in opponents that are bettermatched in terms of skill. The result is that the frequency of 60/06 outcomes should go down each round, and the frequency of 33/42/24 outcomes should go up. Outside of the context of swiss structure, your chance of winning successive matches is not independent; a player who wins a match 60 is more likely to be able to do that again than a player who loses a match 06.
Nonetheless, a neat way to approach the problem of making a ranking threshold.

Rus
United States Mountain View California

Re: can you make it to the final 4 in the OCTGN tournament? (statistical analysis)
Ok, here is the analysis of top 16, along with some predictions.
1. The current #1 player (AndrewRogue) is not guaranteed to stay in the top16. 2. If you're currently in the top16, you have at least a 45% chance of remaining in the top 16. 3. Everybody currently above about rank ~45 has a shot at top 16.
There appears to be a fairly sharp score cutoff for making it to the top16. Most people with score 22 and higher will make it to the top 16, and most people with score 21 and lower will not.
Now for some predictions of US tournament winner.
1. Current leader (AndrewRogue) has about a 30% chance of winning the US Swiss. 2. The next 3 players have about 10% chance each. 3. There is about a 40% chance that it will be somebody else yet. 4. Roughly everybody currently ranked 15 or higher has a shot at winning.
What will be the winner's score?
1. If you want a reasonable shot at winning, your score needs to be at least 30. 2. Most likely winning scores are 30 and 32, with about 30% probability each.

Rus
United States Mountain View California

hollis wrote: In practice, there's at least two important things going on in a swissstyle tournament. First, each successive round results in opponents that are bettermatched in terms of skill. The result is that the frequency of 60/06 outcomes should go down each round, and the frequency of 33/42/24 outcomes should go up. Outside of the context of swiss structure, your chance of winning successive matches is not independent; a player who wins a match 60 is more likely to be able to do that again than a player who loses a match 06.
If somebody wants to quantify these effects, I can fold them into my simulations. (I.e. what I need is the probabilities of different match outcomes as a function of swiss round, and as a function of a player's prior performance.) I suspect these effects won't dramatically change the final results though.

B C Z
United States Reston Virginia

Though I appreciate the effort, I sincerely hope that this does not increase the drop rate or the noncommunication rate due to people figuring it just isn't worth it any more.
As a reminder:
Chance of getting a participation badge if you participate in all 6 rounds: 100%
Chance of getting a participation badge if you voluntarily leave or are forcibly dropped: 0%

Geoff Hollis
Canada Edmonton Alberta

rbelikov wrote: hollis wrote: In practice, there's at least two important things going on in a swissstyle tournament. First, each successive round results in opponents that are bettermatched in terms of skill. The result is that the frequency of 60/06 outcomes should go down each round, and the frequency of 33/42/24 outcomes should go up. Outside of the context of swiss structure, your chance of winning successive matches is not independent; a player who wins a match 60 is more likely to be able to do that again than a player who loses a match 06. If somebody wants to quantify these effects, I can fold them into my simulations. (I.e. what I need is the probabilities of different match outcomes as a function of swiss round, and as a function of a player's prior performance.) I suspect these effects won't dramatically change the final results though.
geekmail me a link to your raw data? I will see if I can add anything. My intuitions are in line with yours. The observations were more academic than pragmatic.
The seasonal nature of the probabilities in your first figure is a bit perplexing. I cannot wrap my head around how it is possible for ER(a, i+j) > ER(b, i+j) if R(a, i) < R(b, i). Where ER(X,Y) is the expected rank of person X in round Y and R(X,Y) is the actual rank of person X in round Y. Thoughts?

Rus
United States Mountain View California

hollis wrote: geekmail me a link to your raw data? My raw data was the results of rounds 13 of OCTGN US tournament 2. Just manually counted. Differences between rounds are not statistically significant there due to low number of games.
hollis wrote: The seasonal nature of the probabilities in your first figure is a bit perplexing. I cannot wrap my head around how it is possible for ER(a, i+j) > ER(b, i+j) if R(a, i) < R(b, i). Where ER(X,Y) is the expected rank of person X in round Y and R(X,Y) is the actual rank of person X in round Y. Thoughts? I noticed that as well, and commented on it in the OP. Basically I think it's because players of different ranks can have the same score, and so things depend on tiebreaker rules. My tiebreaker rules are simply whatever the default MATLAB sorting algorithm uses (which I'm sure is differnet from the octgn tiebreaker rules). I didn't bother implementing octgn tiebreaking rules (does anybody even know what they are?)
edit: by octgn tiebreaker rules I mean challonge tiebreaker rules.



byronczimmer wrote: Though I appreciate the effort, I sincerely hope that this does not increase the drop rate or the noncommunication rate due to people figuring it just isn't worth it any more.
As a reminder:
Chance of getting a participation badge if you participate in all 6 rounds: 100%
Chance of getting a participation badge if you voluntarily leave or are forcibly dropped: 0%
Speaking of which, how come you're sitting with a participation badge already? Can you make your own badge?



Tuism wrote: byronczimmer wrote: Though I appreciate the effort, I sincerely hope that this does not increase the drop rate or the noncommunication rate due to people figuring it just isn't worth it any more.
As a reminder:
Chance of getting a participation badge if you participate in all 6 rounds: 100%
Chance of getting a participation badge if you voluntarily leave or are forcibly dropped: 0% Speaking of which, how come you're sitting with a participation badge already? Can you make your own badge?
I think it's from the first tournament.

Andrew Bartosh
Sunnyvale California

So what I'm gathering from this is I'm about to ruin multiple statistical models at once?

Jeremy Larner
United Kingdom

AndrewRogue wrote: So what I'm gathering from this is I'm about to ruin multiple statistical models at once?
It's called bucking the trend.... I thought you prided your self on doing that...?

Rus
United States Mountain View California

AndrewRogue wrote: So what I'm gathering from this is I'm about to ruin multiple statistical models at once?
Haha
In seriousness though, my model did not assume you were special in any way and so you had a 28% of receiving 0 prestige for your match just like everybody else. So no "ruining the model" or "bucking the trend" has occured (yet)!
However, my above estimates will of course be affected by your match (and everybody else's). I'll update them after round 4 is over.
Edit: actually, only figure 3 will be affected (the one about who has what chances of winning). The other figures are very insensitive to individual match results. Top 16 threshold is still 21 or 22 points, and most likely winning score is still 30 or 32. In fact, these results should hold true for any 6round Swiss Netrunner tournament, including the European one.

Lothar Neu
Germany Sarstedt Niedersachsen

rbelikov wrote: actually, only figure 3 will be affected (the one about who has what chances of winning). The other figures are very insensitive to individual match results. Top 16 threshold is still 21 or 22 points, and most likely winning score is still 30 or 32. In fact, these results should hold true for any 6round Swiss Netrunner tournament, including the European one.
I have 17 points after 4 rounds. Hello intentional draws!

B C Z
United States Reston Virginia

Tuism wrote: byronczimmer wrote: Though I appreciate the effort, I sincerely hope that this does not increase the drop rate or the noncommunication rate due to people figuring it just isn't worth it any more.
As a reminder:
Chance of getting a participation badge if you participate in all 6 rounds: 100%
Chance of getting a participation badge if you voluntarily leave or are forcibly dropped: 0% Speaking of which, how come you're sitting with a participation badge already? Can you make your own badge?
That's from the first event. When I went through the (very lengthy) process of getting the first event's badge created, I had multiples made for the future.
In order to minimize the time the admins have to assign these badges (which is apparently a somewhat manual process), I send them ONE list in a proscribed format at the end of the Swiss period.
The SE portion will have some form of physical prize  I just have to get some time on the local laser cutter, but they will be similar to these  though in a different color due to materials on hand (and will have an '02' and not an '01'):
___ Edits: Picture sizing and missing paren

B C Z
United States Reston Virginia

KingLz wrote: rbelikov wrote: actually, only figure 3 will be affected (the one about who has what chances of winning). The other figures are very insensitive to individual match results. Top 16 threshold is still 21 or 22 points, and most likely winning score is still 30 or 32. In fact, these results should hold true for any 6round Swiss Netrunner tournament, including the European one. I have 17 points after 4 rounds. Hello intentional draws!
Just remember that Intentional Draws leave you with weaker opponents for strength of schedule and, if it comes to that, weaker personal match points in the 3rd string tiebreaker.

Elad David Amir
United States New York New York

Just wanted to point out that while the alt text for tournaments 17 and 9 is of the form "OCTGN Android:Netrunner Tournament #x Participant", the alt text for tournament 8 is "OCTGN Android:Netrunner Tournament Participant" without the tournament number.

B C Z
United States Reston Virginia

IirionClaus wrote: Just wanted to point out that while the alt text for tournaments 17 and 9 is of the form "OCTGN Android:Netrunner Tournament #x Participant", the alt text for tournament 8 is "OCTGN Android:Netrunner Tournament Participant" without the tournament number.
I noticed that, and will ask them to fix it later, most likely when I send the participants list. It's so far into the future that it's not an issue right now.

Lothar Neu
Germany Sarstedt Niedersachsen

byronczimmer wrote: KingLz wrote: rbelikov wrote: actually, only figure 3 will be affected (the one about who has what chances of winning). The other figures are very insensitive to individual match results. Top 16 threshold is still 21 or 22 points, and most likely winning score is still 30 or 32. In fact, these results should hold true for any 6round Swiss Netrunner tournament, including the European one. I have 17 points after 4 rounds. Hello intentional draws! Just remember that Intentional Draws leave you with weaker opponents for strength of schedule and, if it comes to that, weaker personal match points in the 3rd string tiebreaker.
I know. Maybe I won't draw, games are meant to be played, right? But I really want to make the playoffs and the Prestige system really seems to be really unpredictable, because winning does not equal winning in any case.

Aaron Smith
United States Kalamazoo Michigan

Well, I calculate that, barring me getting in on my winning charm alone, I have about a snowball's chance on the sun of making it to the top 16. But I plan on playing all my matches anyway because I just enjoy playing the game...win or lose. I wasn't planning on making it to the final bracket, just wanted to play for the tournament experience.

B C Z
United States Reston Virginia

MinisterOfPropaganda wrote: Well, I calculate that, barring me getting in on my winning charm alone, I have about a snowball's chance on the sun of making it to the top 16. But I plan on playing all my matches anyway because I just enjoy playing the game...win or lose. I wasn't planning on making it to the final bracket, just wanted to play for the tournament experience.
You get a nickel

Scott C
United States Troy OH

MinisterOfPropaganda wrote: I have about a snowball's chance on the sun Hmm. How many subroutines has the Snowball broken on this run?

Rus
United States Mountain View California

rbelikov wrote: hollis wrote: The seasonal nature of the probabilities in your first figure is a bit perplexing. I cannot wrap my head around how it is possible for ER(a, i+j) > ER(b, i+j) if R(a, i) < R(b, i). Where ER(X,Y) is the expected rank of person X in round Y and R(X,Y) is the actual rank of person X in round Y. Thoughts? I noticed that as well, and commented on it in the OP. Basically I think it's because players of different ranks can have the same score, and so things depend on tiebreaker rules. My tiebreaker rules are simply whatever the default MATLAB sorting algorithm uses (which I'm sure is differnet from the octgn tiebreaker rules). I didn't bother implementing octgn tiebreaking rules (does anybody even know what they are?) edit: by octgn tiebreaker rules I mean challonge tiebreaker rules.
It turns out that this argument is insufficient to explain the oscillations on my graphs. I found a minor bug in the code after which these oscillations disappeared (without affecting the overall shape of the curve). I won't bother posting corrected graphs since I think the accuracy of my simulations is not better than those oscillations anyway.


