Tie Breakers in the Tournament Rules

By sWhiteboy, in 2. AGoT Rules Discussion

Posted at
In the tournament rules it says this about tie breakers:


"If a tiebreaker between players with identical win-loss records is needed, any player who has defeated each opponent in the tied group is advanced. If no player has defeated each of the other tied players, strength of schedule is used as the tie-breaker."


Does this mean that each individual grouping of players is checked?


In this example I am seeding for second, third, and fourth place, and I have three players with the same win-loss record:


No player has beaten both of the other players. So, we look at SoS. One player has the highest SoS, and they become the second seed. At that point, do I then check the next two players to see if one of them have beaten the other, or do I just go straight down with SoS after the first time I couldn't tie break with head-to-head stats?



I would really like to hear from KTOM or someone else who has been TO of a major FFG event (like nationals or worlds).

Edited by sWhiteboy
Posted at

In your example, you would just go SoS. No "who beat whom" for 3rd and 4th after seeding 2nd.

If you did revert to looking for head-to-head wins, you would inevitably get to the following situation;

- A beats B

- B beats C

- C beats A

No "beat everyone head-to-head," so you go to SoS. C has the highest SoS, so is seeded 2nd. If you then do head-to-head, B would then be seeded 4th, despite beating the guy in 2nd. It will look like he has been placed below A for the same reason he should have been placed above C.

Any way, that's the long way to say "once you utilize SoS as a tiebreaker for a group, you stick with it."

Posted at

That is what I thought, and it is what I have seen at most tournaments (and from asking some TOs, the same thing that they thought as well).

Then, I got an answer from Damon:

"In your example, you have 3 players who are tied for second place. You use the rule quoted to determine second place.
After the tie has been broken, you are left with a new tie, between 2 players who are now tied for 3rd place. Reapply the rule from the top: If one of those remaining two players has defeated the other, he should advance based on the first part of the rule."
So, I'm left with this question. Which way is right? I mean, does FFG tournament software normally handle this and everyone just assumed it was straight SoS?
Posted at

FFG tournament software. Funny. The .jar they used was written long before they ever introduced head-to-head as the primary tie-breaker. I feel very safe saying it is really only used for seeding the cut round, which is pretty much done by hand. I could be wrong, of course.

I didn't TO any of the major events last year, so apparently they changed their policy when I wasn't looking.

My understanding was that you only "started over" from the first tie-break if two or more were still tied after applying a tie breaker. So, in my above example of A, B and C, you would only go back to the head-to-head tie break (after determining that that C had the highest SoS) if A and B had the SAME SoS. If they didn't, the three-way tie was broken by SoS alone. For what it's worth, that's what the NFL does with 3-way ties, too.

Posted at

I agree with Ktom here.

Once you've gotten to SoS as the tiebreaker, you don't revisit the head-to-head. It's unfair to the remaining players because they were paired head to head by chance and now you can have a case where the highest SoS and the lowest SoS broke pool play by random luck. (This would be Player A SoS 100, no head to head. Player B SoS 80, beats C. Player C SoS 90, but lost to B and eliminated).

I disagree that you would "start over" though. If they were tied in SoS you would actually then go to the next Tiebreaker. Which is "If the strength of schedule of two or more tied players is also tied, calculating and comparing the strength of schedule of each tied player’s set of opponents (in other words, the strength of schedule of each player’s schedule) should be the next tie-breaker."

Tiebreakers are a waterfall.

Check 1. Tie resolved? No.

Check 2. Tie resolved? No.

Check 3. Tie resolved? No.

Check 4. Tie resolved? No.

Check 5. Tie resolved? No.

Play head-to-head game for tie resolution (or tie is unresolved).

This is how I've always had ties resolved when I've played games and sports competitively.

Edited by mdc273
Posted at

I agree with Ktom here.

~ I can't possibly have read that right.

Posted at

Once you've gotten to SoS as the tiebreaker, you don't revisit the head-to-head. It's unfair to the remaining players because they were paired head to head by chance and now you can have a case where the highest SoS and the lowest SoS broke pool play by random luck. (This would be Player A SoS 100, no head to head. Player B SoS 80, beats C. Player C SoS 90, but lost to B and eliminated).

To be fair, SoS is just as much chance as head-to-head.

Edited by sWhiteboy
Posted at

I e-mailed Damon again. This time with a list of three imaginary tied players, their SoS/SoSoS, and the information that the player with the highest SoS/SoSoS had lost to both the other players (but that no one player had beaten both the other players). I then asked him to seed them for me. Apparently he was confused by my initial question, and the new answer lines up with what KTOM said, and is nice and in-depth:

"The question about someone beating all other tied players is asked once. If that is not enough to determine who is at the top rank, we discard that question.

Each player is considered part of the tied pool and you calculate SOS out as far as necessary until the entire order is revealed. In other words, treat SOS as a number with SOSOS as a decimal point, and if somehow against all odds in the universe there is still a tie, SOSOSOS as a further decimal point. Though this is such an edge case that playing an additional round of swiss would have eliminated this problem. Through all of the Gencon's and Worlds we've run I believe we've only once had to move to SOSOS.
So in your example the factoring should have produced a result like this,
Player A has 55.150
Player B has 55.180
Player C has 40.130
Giving us the final order of Player B, Player A, and then Player C.
It does not matter if Player B has lost to A and C. The only thing that matters is that none of the players has beaten every other player they are tied with.
Lets take this one step further
Players A, B, C, and D are all tied for 3rd place with a 4-4 record. Player A has beat every other player, so he gets 3rd place. We now have a new tie for 4th place between players B, C, and D. No player has beaten the other two players in that group so now we go to SOS and we calculate that out as many "decimal points" as is necessary until the placement of all three remaining players has been determined."
And order is restored to the world.
Edited by sWhiteboy
Posted at

Cool. That's how I was thinking it worked, so it's nice to see it confirmed.

#FFGOfficialResponse #Tiebreaker

Edited by mdc273