Tie-breakers

[Cryomaniac]

Originally Posted by Markus Heinsohn

Ah, we're talking about seeding. For me that is something different than tie-breaking, which indeed did get work in OOTP 13. Anyway, I'll have a look at this prior to the OOTP 14 release.

There's also the issue of 3 teams being tied for 2 spots (although I'm not sure how that works in RL (Ping: LGO)).

[The Wolf]

Originally Posted by Markus Heinsohn

Ah, we're talking about seeding. For me that is something different than tie-breaking, which indeed did get work in OOTP 13. Anyway, I'll have a look at this prior to the OOTP 14 release.

Hurray!

[BusterKing]

Originally Posted by oman19

I just hope some more people will join us in our cause.

I am 99% supporting this cause.... BUT, I will not go to the extreme of not buying the game. I have bought it without fail from the first version and to me it's like I have a lifetime subscription.



This is a very high priority glitch that needs to be fixed a.s.a.p. If it's not doing it right, then yes, it is a glitch.

[cody8200]

I put in in as a bug for OOTP 14 beta as a reminder to Markus. I hope this gets looked at as well.

[Le Grande Orange]

Originally Posted by Cryomaniac

There's also the issue of 3 teams being tied for 2 spots (although I'm not sure how that works in RL (Ping: LGO)).

If it's three teams tied for the two wild card spots, the clubs are designated A, B, and C based on their relative head-to-head records. The games played are as follows:

Day 1: B at A (winner is a wild card qualifier; loser plays on Day 2)
Day 2: loser of B/A at C (winner is other wild card qualifier; loser is eliminated)

I am 99% sure OOTP would not handle such a tie situation correctly (or at least, not by MLB's current rules on the subject).

[Cryomaniac]

Thanks. Obviously there are all sorts of other bizarre scenarios I could come up with, but I doubt there are any real life rules for them, so we can ignore them for now lol

[oman19]

Originally Posted by BusterKing

I am 99% supporting this cause.... BUT, I will not go to the extreme of not buying the game. I have bought it without fail from the first version and to me it's like I have a lifetime subscription.



This is a very high priority glitch that needs to be fixed a.s.a.p. If it's not doing it right, then yes, it is a glitch.

Buster...I would never want you to not buy the game just lend us your voice! I don't want Markus and gang to go broke while I wait to get this in!!!! Although by his last response in this thread it looks like I need him to draw up a contract for me so I can be on the hook to pre-order the next 5 versions.

[oman19]

Originally Posted by Le Grande Orange

If it's three teams tied for the two wild card spots, the clubs are designated A, B, and C based on their relative head-to-head records. The games played are as follows:

Day 1: B at A (winner is a wild card qualifier; loser plays on Day 2)
Day 2: loser of B/A at C (winner is other wild card qualifier; loser is eliminated)

I am 99% sure OOTP would not handle such a tie situation correctly (or at least, not by MLB's current rules on the subject).

Do you know if you could break these ties manually by scheduling the games as appropriate?

[oman19]

Originally Posted by cody8200

I put in in as a bug for OOTP 14 beta as a reminder to Markus. I hope this gets looked at as well.

Very much appreciated I finally feel like we're making some headway.

[oman19]

Originally Posted by Markus Heinsohn

Ah, we're talking about seeding. For me that is something different than tie-breaking, which indeed did get work in OOTP 13. Anyway, I'll have a look at this prior to the OOTP 14 release.

Sorry if I wasn't clear in my original post. The issue I'm running into all the time is the issue of seeding when there is a tie which is not handled correctly. According to wikipedia this is the proper way to break a seeding tie.

Breaking ties without playoff games

1.The team with the best record in head to head play.
2.The team with the best overall record in intradivision games.
3.The team with the best record in the final 81 games of the season, ignoring interleague play.
4.The team with the best record in the final 82 games of the season


I can't tell you how happy I am this is getting a look. My pregnant wife jumped when I said YES!!!!! really loud. Markus and the beta guys. Could you make sure to keep me updated as to the status of the work on this because it's killing me not pre-ordering!!!!!

[Le Grande Orange]

Originally Posted by Cryomaniac

Obviously there are all sorts of other bizarre scenarios I could come up with, but I doubt there are any real life rules for them, so we can ignore them for now lol

Not really. Most tie situations boil down to X number of teams tied for Y number of playoff spots. MLB had to address all sorts of potential tie scenarios when it added the second wild card qualifier. See here for the complete list of its rules for various tie scenarios.

One thing to note that's not mentioned there is that, when it comes to seeding teams for match-up purposes in the Division Series round, that seeding only factors in the 162-game regular season record; the results of any tie-breaking games played by a club are disregarded.

Originally Posted by oman19

Do you know if you could break these ties manually by scheduling the games as appropriate?

Most should be, yes, because after you finish with the tie-breaking games clubs will have different winning percentages and thus would no longer be tied.

Using the previously mentioned tie scenario as an example, if A and C won, the three teams' records in tie-breaking games would be: A, 1-0; C, 1-0; and B, 0-1. If A and B won, then the clubs would be: A, 1-0: B, 1-1; C, 0-1. If B and C won, it'd be: B, 1-0; C, 1-0; A, 0-1. No matter the result, one team will clearly have a worse overall record after the tie-breaking games and is therefore eliminated.

The one scenario which doesn't seem to work is the one involving a three-way tie for a division tie, with those clubs being simultaneously tied for the second wild card position. In that case the standard three-way tie-breaker is played: Day 1 is B at A; Day 2 is C at A/B. The winner on Day 2 gets the division title. The loser on Day 2 is declared the second wild card qualifier. If A and C win, it's not a problem. C is 1-0, A is 1-1, and B is 0-1. Similarly, if B and C win, that's fine too: C is 1-0, B is 1-1, and A is 0-1. The clubs cannot have the same winning percentage once the tie-breaking games are added to their regular season record, so one team is clearly eliminated.

There is a problem, however, if A or B wins on both days. Then the results are: A, 2-0; B, 0-1; C, 0-1; or B, 2-0; A, 0-1; and C, 0-1. Technically, since two clubs went 0-1 in the tie-breaker, that means they'd still have the same overall winning percentage. So, really, they're still tied. But the team losing on Day 2 gets the second wild card, meaning the team losing on Day 1 gets shafted. I'm not sure if this is an oversight in MLB's rules or if the rule is intentional. Given that in all the new tie scenarios brought about by the second wild card qualifier, they are all resolved by no more than two days of extra play, it would seem to indicate it is intentional. But it would make for an interesting final standing when two clubs have the same final record but only one of them goes to the post-season.

[Cryomaniac]

Originally Posted by Le Grande Orange

Not really. Most tie situations boil down to X number of teams tied for Y number of playoff spots. MLB had to address all sorts of potential tie scenarios when it added the second wild card qualifier. See here for the complete list of its rules for various tie scenarios.

One thing to note that's not mentioned there is that, when it comes to seeding teams for match-up purposes in the Division Series round, that seeding only factors in the 162-game regular season record; the results of any tie-breaking games played by a club are disregarded.

I was thinking more of non-MLB setups causing odd things (such as say 4 teams tied for 3 wildcard slots).

Clearly we are never going to be able to come up with a solution to every scenario (although for "x teams tied for 1 playoff place", it's pretty trivial, even up to the max number of teams OOTP allows, even if it would take a lot of play off games).

[Le Grande Orange]

Originally Posted by oman19

Breaking ties without playoff games

1.The team with the best record in head to head play.
2.The team with the best overall record in intradivision games.
3.The team with the best record in the final 81 games of the season, ignoring interleague play.
4.The team with the best record in the final 82 games of the season

Actually, that procedure is for determining which club gets to host the tie-breaking playoff game between two clubs tied for a division title.

For determining the seeding order of the three division winners for the Division Series round, you need to insert one additional step:

(1) The team with the better record in head-to-head play;
(2) The team with the better record in games within its division;
(3) The team with the better record in games within its league;
(4) The team with the better record in games within its league over the last half of the season (81 games);
(5) The team with the better record in games within its league over the last half of the season plus one game (82 games). If still tied, then continue to go back one intraleague game at a time until such time as the clubs are no longer tied.

Note also that the results of any tie-breaking playoff games are disregarded for seeding purposes. Only results from the 162-game regular season are considered.

[Le Grande Orange]

Originally Posted by Cryomaniac

I was thinking more of non-MLB setups causing odd things (such as say 4 teams tied for 3 wildcard slots).

Four teams tied for three spots? Not a problem. Designate the teams as A, B, C, and D based on their aggregate head-to-head records. The tie-breaking games are:

Day 1: B at A; D at C (winners advance to post-season)
Day 2: loser of D/C at loser of B/A (winner advances to post-season; loser is eliminated)

Three teams advance from four tied clubs. Piece of cake.

It only gets a bit tricky if you want to factor in a division tie as well as a tie for wild card spots. But MLB has a procedure for it, and it's similar to the above.

[Cryomaniac]

Originally Posted by Le Grande Orange

Four teams tied for three spots? Not a problem. Designate the teams as A, B, C, and D based on their aggregate head-to-head records. The tie-breaking games are:

Day 1: B at A; D at C (winners advance to post-season)
Day 2: loser of D/C at loser of B/A (winner advances to post-season; loser is eliminated)

Three teams advance from four tied clubs. Piece of cake.

It only gets a bit tricky if you want to factor in a division tie as well as a tie for wild card spots. But MLB has a procedure for it, and it's similar to the above.

Okay, now I'm trying to come up with something that isn't straight forward just to be difficult ( ) ... 7 teams tied for 4 slots?

(I've also just been working out "x teams for one place" and writing it in notepad, I'm up to 10)

Edit: Here's the list for those, after 4 is non-official, and me making things up, but it's how I'd do it:

tie breaking for one place:

2 teams:

day 1: A v B

3 teams:

day 1: A v B
day 2: (AB) v C

4 teams: 

day 1: A v B; C v D
day 2: (AB) v (CD)

5 teams:

day 1: A v B; C v D
day 2: (CD) v E
day 3: (AB) v (CDE)

6 teams:

day 1: A v B; C v D;
day 2: (AB) v E; (CD) v F
day 3: (ABE) v (CDF)

7 teams:

day 1: A v B; C v D; E v F
day 2: (AB) v (CD); (EF) v G
day 3: (ABCD) v (EFG)

8 teams:

day 1: A v B; C v D; E v F; G v H
day 2: (AB) v (CD); (EF) v (GH)
day 3: (ABCD) v (EFGH)

9 teams:

day 1: A v B; C v D; E v F
day 2: (AB) v G; (CD) v H; (EF) v I
reallocate - ABG = J; CDH = K; EFI = L
day 3: J v K
day 4: (JK) v L

10 teams:

day 1: A v B; C V D; E v F; G v H; I v J
day 2: (AB) v (CD); (EF) v (GH);
day 3: (ABCD) v (EFGH)
day 4: (ABCDEFGH) v (IJ)

[Le Grande Orange]

Originally Posted by Cryomaniac

Okay, now I'm trying to come up with something that isn't straight forward just to be difficult ( ) ... 7 teams tied for 4 slots?

(I've also just been working out "x teams for one place" and writing it in notepad, I'm up to 10)

Every situation is ultimately solvable. It's just a matter of whether the process is convenient or not. MLB's practice currently is that all tie scenarios have to be resolved by no more than two days of extra play, since the realities of the post-season schedule and Wild Card Games don't allow more time than that.

I'd say anything involving up to four-way ties should be included, particularly since MLB has rules which cover that. Five-way ties are exceedingly unlikely, and probably don't need to be included, but I did work out a process for a five-way tie for two and three playoff spots.

For anything over that, a simple system, similar to that used in the minor leagues, could be put in place where non-game tie-breaking methods are used to eliminate the lowest club(s) in a tie situation, leaving only the number of tied teams for which there is a more detailed procedure set out. For example, if OOTP topped out at four-way ties, then in the case of a five-way tie, the standard non-game tie-breakers (e.g. head-to-head record, divisional record, etc.) are used to determine the bottom club of the five, and it is eliminated outright. That leaves four clubs who can then engage in the four-way tie-breaking playoff process.

[Cryomaniac]

Originally Posted by Le Grande Orange

For anything over that, a simple system, similar to that used in the minor leagues, could be put in place where non-game tie-breaking methods are used to eliminate the lowest club(s) in a tie situation, leaving only the number of tied teams for which there is a more detailed procedure set out. For example, if OOTP topped out at four-way ties, then in the case of a five-way tie, the standard non-game tie-breakers (e.g. head-to-head record, divisional record, etc.) are used to determine the bottom club of the five, and it is eliminated outright. That leaves four clubs who can then engage in the four-way tie-breaking playoff process.

That's certainly the most logical answer. Are non-game tiebreakers in OOTP at all at the moment?

[oman19]

Originally Posted by Cryomaniac

That's certainly the most logical answer. Are non-game tiebreakers in OOTP at all at the moment?

Was wondering the same thing myself.

[hefalumps]

Whoops... I posted my thought before I finished completely reading the thread, so never mind.

But I agree with everything said so far. And glad Markus is looking into seeding and home field advantage taking head-to-head record into account.

[oman19]

Have my fingers crossed we see an update about this in the next news letter.