The TQB will be on everyone’s mind during the final matchday of the French D1 championship.
Indeed, ties are possible among the top teams, and calculators will be needed to separate them.
For a few years now, this mathematical formula has been adopted to separate teams tied in the standings during the round-robin phase of a competition, especially when the matches between the tied teams don’t allow for a clear separation, and when playing an elimination match (or matches) isn’t possible due to the schedule.
The pure point differential (points scored – points allowed) doesn’t apply to baseball, as the number of innings batted can vary depending on the game’s progression: the home team might not bat in the 9th inning, thus missing out on a chance to score.
Therefore, a mathematical formula—a mean—was chosen to establish the standings.
The TQB (Team Quality Balance) boils down to a simple formula: (points scored divided by innings batted) – (points allowed divided by innings fielded). The team with the best TQB is ranked higher.
We could use the offensive average, the earned run average, the number of errors, or hits, but these statistics can be questionable, as scoring can vary based on the interpretation of a play. Here, with points, a certain accounting logic is applied.
But how does this work in practice?
Let’s start with a simple example.
In Group A, let’s say Rouen loses twice to PUC on July 24th (this won’t happen, it’s just an example. And if it does, I’ll quit baseball). Rouen and Montigny would then be tied with a 10-6 record.
The first tiebreaker is the head-to-head matches. Two wins each. On the first matchday, 6-0 for Montigny and 8-1 for Rouen, and on the second matchday, 2-1 for Rouen and 4-3 for Montigny.
There were two wins each: 6-0 and 4-3 for Montigny, 8-1 and 2-1 for Rouen. Impossible to make a difference.
That’s where the TQB comes into play.
Let’s break down the calculation.
So Rouen, with a TQB of +0.032, finishes ahead of Montigny.
It came down to just two points. Two extra points scored by the Rouen team made the difference.
So, whatever happens on July 24th, Savigny and Rouen are qualified for the semifinals.
Now let’s look at Group B.
And this is a bit more complicated.
Recapping the situation:
Montpellier and Sénart have 11 wins and 3 losses, La Rochelle has 12 wins and 4 losses.
And there’s still a Montpellier vs. Sénart match to play.
If one of the two teams wins both matches, no problem, they’ll advance to the semifinals with the Boucaniers.
But if there’s a split, we’ll have a three-way tie at 12-4.
So we consider the matches between these three teams. The games against Nice and Toulouse no longer count.
But it turns out that at every matchday, there was a split:
- 1st matchday: Montpellier – La Rochelle 11-4; 1-9
_ 4th matchday: La Rochelle – Sénart 1-3, 10-2
_ 5th matchday: Sénart – Montpellier 1-4, 7-3
_ 6th matchday: La Rochelle – Montpellier 2-12, 6-4
_ 9th matchday: Sénart – La Rochelle 3-11, 9-5
_ 10th matchday: Montpellier – Sénart 1-1
So everyone is at 4 wins – 4 losses.
Therefore, we must resort to the TQB.
Montpellier has 35 points scored in 53 innings and 27 points allowed in 52 innings, for a TQB of +0.103.
La Rochelle has 48 points scored in 70 innings and 45 points allowed in 70 innings, for a TQB of +0.043.
Sénart has 25 points scored in 52 innings and 34 points allowed in 53 innings, for a TQB of –0.161.
We won’t apply all possible scenarios, but to simplify, Sénart must win by at least 8 runs across both matches to qualify. For example, with a 15-5, 2-4, or a 9-0, 0-1, it’s the Templiers who advance.
In any case, La Rochelle is qualified.
Because if Sénart beats Montpellier by 8 runs, their TQB becomes –0.014 and Montpellier’s becomes –0.029. Every additional point by the Templiers lowers the Barracudas’ TQB.
And if Montpellier limits the damage to 7 runs, their TQB becomes –0.014 and Sénart’s becomes –0.029. Same logic, every additional point by the Barracudas lowers the Templiers’ TQB.
So, with their 0.043, La Rochelle is safe.
Now, determining the final standings of the group might get a bit complicated.
- Case #1. Montpellier loses with a differential over the two matches
less than three runs (for example, 5-8, 6-4).
The Barracudas are 1st, the Boucaniers 2nd.
_ Case #2. Montpellier loses with a differential over the two matches
more than three runs and less than 7 (for example, 0-10, 9-5)
The Boucaniers are 1st, the Barracudas 2nd.
_ Case #3, Sénart wins with a differential over the two matches more
than eight runs and less than 12 runs (for example, 15-3, 4-7),
The Boucaniers are 1st, the Templiers 2nd.
_ Case #4: Sénart wins with a differential over the two matches more
than 12 runs (for example, 15-1, 0-1),
the Templiers are 1st and the Boucaniers 2nd.
The most observant among you will have noticed that two cases weren’t mentioned: those where Montpellier has a three-run lead and where Sénart has a twelve-run lead.
Well, in these two cases, there’s a double tie with exactly the same TQB between Montpellier and La Rochelle in the first case and Sénart and La Rochelle in the second case.
And in these two cases, we only consider the results between the teams in question. And it will be in favor of Montpellier (TQB 0.200) in the first case (so Montpellier 1st and La Rochelle 2nd) and La Rochelle (TQB 0.286) in the second case (so La Rochelle 1st and Montpellier 2nd).
To be completely thorough, here’s what Annex 21 of the RGES says, which details all the tiebreaker methods:
In case of impossibility to apply the provisions of Article 36.03.02 of these regulations, the following provisions shall apply:
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The team that has won the most matches between the tied teams will be awarded the best ranking.
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The team with the best TQB (team’s quality balance) (points scored divided by innings batted) – (points allowed divided by innings fielded).
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The team with the best « TQB applied to earned runs » (earned runs scored divided by innings batted) – (earned runs allowed divided by innings fielded).
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The highest batting average calculated on the matches between the tied teams.
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Draw.
When criteria 2, 3, or 4 are applied and a team is designated for the first or last place and the other teams are still tied with the same TQB result (2), earned runs TQB (3), or batting average (4), the order to separate these teams restarts at criterion 1 (number of wins between these tied teams).
That’s it, I hope we’ve been clear. On my end, I’m going to take a pill for the headache.
François Colombier
