Difference between revisions of "Tiebreaker"
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| − | {{ | + | {{Rules of the game}} |
| − | *[[tiebreaker game]] | + | <onlyinclude>A '''tiebreaker''' may refer to one of several different things: |
| − | *the extra question | + | * an [[tiebreaker game|extra game(s)]] used to resolve a tie in either overall or pool standings at a [[tournament]] ('''playing it out''') |
| + | * a statistical measure used in lieu of an extra game for the same purpose | ||
| + | * an extra question(s) used during [[overtime]] to resolve a tie in the score of a [[game]]</onlyinclude> | ||
| + | |||
| + | ==Tiebreaker games== | ||
| + | :''Main article: [[tiebreaker game]]'' | ||
| + | |||
| + | A tiebreaker game is a game played on a whole or half packet to split a tie between two teams. Tiebreakers between teams tied for first overall are simply [[finals]]. | ||
| + | |||
| + | The need to play a tiebreaker can arise at the end of a tournament to determine final placement, but most tournaments will not split ties for any place other than first. National tournaments will generally determine second or third if necessary then either lump teams into tiers (e.g. t-7th at [[HSNCT]]) or order them by stats like win-loss record or [[PPG]]. | ||
| + | On rare occasions a tie for first will not be resolved either, ending the tournament in a tie. | ||
| + | |||
| + | It may also be necessary to do tiebreakers when rebracketing between stages of play like prelims, playoffs, or superplayoffs. For example splitting an 8x8 [[round robin]] into equally sized upper/lower playoff brackets requires selecting four teams from each bracket, meaning a tiebreaker is necessary if there is a tie for fourth. This is quite common: if all teams have an equal chance of winning there's a ~90% chance that there's a tie for fourth, with the majority of those being [[circle of death|circles of death]] in which three or more teams are tied. The slightly more realistic scenario where team i has an i/(i+j) chance to beat team j has a slightly lower chance of a tie for fourth of ~75%.<ref>Value calculated empirically with the following Python code:<div class="mw-collapsible mw-collapsed"> | ||
| + | import random | ||
| + | from collections import Counter | ||
| + | |||
| + | def simulate_tournament(): | ||
| + | wins = [0] * 8 | ||
| + | for i in range(8): | ||
| + | for j in range(i + 1, 8): | ||
| + | winner = random.choice([i, j]) | ||
| + | # The following line is used to weight the strength of teams by their index | ||
| + | # winner = random.choice([i] * i + [j] * j) | ||
| + | wins[winner] += 1 | ||
| + | sorted_wins = sorted(wins, reverse=True) | ||
| + | |||
| + | # Determine how many teams share the 4th-highest win total | ||
| + | fourth_score = sorted_wins[3] | ||
| + | tie_count = sum(1 for w in sorted_wins if w == fourth_score) | ||
| + | |||
| + | if tie_count == 1: | ||
| + | return "no tie for 4th" | ||
| + | else: | ||
| + | return f"{tie_count}-way tie" | ||
| + | |||
| + | def run_simulation(num_trials=1_000_000): | ||
| + | results = Counter(simulate_tournament() for _ in range(num_trials)) | ||
| + | total = sum(results.values()) | ||
| + | for outcome in sorted(results.keys()): | ||
| + | percent = 100 * results[outcome] / total | ||
| + | print(f"{outcome:<15}: {percent:6.2f}%") | ||
| + | |||
| + | if __name__ == "__main__": | ||
| + | run_simulation() | ||
| + | </div></ref> | ||
| + | |||
| + | ==Tiebreaker statistics== | ||
| + | Tiebreaker games (or half-games) are the default way to resolve ties and should be used if at all possible. If it is not possible to play them out, a statistical measure may be used instead. | ||
| + | |||
| + | The most common reasons why a tie will not be played out are "there isn't enough time" and "there are not enough questions". All tournaments are subject to this first scenario but the second is a constraint only imposed on the [[tournament director]]s of regular-season [[mirror]]s, as they cannot control the writing team. In contrast, tournaments with tailor-made sets like nationals have the opportunity to write extra packets to allow more tiebreakers in their schedule. The number of questions required remains a bottleneck, though: it is impractical to write an arbitrary number of extra questions that may not be seen and even though tiebreakers can be reused across disjoint [[bracket]]s the number required grows with schedule complexity and consequently with field size. As such it is common to both have more tiebreaker packets and employ question-saving measures like using half-packets. | ||
| + | |||
| + | Both [[ACF]] and [[NAQT]] give recommendations on the procedure for breaking ties in record but these are optional for tournaments using their ruleset but not under their direct supervision. | ||
| + | |||
| + | There are only two statistical tiebreakers with wide support: | ||
| + | # [[PPG]] when teams have played common opponents, such as within a bracket | ||
| + | # [[PPB]] when teams have not played common opponents, such as between brackets | ||
| + | |||
| + | PPG is a better choice because it incorporates tossup performance while also including PPB; however, it is very susceptible to the [[strength of schedule]] and thus should only be used when there are common opponents. PPB is more robust to this but only captures one facet of a team's strength. This is the same fundamental dilemma that the [[A-value]] and [[D-value]] metrics for developed to handle, but these are impractical to use while running a tournament and need a large sample size. | ||
| + | |||
| + | There are almost {{u|no scenarios where [[head-to-head]] record should be used}} because it double-counts the results of the match that these two teams played. [https://en.wikipedia.org/wiki/Without_loss_of_generality Without loss of generality], let's imagine that team A beat team B. Besides the inherent unfairness in having any one match be worth more, it is especially unfair to team B for that to be the head-to-head game. The argument for team A winning the tiebreaker is that they won that one match; however, in this scenario it is also necessarily true that A lost at least one game to a team that B beat.<ref>Statement (a): Both team A and B have an equal number of losses ''N''. It is known that one of the losses of team B is to team A. This means that A's losses are strict subset of B's losses as A cannot lose to themselves. Since A and B have an equal number of losses A has at least one loss to a team that B beat (not in B's losses).</ref> | ||
| + | If each team's game against this common opponent was the one double-counted instead, then B would win the tiebreaker instead. | ||
| + | |||
| + | Imagine another scheme where random games are double-counted until one team wins. Though straightforwardly terrible its symmetry actually gives both teams equal odds to win, which makes it clear that it is only the choice to select the head-to-head game that produces an effect in this tiebreaker. | ||
| + | |||
| + | ==Tiebreaker questions== | ||
| + | {{Rules disclaimer|section}} | ||
| + | |||
| + | It is possible for any game to end in a tie, either at the end of regulation gameplay or after [[protest]]s are resolved. These are always broken by entering [[overtime]] and reading additional questions, meaning that these are also the only kinds of tiebreakers with an explicit procedure outlined in the rules. | ||
| + | |||
| + | ACF rule C6 states that a sudden-death tossup is played if the game ends in a tie, with additional ones being read if there is no change in score. Addendum CX also allows for the options of using an "NAQT-style" tiebreaker or letting the match end as a tie. | ||
| + | |||
| + | NAQT rule F7 outlines a three-tossup overtime period that is played in its entirety, with the affordance for more sudden-death tossups if the score remains tied. | ||
| + | |||
| + | The questions used for tiebreakers are generally drawn from an unused packet that may or may not be subdistributed normally. There are usually few enough of these packets that a team can expect to not hear two tiebreakers in the same category, though it is unlikely that they will be in enough ties for this to be relevant. | ||
| + | |||
| + | {{Refs}} | ||
Revision as of 21:44, 13 July 2025
Before the game
During the game
A tiebreaker may refer to one of several different things:
- an extra game(s) used to resolve a tie in either overall or pool standings at a tournament (playing it out)
- a statistical measure used in lieu of an extra game for the same purpose
- an extra question(s) used during overtime to resolve a tie in the score of a game
Tiebreaker games
- Main article: tiebreaker game
A tiebreaker game is a game played on a whole or half packet to split a tie between two teams. Tiebreakers between teams tied for first overall are simply finals.
The need to play a tiebreaker can arise at the end of a tournament to determine final placement, but most tournaments will not split ties for any place other than first. National tournaments will generally determine second or third if necessary then either lump teams into tiers (e.g. t-7th at HSNCT) or order them by stats like win-loss record or PPG. On rare occasions a tie for first will not be resolved either, ending the tournament in a tie.
It may also be necessary to do tiebreakers when rebracketing between stages of play like prelims, playoffs, or superplayoffs. For example splitting an 8x8 round robin into equally sized upper/lower playoff brackets requires selecting four teams from each bracket, meaning a tiebreaker is necessary if there is a tie for fourth. This is quite common: if all teams have an equal chance of winning there's a ~90% chance that there's a tie for fourth, with the majority of those being circles of death in which three or more teams are tied. The slightly more realistic scenario where team i has an i/(i+j) chance to beat team j has a slightly lower chance of a tie for fourth of ~75%.[1]
Tiebreaker statistics
Tiebreaker games (or half-games) are the default way to resolve ties and should be used if at all possible. If it is not possible to play them out, a statistical measure may be used instead.
The most common reasons why a tie will not be played out are "there isn't enough time" and "there are not enough questions". All tournaments are subject to this first scenario but the second is a constraint only imposed on the tournament directors of regular-season mirrors, as they cannot control the writing team. In contrast, tournaments with tailor-made sets like nationals have the opportunity to write extra packets to allow more tiebreakers in their schedule. The number of questions required remains a bottleneck, though: it is impractical to write an arbitrary number of extra questions that may not be seen and even though tiebreakers can be reused across disjoint brackets the number required grows with schedule complexity and consequently with field size. As such it is common to both have more tiebreaker packets and employ question-saving measures like using half-packets.
Both ACF and NAQT give recommendations on the procedure for breaking ties in record but these are optional for tournaments using their ruleset but not under their direct supervision.
There are only two statistical tiebreakers with wide support:
- PPG when teams have played common opponents, such as within a bracket
- PPB when teams have not played common opponents, such as between brackets
PPG is a better choice because it incorporates tossup performance while also including PPB; however, it is very susceptible to the strength of schedule and thus should only be used when there are common opponents. PPB is more robust to this but only captures one facet of a team's strength. This is the same fundamental dilemma that the A-value and D-value metrics for developed to handle, but these are impractical to use while running a tournament and need a large sample size.
There are almost no scenarios where head-to-head record should be used because it double-counts the results of the match that these two teams played. Without loss of generality, let's imagine that team A beat team B. Besides the inherent unfairness in having any one match be worth more, it is especially unfair to team B for that to be the head-to-head game. The argument for team A winning the tiebreaker is that they won that one match; however, in this scenario it is also necessarily true that A lost at least one game to a team that B beat.[2] If each team's game against this common opponent was the one double-counted instead, then B would win the tiebreaker instead.
Imagine another scheme where random games are double-counted until one team wins. Though straightforwardly terrible its symmetry actually gives both teams equal odds to win, which makes it clear that it is only the choice to select the head-to-head game that produces an effect in this tiebreaker.
Tiebreaker questions
Disclaimer: This section is about rules. Its contents are not authoritative. Please consult official rules for up-to-date information.
| NAQT | gameplay rules • eligibility rules • correctness guidelines |
|---|---|
| ACF | gameplay rules • eligibility rules |
| PACE | gameplay and eligibility rules |
It is possible for any game to end in a tie, either at the end of regulation gameplay or after protests are resolved. These are always broken by entering overtime and reading additional questions, meaning that these are also the only kinds of tiebreakers with an explicit procedure outlined in the rules.
ACF rule C6 states that a sudden-death tossup is played if the game ends in a tie, with additional ones being read if there is no change in score. Addendum CX also allows for the options of using an "NAQT-style" tiebreaker or letting the match end as a tie.
NAQT rule F7 outlines a three-tossup overtime period that is played in its entirety, with the affordance for more sudden-death tossups if the score remains tied.
The questions used for tiebreakers are generally drawn from an unused packet that may or may not be subdistributed normally. There are usually few enough of these packets that a team can expect to not hear two tiebreakers in the same category, though it is unlikely that they will be in enough ties for this to be relevant.
References
- ↑ Value calculated empirically with the following Python code:
import random from collections import Counter def simulate_tournament(): wins = [0] * 8 for i in range(8): for j in range(i + 1, 8): winner = random.choice([i, j]) # The following line is used to weight the strength of teams by their index # winner = random.choice([i] * i + [j] * j) wins[winner] += 1 sorted_wins = sorted(wins, reverse=True) # Determine how many teams share the 4th-highest win total fourth_score = sorted_wins[3] tie_count = sum(1 for w in sorted_wins if w == fourth_score) if tie_count == 1: return "no tie for 4th" else: return f"{tie_count}-way tie" def run_simulation(num_trials=1_000_000): results = Counter(simulate_tournament() for _ in range(num_trials)) total = sum(results.values()) for outcome in sorted(results.keys()): percent = 100 * results[outcome] / total print(f"{outcome:<15}: {percent:6.2f}%") if __name__ == "__main__": run_simulation() - ↑ Statement (a): Both team A and B have an equal number of losses N. It is known that one of the losses of team B is to team A. This means that A's losses are strict subset of B's losses as A cannot lose to themselves. Since A and B have an equal number of losses A has at least one loss to a team that B beat (not in B's losses).