How To Keep Round Robin Matches Interesting

Italian translation at settesei.it

Round robins–such as the formats used by the ATP and WTA Tour Finals–have a lot going for them. Fans are guaranteed at least three matches for every player, and competitors can recover from one (or even two) bad outings. Best of all, when compared to a knockout-style draw, it’s twice as much tennis.

On the other hand, round robins have one major drawback: They can result in meaningless matches. It’s fairly common that, after two matches, a player is guaranteed a spot in the semifinals (sometimes even a specific seed) or eliminated from contention altogether. At a high-profile event such as the Tour Finals, with sky-high ticket prices, do we really want to run the risk of dead rubbers?

I don’t claim to have the answer to that question. However, we can take a closer look at the round robin format to answer several relevant questions. What is the probability that the final day of a four-player group will include at least one dead rubber? What about the final match? And most importantly, before the event begins, can we set the schedule in such a way to minimize the likelihood of dead rubbers?

The range of possibilities

As a first step, let’s determine all of the possible outcomes of the first four matches in a four-player round robin group. For convenience, I’ll refer to the players as A, B, C, and D. The first day features two matches, A vs B and C vs D. The second day is A vs C and B vs D, leaving us with a final day of A vs D and B vs C.

Each match has four possible outcomes: the first player wins in two sets, the first player wins in three, the second wins in two, or the second wins in three. (Sets won are important because they are used as a tiebreaker when, for instance, three players win two matches apiece.) Thus, there are 4 x 4 x 4 x 4 = 256 possible arrangements of the group standings entering the final day of round robin play.

Of those 256 permutations, 32 of them (12.5%) include one dead rubber on the final day. In those cases, the other match is played only to decide semifinal seeding between the players who will advance. Another 32 of the 256 permutations involve one “almost-dead” match, between a player who has been eliminated and a player who is competing only to determine semifinal seeding.

In other words, one out of every four possible outcomes of the first two days results in a day three match that is either entirely or mostly meaningless. Later on, we’ll dig into the probability that these outcomes occur, which depends on the relative skill levels of the four players in the group.

Before we do that, let’s take a little detour to define our terms. Because of the importance of semifinal seeding, some dead rubbers are less dead than others. Further, it is frequently the case that one player in a match still has a shot at the semifinals and the other doesn’t. Altogether, from “live” to “dead,” there are six gradations:

  1. live/live — both players are competing to determine whether they survive
  2. live/seed — one player could advance or not; the other will advance, and is playing to try to earn the #1 group seed
  3. live/dead — one player is trying to survive; the other is eliminated
  4. seed/seed — both players will advance; the winner gets the #1 group seed
  5. seed/dead — one player is in the running for the #1 seed; the other is eliminated
  6. dead/dead — both players are eliminated

All else equal, the higher a match lies on that scale, the more engaging its implications for the tournament. For the remainder of this article, I’ll refer only to the “dead/dead” category as “dead rubbers,” though I will occasionally discuss the likelihood of “dead/seed” matches as well. I’ll assume that the #1 seed is always more desirable than #2 and ignore the fascinating but far-too-complex ramifications of situations in which a player might prefer the #2 spot.

The sixth match

As we’ve seen, there are many sequences of wins and losses that result in a dead rubber on day three. Once the fifth match is played, it is even more likely that the seedings have been determined, making the sixth match meaningless.

After five matches, there are 1,024 possible group standings. (256 permutations after the first four matches, multiplied by the four possible outcomes of the fifth match.) Of those, 145 (14.1%) result in a dead sixth rubber, and another 120 (11.7%) give us a “dead/seed” sixth match.

We haven’t yet determined how likely it is that we’ll arrive at the specific standings that result in dead sixth rubbers. So far, the important point is that dead rubbers on day three aren’t just flukes. In a four-player round robin, they are always a real possibility, and if there is way to minimize their likelihood, we should jump at the chance.

Real scenarios, really dead rubbers

To figure out the likelihood of dead rubbers in practical situations, like the ATP and WTA Tour Finals, I used a hypothetical group of four players with Elo ratings spread over a 200-point range.

Why 200? This year’s Singapore field was very tightly packed, within a little bit more than 100 points, implying that the best player, Angelique Kerber, had about a 65% chance of beating the weakest, Svetlana Kuznetsova. By contrast, the ATP finalists in London are likely to be spread out over a 400-point range, giving the strongest competitor, Novak Djokovic, at least a 90% edge over the weakest.

I’ve given our hypothetical best player a rating of 2200, followed by a field of one player at 2130, one at 2060, and one at 2000. Thus, our favorite has a 60% chance of beating the #2 seed, a 69% chance of defeating the #3 seed, and a 76% chance of besting the #4 seed.

For any random arrangement of the schedule, after the first two days of play, this group has a 17% chance of giving us a dead rubber on day three, plus a 23% chance of a “dead/seed” match on day three.

After the fifth match is contested, there is a 16% chance of that the sixth match is meaningless, with an additional 12% chance that the sixth match falls into the “dead/seed” category.

The wider the range of skill levels, the higher the probability of dead rubbers. This is intuitive: The bigger the range between the top and bottom, the more likely that the best player will win their first two matches–and the more likely they will be straight-setters. Similarly, the chances are higher that the weakest player will lose theirs. The higher the probability that players go into day three with 2-0 or 0-2 records, the less likely that day three matches have an impact on the outcome of the group.

How to schedule a round robin group

A 17% chance of a dead rubber on day three is rather sad. But there is a bright spot in my analysis: By rearranging the schedule, you can raise that probability as high as 24.7% … or drop it as low as 10.7%.

Remember that our schedule looks like this:

Day one: A vs B, C vs D

Day two: A vs C, B vs D

Day three: A vs D, B vs C

We get the lowest possible chance of a day three dead rubber if we put the players on the schedule in order from weakest to strongest: A is #4, B is #3, and so on:

Day one: #4 vs #3, #2 vs #1

Day two: #4 vs #2, #3 vs #1

Day three: #4 vs #1, #3 vs #2

There is a small drawback to our optimal arrangement: It increases the odds of a “dead/seed” match. It turns out that you can only optimize so much: No matter what the arrangement of the competitors, the probability of a “dead/dead” or “dead/seed” match on day three stays about the same, between 39.7% and 41.7%. While neither type of match is desirable, we’re stuck with a certain likelihood of one or the other, and it seems safe to assume that a “dead/seed” rubber is better than a totally meaningless one.

Given how much is at stake, I hope that tournament organizers heed this advice and schedule round robin groups in order to minimize the chances of dead rubbers. The math gets a bit hairy, but the conclusions are straightforward and dramatic enough to make it clear that scheduling can make a difference. Over the course of the season, almost every tennis match matters–it would be nice if every match at the Tour Finals did, too.

(I wrote more about this, which you can read here.)

The Pointlessness of Playing the Lets

Italian translation at settesei.it

Some people always want tennis matches to be shorter. Among the many recurring proposals to accomplish that, one that has been implemented in some places is eliminating service lets. In other words, serves are treated the same way as any other shot: If the serve clips the net and lands in the box, it’s in play.

“No-let” rules have been adopted by World Team Tennis and American university tennis. In the latter case, eliminating lets has more to do with ensuring fair play in the absence of an umpire. In 2013, the ATP experimented with no lets on the Challenger tour for the first three months of the year.

With an umpire on every professional court and machines that detect service lets at tour-level events, fairness (or avoiding cheating) is not the issue here. The reason we’re talking about this is that service lets take time, and apparently time is the enemy.

How much time?

The Match Charting Project has tracked lets in most of the 2,500-plus matches it has logged. Thus, we have some real-life data on the frequency of service lets. For today, I’ve limited our view to matches since 2010, which still gives us more than 2,000 matches to work with.

The average men’s match in the database, which consists of 151 total points, had six first-serve lets and fewer than one (0.875) second-serve let. Women’s matches are similar: Of the typical 139 points, there were 4.5 first-serve lets and 0.8 second-serve lets.

Let’s estimate the extra time all those lets are taking. After a first-serve let, most players restart their preparations, so let’s say a first-serve let is an extra 20 seconds. When the second serve is a let, most players are quicker to try again, so call that 10 seconds.

For the average men’s match in the database, that’s an extra 128 seconds–just over two minutes. For women, that’s 99 extra seconds per match. In both cases, the time consumed by service lets is less than one second per point.  Just about any other rule change aimed at speeding up the game would be more effective than that.

Even at the extremes, it’s tough to argue that service lets are taking too much time. Of all the matches in the charting database, none had more than 24 service lets, and that was in the 2012 London Olympics marathon between Roger Federer and Juan Martin Del Potro. Using the estimates I gave above, those 20 first-serve and four second-serve lets accounted for just over seven minutes of the total match time of 4:26.

Only one of the 1,000 women’s matches in the database featured more than 17 service lets or more than five let-attributable minutes: Petra Cetkovska‘s three-set upset of Angelique Kerber at the 2014 Italian Open. That outlier included 22 lets, which we would estimate at a cost of just under seven minutes.

Playing service lets wouldn’t destroy the very fabric of tennis as we know it, but it also wouldn’t substantially shorten matches. By changing the let rule, tennis executives would needlessly annoy players and fans for no noticeable benefit.

What Would Happen If the WTA Switched to Super-Tiebreaks?

Italian translation at settesei.it

It’s in the news again: Some tennis execs think that matches are too long, fans’ attention spans are too short, and the traditional format of tennis matches needs to change. Since ATP and WTA doubles have already swapped a full third set for a 10-point super-tiebreak, something similar would make for a logical proposal to cap singles match length.

Let’s dig into the numbers and see just how much time would be saved if the WTA switched from a third set to a super-tiebreak. It is tempting to use match times from doubles, but there are two problems. First, match data on doubles is woefully sparse. Second, the factors that influence match length, such as average point length and time between points, are different in doubles and singles.

Using only WTA singles data, here’s what we need to do:

  1. Determine how many matches would be affected by the switch
  2. Figure out how much time is consumed by existed third sets
  3. Estimate the length of singles super-tiebreaks
  4. Calculate the impact (measured in time saved) of the change

The issue: three-setters

Through last week’s tournaments on the WTA tour this year, I have length (in minutes) for 1,915 completed singles matches.  I’ve excluded Grand Slam events, since third sets at three of the four Slams can extend beyond 6-6, skewing the length of a “typical” third set.

The average length of a WTA singles match is about 97 minutes, with a range from 40 minutes up to 225 minutes. Here is a look at the distribution of match times this year:

histo1

The most common lengths are between 70 and 90 minutes. Some executives may wish to shorten all matches–switching to no-ad games (which I’ve considered here) or a more radically different format such as Fast4–but for now, I think it’s fair to assume that those 90-minute matches are safe from tinkering.

If there is a “problem” with long matches–both for fan engagement and scheduling–it arises mostly with three-setters. About one-third of WTA matches go to a third set, and these account for nearly all of the contests that last longer than two hours. 460 matches have passed the two-hour mark this season. Of those, all but 24 required a third set.

Here is the distribution of match lengths for WTA three-setters this season:

histo3

If we simply removed all third sets, nearly all matches would finish within two hours. Of course, if we did that, we’d be left with an awful lot of ties. Instead, we’re talking about replacing third sets with something shorter.

Goodbye, third set

Third sets are a tiny bit shorter than the first and second sets in three-setters. If we count sets that go to tiebreaks as 14 games, the average number of games in a third set is 9.5, while the typical number of games in the first and second sets of a three-setter is 9.7.

Those counts are close enough that we can estimate the length of each set very simply, as one-third the length of the match. There are other considerations, such as the frequency of toilet breaks before third sets and the number of medical timeouts in different sets, but even if we did want to explore those minor issues, there is very little available data to guide us in those areas.

The length of a super-tiebreak

The typical WTA three-setter involves about 189 individual points, so we can roughly estimate that foregoing the third set saves about 63 points. How many points are added back by playing a super-tiebreak?

The math gets rather involved here, so I’ll spare you most of the details. Using the typical rate of service and return points won by each player in three-setters (58% on serve and 46% on return for the better player that day), we can use my tiebreak probability model to determine the distribution of possible outcomes, such as a final score 10-7 or 12-10.

Long story short, the average super-tiebreak would require about 19 points, less than one-third the number needed by the average third-set.

That still doesn’t quite answer our question, though. We’re interested in time savings, not point reduction. The typical WTA third set takes about 44 minutes, or about 42 seconds per point. Would a super-tiebreak be played at the same pace?

Tiebreak speed

While 10-point breakers are largely uncharted territory in singles, 7-point tiebreaks are not, and we have plenty of data on the latter. It seems reasonable to extend conclusions about 7-pointers to their 10-point cousins, and they are played with similar rules–switch servers every two points, switch points every six–and under comparable levels of increased pressure.

Using IBM’s point-by-point data from this year’s Grand Slam women’s draws, we have timestamps on about 700 points from tiebreaks. Even though the 42-seconds-per-point estimate for full sets includes changeovers, tiebreaks are played even more slowly. Including mini-changeovers within tiebreaks, points take about 54 seconds each, almost 30% longer than the traditional-set average.

The bottom line impact of third-set super-tiebreaks

As we’ve seen, the average third-set takes about 44 minutes. A 19-point super-tiebreak, at 54 seconds per point, comes in at about 17 minutes, chopping off more than 60% off the length of the typical third set, or about 20% from the length of the entire match.

If we alter this year’s WTA singles match times accordingly, reducing the length of all three-setters by one-fifth, we get some results that certain tennis executives will love. The average match time falls from 97 minutes to 89 minutes, and more importantly, far fewer matches cross the two-hour threshold.

Of the 460 matches this season over two hours in length, we would expect third-set super-tiebreaks to eliminate more than two-thirds of them, knocking the total down to 147. Here is the revised match length distribution, based on the assumptions I’ve laid out in this post:

histo4

The biggest benefit to switching to a third-set super-tiebreak is probably related to scheduling. By massively cutting down the number of marathon matches, it’s less likely that players and fans will have to wait around for an 11:00 PM start.

Of the various proposals floating around to shorten matches–third-set super-tiebreaks, no-ad scoring, playing service lets, and Fast4–changing the third-set format strikes the best balance of shortening the longest matches without massively changing the nature of the sport.

Personally, I hope none of these changes are ever seen on a WTA or ATP singles court. After all, I like tennis and tend to rankle at proposals that result in less tennis. If something must be done, I’d prefer it involve finding new executives to replace the ones who can’t stop tinkering with the sport. But if some rule needs to be changed to shorten matches and make scheduling more TV-friendly, this is likely the easiest one to stomach.

Elina Svitolina and Multiple #1 Upsets

Last week in Beijing, Elina Svitolina beat new WTA #1 Angelique Kerber. It was the first time the Ukrainian defeated Kerber this season, but it wasn’t her first 2016 triumph over a player ranked #1. At the Rio Olympics in August, Svitolina upset then-top-ranked Serena Williams.

It’s unusual for a player to face two (or more) different #1-ranked opponents in the same season. Since 1985, it has happened 136 times on the WTA tour and 148 times on the ATP tour. That’s less than five times per season per tour.

Of course, it’s much less common to upset multiple #1-ranked opponents, as Svitolina did. This was only the 16th time a woman did so (again, since 1985), while it has happened on the men’s side 18 times.

Here is a full list of WTA player-seasons that featured defeats of more than one top-ranked player:

Year  Player               Upsets                      
2016  Elina Svitolina      Kerber; Serena              
2010  Samantha Stosur      Serena; Wozniacki           
2009  Venus Williams       Serena; Safina              
2008  Dinara Safina        Henin; Sharapova; Jankovic  
2006  Justine Henin        Davenport; Mauresmo         
2003  Justine Henin        Serena; Clijsters           
2002  Kim Clijsters        Serena; Venus               
2002  Serena Williams      Capriati; Venus             
2001  Lindsay Davenport    Capriati; Hingis            
1999  Amelie Mauresmo      Hingis; Davenport           
1999  Venus Williams       Davenport; Hingis           
1997  Amanda Coetzer       Hingis; Graf                
1996  Jana Novotna         Graf; Seles                 
1996  Kimiko Date Krumm    Graf; Seles                 
1991  Martina Navratilova  Graf; Seles                 
1991  Gabriela Sabatini    Graf; Seles

It’s quite an accomplished list. As we might expect, there’s a lot of overlap between the players who achieved these upsets and past and future #1-ranked players. The real standouts here are Justine Henin and Venus Williams, who managed the feat twice, and Dinara Safina, who faced three different #1s in 2008, going undefeated against them.

Here are the men who beat multiple #1s in the same season:

Year  Player                 Upsets             
2013  Juan Martin Del Potro  Nadal; Djokovic    
2012  Andy Murray            Federer; Djokovic  
2011  David Ferrer           Nadal; Djokovic    
2011  Jo Wilfried Tsonga     Nadal; Djokovic    
2010  Marcos Baghdatis       Nadal; Federer     
2009  Juan Martin Del Potro  Nadal; Federer     
2008  Andy Murray            Nadal; Federer     
2008  Gilles Simon           Nadal; Federer     
2003  Rainer Schuettler      Roddick; Agassi    
2003  Fernando Gonzalez      Hewitt; Agassi     
2001  Greg Rusedski          Safin; Kuerten     
2001  Max Mirnyi             Safin; Kuerten     
1995  Michael Chang          Agassi; Sampras    
1992  Richard Krajicek       Courier; Edberg    
1991  Guy Forget             Edberg; Becker     
1991  Andrei Cherkasov       Edberg; Becker     
1990  Boris Becker           Lendl; Edberg      
1988  Boris Becker           Wilander; Lendl

This list isn’t quite as impressive, though it does capture several very good players at their best.  It also highlights the world-beating potential of Max Mirnyi, who–despite never reaching the top 15 himself–finished the 2001 season with a 3-1 record against ATP #1s.

The rarity of facing multiple #1s in the same season–let alone beating them–stops us from drawing any meaningful conclusions about what Svitolina’s feat indicates for her future. At the very least, however, it reminds us of the Ukrainian’s potential as a future star, and puts her among some very good historical company.