Jerzy Janowicz and the Frequency of Tiebreak Shutouts

In Marseille this week, Jerzy Janowicz played two dominant tiebreaks.  In his second-round win over Julien Benneteau, he put away the first set with a 7-0 breaker en route to a straight-set victory.  In the quarterfinals, he won another 7-0 tiebreak to even his match with Tomas Berdych before falling in three.

Amazingly, this is not the first time anyone on the ATP tour has won two tiebreaks by a score of 7-0 in back-to-back matches.  It is, however, the first time it’s been done in best-of-3 matches.  In 1992, Brad Gilbert won both his 2nd- and 3rd-round contests at the US Open in five sets, winning 7-0 tiebreaks in the 5th set both times.  If that’s not a case for fifth-set tiebreaks at slams, I don’t know what is.

Janowicz’s accomplishment and Gilbert’s feat are the only two times anyone on tour has won two shutout breakers in the same event.  That’s not much of a surprise, since there are typically fewer than 25 such tiebreaks at tour level per year.

What’s particularly odd here is that Jerzy’s two shutouts weren’t the only ones in Marseille.  In the first round, wild card Lucas Pouille was 7-0’d by Benneteau, the same guy who Janowicz victimized first. Weirdly, both losing and winning 7-0 breakers in the same event is slightly more common than winning two.  It has happened three times before, most recently at the 2009 Belgrade event by Lukasz Kubot, who shut out Karlovic in a semifinal tiebreak then got 7-0’d by Novak Djokovic in the final.

Finally, while we’re wallowing in trivia, here’s one more.  Only once has a player lost two 7-0 tiebreaks at the same event.  This is quite the feat, because to pull it off, you have to win the first match despite losing a set in painful fashion.  The only man to do it is Simone Bollelli, who beat Dmitri Tursunov in the 2nd round of the 2007 Miami Masters despite losing the first set in a 7-0 tiebreak, then lost in the 3rd to David Ferrer, who threw in another tiebreak bagel on the way to straight-set win.

Rare, but not rare enough

Shutout tiebreaks don’t occur very often, but they occur more often than we might expect.  On tour since 1991, there have been 30,259 tiebreaks, and 524 of them–about 1.7%–have been by the score of 7-0.  That’s barely more than the number that end 11-9.

However, if we assume that players who reach a tiebreak are reasonably equal, that’s almost double the frequency we would expect.  A discrepancy like that has serious implications about player consistency.

The arithmetic here is simple.  Say that both players have a 70% chance of winning a point on serve.  In order to win a tiebreak 7-0, the player who serves first must win three points serving and four points returning.  The probability of pulling that off is about (0.7^3)(0.3^4) = 0.28%.  It’s easier if you serve second.  You must win four points serving and three returning: (0.7^4)(0.3^3) = 0.65%.  In this scenario, both players have equal skills, so each one has the same chance of winning 7-0, and the chance of the breaker ending in a shutout is the sum of those two probabilities, 0.93%.

Of course, this simple model obscures a lot of things.  First, players who reach a tiebreak aren’t necessary equal.  Just last month, Bernard Tomic got to 6-6 against Roger Federer, and even more recently, Martin Alund played a tiebreak against Rafael Nadal.  Second, any competitor’s level of play fluctuates, and some guys seem to fluctuate quite a bit when the pressure is on.

Still, the gap between predicted (no more than 0.93%) and observed (1.7%) is enormous.  To predict that 1.7% of tiebreaks would end in a 7-0, we’d need to start with much more extreme assumptions.  For instance, if one player is likely to win 77% of serve points and the other only 64% of serve points, the likelihood of a 7-0 tiebreak is 1.7%.  Those assumptions also imply that, if each man kept up the same level of play all day, the better player has a 93% chance of winning the match.  Perhaps true of Nadal/Alund or even Federer/Tomic, but certainly not Janowicz/Benneteau or Janowicz/Berdych, or most of the other matches that reach a tiebreak.

This is all a roundabout way of saying that–breaking news!–players are inconsistent. Or streaky, or clutch, or unclutch … pick your favorite.  Were players machines, 7-0 tiebreaks wouldn’t come around nearly as often.  As it is, we shouldn’t expect more from Jerzy for a while … unless Brad Gilbert is planning a comeback.

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3 Comments

Filed under Hot Hand, Jerzy Janowicz, Records, Tiebreaks

3 responses to “Jerzy Janowicz and the Frequency of Tiebreak Shutouts

  1. Didn’t occur to me to check for this before publishing, but now that I’ve been reminded…

    One time, there were two 7-0s in the same match! 2002 Rotterdam R32, Martin Lee beat Sjeng Schalken 7-6(0) 7-6(0).

    You won’t be surprised to learn that Schalken had a career losing record in tiebreaks.

  2. One more. Hana Mandlikova won 7-0 tiebreaks in R128 and R64 of the 1989 US Open:
    http://www.tennisabstract.com/cgi-bin/wplayer.cgi?p=HanaMandlikova&f=A1989qqDUS_Openqq

  3. i think he is training special for tie-breaks

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