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Wednesday, October 6, 2010

Some Thoughts on the American Ryder Cup Team

The American Team came up a half-point short of defending the Ryder Cup, and of course the big question after each loss is why? Why can't the US (usually the better team on paper) pull it together against the Euros? I thought I'd throw out a few of my thoughts, based on what I saw and some of the things I've already heard. Sorry this post is so long, but sorting out the points took a lot of explanation.

Monty implied that one of the reasons the Euros won was that everybody contributed at least a half-point. At least, many critics will take it that way. It may not get noticed, but the same can be said for the American team. Here is the scoring breakdown, with the Americans in bold:
  • 3.0 -- Woods, Stricker, Donald, Poulter
  • 2.5 -- Cink, Kaymer, McDowell, Westwood
  • 2.0 -- Overton, Z-John, Kuchar, McIlroy, Jiminez, Fisher, Harrington
  • 1.0 -- Fowler, Mickelson, Watson, Mahan, D-John, E-Mol, Hanson
  • 0.5 -- Furyk, F-Mol
Now, the first thing you'll notice is that the Euros had 3 more players with 2 points or better than the Americans... but the math of team play isn't that straightforward. That looks like 7 more points (2.5 + 2.5 + 2.0). But since it takes two players to get a team point, this is very misleading. Bear in mind, the Euros won the Cup by only a single point!

The problem isn't whether everybody contributes, although that helps. The problem is point distribution. One frequently-sited difference between Euro and US strategy is that the Euros tend to put a team together and ride them as hard as possible, while the US tends to split successful teams in hopes of jumpstarting less-successful players. But is this really what's happening?

Let's do some admittedly simplified math to see how this works. Since there are a total of 28 points available at the Ryder Cup, we'll just use a tie game -- 14 points -- for our calculations. When I talk about "team points" I'm just talking about points won during the paired sessions; "total points" refers to actual points posted toward the final team total, whether they were made in pairs or singles.

If one side could sweep the teamplay, they'd have 16 points and win without any singles at all. If they could sweep the singles, they'd only need 2 teamplay points to tie. Neither scenario is likely, so let's move on.

At the simplest, we could halve every match to get the tie. That's easy to figure for the singles -- each player gets a half-point. But for the team matches, which count for 4 points in each round, it's more complicated. Two players need to get a half-point to halve a match, so eight players get halves in their individual totals if all four matches get halved. To get the most evenly-distributed points, you end up with eight players playing 3 matches and four players playing 2 matches. (That's 16 matches * two players = 32 players. And 12 + 12 + 8 = 32.) Add in the singles halves, and you need eight players to get 2 points and four players to get 1.5 points in order to split the matches at 14.

Do you need 14.5 points to win? Either two match players have to win 1 match, which means two players get an extra half-point on their individual totals, or you need a singles win, which means one player gets an extra half-point on his individual total.

Essentially, "spreading the wins around" like this has been the US approach to winning. What happens if you "ride your horses" the way we think the Euros do?

Let's assume you build a powerhouse duo capable of winning at both foursomes and fourballs. This team is capable of getting you 6 points -- 2 points in foursomes, 2 points in fourballs, and 2 individual singles matches. You have nearly half your points already! If you can create two teams like this, then you have four players who can get you 12 points -- a mere 2 points short of your tie! Of course, each of these four players will end up with a perfect 5-0-0 record, which isn't very likely. Let's look at a more likely scenario for this "horses" attack.

With 12 players we can create 6 different teams. (Remember, we're approaching this simply. In this example, each player plays on only one team.) I'm going to create "specialist" teams -- each pair is better in foursomes or fourballs, but not both. Because of this, I don't really have a team capable of winning all four matches... but they could conceivably win 2 matches, plus their two singles. That's 4 possible points for each team -- and again, it's unlikely you'd see many sweeps. But just for you to think about, note that only four teams (eight players) would need sweeps to get 16 points and the win.

It's more likely that teams could get 2.5 to 3 points. Five teams winning 3 points is a 15 point win, and six teams getting 2.5 points is as well. But is that what we see from the Euros? No way!

So what's really happening at the Ryder Cup?

Let's look at the two 3-point winners from each side. (Because of the format change, no one could win more than 4 points, so these were outstanding performances... and possibly why the matches were so close.) Since we want to view their team strategies, eliminate the points for singles (which all four won -- again, outstanding play on both sides) and we have four leaders who each won 2 out of 3 possible points. BUT... Stricker and Woods played together for both points; Poulter and Donald won only 1 of their points together.

That means Stricker and Woods won 4 total points toward the 14, but Poulter and Donald won 5 total points. This isn't what you would expect, given how the commentators tell us that the Euros create teams and don't change them much. Let's check the 2.5 winners and see if it fits the same pattern.

The US only had Stewart Cink in this group... and he played with Matt Kuchar all 3 team matches. Cink halved his singles, so he put up 2 points (1 win and 2 halves) in teams. All of Cink's 2.5 points were team points.

The Euros had Kaymer, McDowell, and Westwood. Kaymer and Westwood both lost their singles, while McDowell won his (what an understatement!) so let's figure him at 1.5 team points. So....
  • Kaymer played with Westwood twice (win and halve) and Poulter once (win).
  • McDowell played with McIlroy all three times (halve, win, and loss).
  • Westwood played with Donald (win) the one time he didn't play with Kaymer.
Since we already counted Poulter and Donald's points, we need to subtract the duplicates from these three players. So give Kaymer 1.5 team points (same for total points), McDowell 1.5 team points (2.5 total points -- don't forget his single!), and 0 points for Westwood since he was teamed with players we've already counted. That means the three Euro players added 4 total points to the team.

So here's what happened:
  • The US got 6.5 total points from three players with 2.5+ individual points. These three players played in two team combinations. (Kuchar was the fourth member.) This means the other nine players needed to get 7.5 points to reach the magic 14 number. (They got 4.5 of those in singles, where they're usually stronger, and only 2.5 in teamplay to come up .5 short.)
  • The Euros got 9 total points from five players with 2.5+ individual points. These five players played in seven team combinations.(Fisher, McIlroy, and Harrington were also in the pairings.) This means the other seven players needed to get 5.5 points to reach the magic 14.5 number. (They got 2 of those in singles.)
I should add that the trio of Fisher, McIlroy, and Harrington were able to add another 2.5 points (Fisher aand Harrington added 2 team points together and McIlroy a halve in singles), while Kuchar was unable to add any points at all. That means 8 Euro players put up 11.5 total points versus 4 US players with 6.5 total points.

Granted, the format change caused by the weather altered the normal strategies somewhat, but it looks to me like the real key here is having several players who can:
  • score 2.5 individual points (or more) and
  • play together more or less interchangeably. But they don't use a single approach -- sometimes one pairing stays together the whole time, while other groups seem to play mix-and-match, depending on how they're playing.
The US just can't pull that off yet. Our best players just don't play particularly well in the Ryder Cup format, although they do fine in Presidents Cup. I'll tell you why I think that is in tomorrow's post.


  1. What do you think of the possibility of having everybody play in three rounds before the singles ? Maybe a round of Four Balls on Friday, then a round of Foursomes on Saturday morning and Four Balls again in the afternoon, then singles on Sunday.

    I know that makes for a total of 30 points instead of the traditional 28, but players only play a possibility of four rounds instead of 5, and everybody is in the mix all 3 days.

  2. I actually like being able to hide players during the "paired" matches for two reasons:

    1) In a real sense, players qualify for the Ryder Cup by playing singles. You should expect all of your players to make a good showing in singles, so everybody plays in those matches -- no hiding. Pairs is a different animal, requiring different skills; as it stands, qualifying for the team doesn't mean you have those team skills.

    2) Choosing pairs is the biggest part of Ryder Cup strategy. The simple fact is that not everyone is a team player, and some players are better in one style than the other. Captains should not have to play everyone during the pairs matches; as it stands, captains can create the equivalent of "special teams" for the competition.

    I think the Ryder Cup finishes would become too lopsided if every player had to play in every session. (Some people probably think it's too lopsided now ;-)