The Next 5 Games

There was something I really wanted to show you
But I just can’t find it
— “Can’t Find It”
Smoking Popes

I was reading the newspaper this morning (online of course), and one of the lines caught my eyes:

SOFT SCHEDULE: All of the Knicks’ next five games are against teams with losing records.

The sentence implies that the Knicks will be at an advantage their next few games, since they are playing “bad” teams. But is this really true? There is a well known formula that you can use to try to predict who will win a certain matchup. According to Dean Oliver:

In a 0.500 league, i.e., where all we have are the overall records and no information about home court advantage, etc.:

where Win%A_B is the chance that A will beat B, Win%A is A’s winning percentage against the league, and Win%B is B’s winning percentage against the league.

So I took the Knicks next 5 opponents, and plugged their records into a spreadsheet. Using the above equation, I was able to figure out what the Knicks’ chances are to win each game (KN_w%).

Knicks Opponent	W	L	Pct	KN_w%
 Philadelphia	24	36	.400	.527
 Toronto Raps	25	34	.424	.503
 Washington	19	39	.328	.604
 Boston Celts	26	36	.419	.507
 Philadelphia	24	36	.400	.527

According to this equation, the Knicks have about an even chance at beating the Raptors and the Celtics. They have slightly better odds against the Sixers (twice), and pretty good odds against the Wizards. In fact according to this, the Knicks should be favorites in 3 of the five games, and at least even in the other two. However, a keen eye might notice that this equation doesn’t care who is at home or away. If teams do better at home, wouldn’t we want to take account of this?

The answer is yes. Since I already had the NBA standings in my spreadsheet, I decided to calculate the home records for whole league. The home team in the NBA this year wins 64% of the time. That seems to be a huge advantage, so can’t we account for this in our matchup equation? Luckily someone already thought of this as well. Back to Dean’s web page:

For example, if Team A is the home court team and Win%H is the percentage of times the home team wins, we have

Win%A_B = [Win%A*(1-Win%B)*Win%H]/[Win%A*(1-Win%B)*Win%H+(1-Win%A)* Win%B*(1-Win%H)]

So recalculating:

H?	Knicks Opponent	W	L	Pct	KN_w%	KN_w%lgw
H	 Philadelphia	24	36	.400	.527	.667
A	 Toronto Raps	25	34	.424	.503	.360
A	 Washington	19	39	.328	.604	.459
H	 Boston	Celts	26	36	.419	.507	.649
A	 Philadelphia	24	36	.400	.527	.382
Home teams in bold

Now things have radically changed. The Knicks are substantial favorites in their two home games against the Sixers and the Celts. They are underdogs against the other three teams, and substantially more against the Raptors and Sixers. What’s especially noteworthy is that they play the Sixers twice. Against Philly, they go from being a 67% team at home to a 38% team on the road. Also in a neutral site, they are most likely to beat Washington, but accounting for home court advantage, they are no longer the favorite.

Of course these are just percentages. The Knicks could win all 5 games (1 in 37 chance), or they could loose all 5 (1 in 40 chance). Going back to the newspaper quote, you might expect the Knicks to win 3 or 4 of their next 5 games, but in reality they’re expected to only win 2 or 3.

On a final note, this equation doesn’t take into account many factors. The two most important I can think of are injuries, and whether the records used are indicative of a team’s true strength. For example this Knick team is radically different from the one that started 2-8. Let’s say with the additions of Marbury, Hardaway, Nazr, Tim Thomas, and Lenny Wilkens the Knicks are better than their record. If they were let’s say a .500 team, things would change even more. The Knicks would become heavy favorites at home against Philly (73%) and Boston (71%), slight favorites away against Washington (53%), and slight underdogs against Philly (45%) and Toronto (43%).