playoff suspensions

Check out these videos and see if you can see any consistency in some of Stu Jackson's decisions for NBA playoff fouls.

Robert Horry body checks Steve Nash into the scorer's table


Punishment: suspended 2 games (Amare and Diaw were each suspended for leaving the bench area).


Rajon Rondo throws Hinrich into the scorer's table then tries to elbow him in the head


Punishment: nothing


Kenyon Martin shoves Dirk Nowitzki


Punishment: $25,000 fine


Dwight Howard elbows Samuel Dalembert in the head


Punishment: suspended one game


Kobe Bryant elbows Ron Artest somewhere between his chest and throat


Punishment: Essentially nothing (upgraded to a flagrant 1)


Derek Fisher doing an NFL hit on Luis Scola


Punishment: Suspended one game


Rafer Alston slaps Eddie House in the back of the head


Punishment: Suspended one game


Rajon Rondo slaps Brad Miller in the face


Punishment: nothing


And most recently, Kendrick Perkins elbows Mickael Pietrus in the throat


Punishment: nothing


For the record, the Celtics are 3 for 3 on questionable plays reviewed by the league in this post season.

I wanted to look at the numbers to see if having a player that is suspended helps or hurts a team's chances of winning. This year the Magic won Game 6 vs Philadelphia when Howard was suspended and last night they won Game 3 vs Boston when Alston was suspended. The Lakers also won last night without Fisher. The Spurs won Games 5 and 6 when Robert Horry was suspended (although in Game 5 Phoenix didn't have Amare or Diaw due to suspension). I also remember Phoenix winning a Game 5 over the Lakers in 2006 after Raja Bell choke-slammed Kobe in Game 4. I haven't been able to find historical suspension data in the playoffs to fully investigate but it appears that no suspension for Perkins may be a bad thing for Boston.

overtime madness

The most talked about matchup so far has been the Boston/Chicago series. Four out of six of their games have gone to overtime, one of those went to double overtime, and the game last night went to triple overtime. To put it another way, out of 13 times that the buzzer has sounded at the end of an overtime period or at the end of a 4th quarter, 7 times the score has been tied.

To emphasize just how unlikely a series is to have 4 overtime games out of the first 6 played, I estimated the probability of any one game going to overtime using this year's regular season data. Since I couldn't easily find the total number of games that went to overtime, I estimated it by using team data and looking at total minutes, subtracting at the total number of minutes in regulation, then dividing by 5 to see how many overtime periods were played in total. The minute statistic is subject to rounding error and I ended up getting 81.76 overtime periods played in the regular season. I use this exact number because without additional information there's no way to know if the true number is really 82, 81, or some other number close to that. Our estimate for the probability of a game going to overtime is then the number of overtime periods divided by the total number of games played, or 81.76/1230 = 0.06647. This is most likely a high estimate since it's assuming that every overtime game had one overtime period, which is almost certainly false but is still a very close estimate. So if any one game has a 6.647% chance of going to overtime, the chances of 4 out of 6 games in a row going to overtime can be calculated using this equation:

(n!/((k!)*(n-k)!))*(p^k)*(1-p)^(n-k)

where:

n=the number of games (6)

k=the number of games that go into overtime (4)

p=the probability any one game goes into overtime (0.06647)

After plugging in the values we get the result of 0.00025518 or 0.025518%. Basically 2.5 times out of every 10,000 series that go at least 6 games will have 4 overtime games in the first 6 games played. If you add in the possibility of having 5 or 6 overtime games the result is only slightly higher at 0.00026254. Also, we can calculate the chances of out of 13 possible times to close out a game that 7 of those times end up tied. The value for p will change slightly since we are including overtime periods, so instead of 81.76/1230 we use 81.76/(1230+81.76) = 0.06233. The value for n changes to 13 and the value for k changes to 7. In this scenario the probability ends up being 0.00000426282 or 0.000426282%. Essentially, just over 4 times out of every 10 million. Again, if we add in the possibility of having more than 7 ties out of 13 attempts, the probability increases slightly to 0.00000448341.

As a result, I don't think we're going to see a series like this ever again.

updated series odds

Eventually I want this table to be on a sidebar, but until then I'll just keep updating it with new posts.

team	games won	% chance of winning series	frequency
CLE	4	100.0	33
DET	0	0.00
BOS	2	63.64	11
CHI	2	36.36
ORL	2	60.00	10
PHI	2	40.00
ATL	1	40.00	20
MIA	2	60.00
LAL	3	97.73	44
UTH	1	2.27
DEN	2	91.67	84
NOR	1	8.33
SAS	1	18.18	22
DAL	3	81.82
POR	1	18.18	22
HOU	3	81.82

The most influential game over the weekend was Miami winning Game 3 of their series. While the more difficult task may have been winning Game 2 in Atlanta, coming off that win and winning Game 3 bumped them up to a 60% chance of winning the series from 31.11% before that game. After a team without home court advantage wins Game 2 to tie the series they are likely to lose Game 3 (55.56% chance) even though Game 3 is on their home floor.

Other notes:

• Orlando swings the series back in their favor with a win at Philadelphia


• While New Orleans doesn't improve their odds a whole lot they keep themselves from essentially losing the series since no one has come back in a best-of-7 series in the NBA down 0-3.


• Chicago's exciting win only increases their odds by 7.79%, although a loss would've been devastating. In 5 tries no team without home court advantage has come back down 1-3 after leading 1-0.


• Not statistically interesting, but Cleveland is the first team onto the next round

playoff odds

As a follow up to my earlier post, I’m going to keep updating the odds of each team winning the series. (I also sent a brief email to Henry Abbott discussing the playoffs along these same lines, which he posted on TrueHoop earlier this week). In the table below I calculated the current odds based on historical data for all best-of-7 series going back to 1977, excluding the NBA finals because the format in that series switches from 2-2-1-1-1 (i.e. 2 home, 2 away, 1 home, 1 away, 1 home) to 2-3-2. My data goes back only to 1977 because prior to that year, the 7 game formats were not consistently the same as the current format.

team	games won	% chance of winning series	frequency
CLE	2	94.81	135
DET	0	5.19
BOS	2	71.43	14
CHI	1	28.57
ORL	1	45.83	48
PHI	1	54.17
ATL	1	68.89	45
MIA	1	31.11
LAL	2	91.67	84
UTH	1	8.33
DEN	2	94.81	135
NOR	0	5.19
SAS	1	31.25	32
DAL	2	68.75
POR	1	45.83	48
HOU	1	54.17

As we break down these numbers, there are a few interesting things to note. One is that order matters. Orlando, Portland and Atlanta all have home court advantage and all are tied 1-1 in their respective series. However, Orlando and Portland each have a 45.83% chance of winning the series while Atlanta has a 68.89% chance—the difference being that Atlanta lost Game 2 of their series while Orlando and Portland each lost Game 1. While it would seem logical that no matter how you got there, being tied 1-1 is the same, history suggests otherwise. We can also run a test of significance to see if the difference is more than just by random chance. To do this, we set up our test by setting up our null hypothesis, which states that a team with home court advantage that loses Game 1 and wins Game 2 is just as likely to win the series as a team with home court advantage that wins Game 1 and loses Game 2. The alternative hypothesis is that these two situations are not equally likely. We can write these hypotheses as follows:

H₀: P₁ = P₂
H₁: P₁ ≠ P₂
Where H₀ is our null hypothesis and H₁ is our alternative hypothesis. P₁ is the probability of a team winning the series that has home court advantage and has won Game 1 and lost Game 2. P₂ is the probability of a team winning the series that has home court advantage and has lost Game 1 and won Game 2. From our data table above, we can form an equation to calculate the probability of observing these values or more extreme ones based on the assumption that P₁=P₂. Or in other words, assuming that P₁ and P₂ have the same value we calculate the probability of observing a difference of 23.06% (68.89% - 45.83%) or more. This probability is called a p-value. In order to calculate this we use the equation for Z-score, which we can then convert to a probability.

Test statistic

Z = (p1-p2)/(SE)

Where: 

SE = sqrt((p)*(1-p))*sqrt((n1+n2)/(n1n2))

And:

p=(n1p1+n2p2)/(n1+n2)

So in our case the values are as follows:

p1 = 0.6889
p2 = 0.4583
n₁ = 45
n₂ = 48

After plugging in the values we get a Z-score of 2.2447, giving us a p-value of 0.0248. In other words, there is a 2.48% chance of observing results with at least a difference of 0.2306 if P₁ and P₂ were equal. Since this is such a low percentage we can conclude that P₁ and P₂ are not equal. Therefore a team with home court advantage that wins Game 1 and loses Game 2 is more likely to win its series than a team that loses Game 1 and wins Game 2.