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.