winning games 1 and 2

I was reading this post over at thirdqquartercollapse.com (an Orlando Magic blog) and thought a statistic that erivera7 quoted was worth some additional analysis.

[W]hen an NBA team nets a 2-0 series lead, the series victory probability is
93.5% (203-14).

This got me wondering if it matters whether the winner of Games 1 and 2 have home court advantage (meaning they won Games 1 and 2 at home) or not (meaning they won Games 1 and 2 on the road), especially since both the Magic and Rockets are in that situation. It seems that a team would be more likely to win the series if they won Games 1 and 2 on the road than a team that won Games 1 and 2 at home. If you win on the road you have 3 out of the next 5 possible games at home whereas if you win games 1 and 2 at home only 2 out of the next 5 are at home.

Using data from best-of-7 series going back to 1977 and excluding the NBA Finals, I got roughly the same overall statistic as erivera7. Out of 147 series where a team has led 2-0, that team has won 138 times or 93.88% of the time. However, the team with home court advantage won the series 128 out of 135 times (94.81%) when leading 2-0 while the team without home court advantage won 10 out of 12 times (83.33%). Interestingly, the team with home court advantage appears to have a better chance of winning a series when up 2-0 than a team without home court advantage that also leads 2-0 even though they would have more home games left. However, this is not statistically significant as it results in a p-value of 0.336. For a review on proportion tests you can look at my earlier post here or you can use a proportion test calculator found here. Since there are very few times in history when a road team has won Games 1 and 2 it is likely that the difference observed is due to random chance. So if Orlando and/or Houston win tonight there's no evidence that Boston or LA have any better chance of winning their series than Dallas has of beating Denver.