A few months ago, I wrote about the possibility of modeling basketball games by estimating the probability of points per possession: namely, that if one started with data about field goal shooting, free throw shooting, 3-point shooting, rebounding and turnovers one could model the outcome of any individual possession, and determine how likely it was for a team to score zero, one, two, or three points with that possession. The model was then applied to the 2010 NCAA Women's Basketball Finals - Connecticut vs. Stanford - and the results were reported.
With the WNBA Finals getting ready to start, most observers won't even give the Atlanta Dream a chance. This led me to return to the model in hopes of predicting how the Finals would turn out. Could Atlanta win? And if so, how likely were the Dream to win?
It was quite simple to determine the probabilities per possession. I used all 38 games played by each team - the 34 regular-season games and the four playoff games. For each of the possible outcomes of an Atlanta Dream possession, I'll list the values the percentage probabilities the model returns both for the Atlanta Dream and for a typical Atlanta Dream opponent. The values are separated with a slash:
zero points: 51.3 percent/54.1 percent
one point: 5.0 percent/4.5 percent
two points: 38.0 percent/34.7 percent
three points: 5.6 percent/6.7 percent
The numbers seem to make sense: since the Dream is a winning team and not a losing one, it makes more sense that it's more likely for the opponent to end up with zero points on a possession in a Atlanta Dream game (54.1 percent) than the Dream themselves (51.3 percent). However, the Dream aren't a great 3-point shooting team. It's more likely that a Dream opponent will have a possession end with a 3-pointer (6.7 percent) than the Dream themselves will (5.6 percent).
Here are the Seattle Storm's numbers.
zero points: 52.1 percent/56.1 percent
one point: 3.9 percent/3.9 percent
two points: 33.4 percent/31.7 percent
three points: 10.6 percent/8.4 percent
The model also determines an average possessions per game: For the Dream, it's 83.6; for the Storm it's 75.5. I decided to give each team 79 possessions, and for any particular possession for a team, I used one of the two sets of probabilities. For example, based on random chance for an Atlanta possession, if Atlanta controlled the tempo I'd use the probabilities for an Atlanta possession. However, if Seattle controlled the tempo during Atlanta's possession, Atlanta would use the "typical Storm opponent" probabilities, which are typically worse. The same rule was used with the Storm.
I generated 100 games randomly. Each game is played on a hypothetical neutral court. Here were the outcomes:
60 games end in a Seattle victory
37 games end in an Atlanta victory
3 games end in a tie
Seattle's average win: 3.31 points
Biggest Seattle win: Seattle 97, Atlanta 53
Biggest Atlanta win: Atlanta 85, Seattle 53
In this case, the probability to determine if Seattle wins a game against Atlanta should be simple: give Seattle full credit for each game it wins and 1/2 credit for each time game. (60 + 3 x (1/2))/100 = 0.615, which should be the probability that Seattle wins a given match. There you go.
There is, however, a problem with our admittedly primitive model. It does not build in home court advantage, and supposedly there is something to this "Key Magic". Is there a way to model this? Lucky for us, we have masseyratings.com, which uses its own models to determine the chances of a WNBA team winning.
The Massey Ratings webpage claims that home court advantage is 2.76 points. But is the advantage at Key Arena equal to the one at Philips Arena? Really? The great thing about the Massey Ratings web page is that it gives an individualized home court advantage for each team. Seattle's home court advantage is the best in the WNBA: it is listed at 6.28 points. (Connecticut is second with 5.71.)
As for Atlanta? According to Massey Ratings, Atlanta has the worst home field advantage: one of -0.63. All in all, Atlanta's opponents gained one point playing at Philips Arena. Only Los Angeles at -0.35 and Chicago at -0.30 had negative home field advantages.
I then replayed all of the games above at Philips Arena. This was done by subtracting one point from Atlanta's point total.
63 games end in a Seattle victory
36 games end in an Atlanta victory
1 game ends in a tie
The probability assigned to Seattle winning at Philips Arena against Atlanta becomes 0.635 in the model. Now, let's play those 100 games at Key Arena and give Seattle six extra points in each game.
78 games end in a Seattle victory
20 games end in an Atlanta victory
2 games end in a tie.
The probabilty assigned to Seattle winning at Key Arena against Atlanta becomes 0.79 in the model.
Given those probabilities, we can now simulate each game of the 2-2-1 series of the 2010 WNBA finals. I ran 1000 simulations of the finals, and here's how the series turns out.
15 end with the Dream winning in 3 games (1.5 percent)
51 end with the Dream winning in 4 games (5.1 percent)
43 end with the Dream winning in 5 games (4.3 percent)
195 end with the Storm winning in 5 games (19.5 percent)
297 end with the Storm winning in 4 games (29.7 percent)
399 end with the Storm winning in 3 games (39.9 percent)
Almost 40 percent of the time the Storm sweep the series. The chances of Seattle winning - no matter how long it takes - is 89.1 percent. If the Storm win, then they are 78.1 percent likely to celebrate a championship in Atlanta, and only 21.9 percent likely to clinch in Seattle.
As for Atlanta, the Dream's chances of winning are only 10.9 percent. It's not impossible, but it's very difficult. Note that the chances of the Dream winning in four games (5.1 percent) is greater than the chance of winning in five (4.3 percent). This is because if the Dream don't win in four, they have to go to Seattle to try to win in five and face a very tough prospect.
So if you're an Atlanta Dream fan, should you be disheartened? I don't think so. It's going to make victory taste that much sweeter. The Dream will shock the planet!