NBA Eastern Conference Playoffs

With the NBA playoffs starting today, I wanted to use the hazard rates model described in this great post and apply it to the NBA Playoffs. Instead of finding the rate of losing and subtracting that rate from one, I used the probability of a team winning directly. To calculate the odds of a team advancing, I used a binomial probability distribution function. In a binomial distribution, the result of every trial is either 1 or 0, in this case a win or a loss. When a successful trial is a win for the given team.

In essence, the probability of a team advancing is the probability that it can win four games. I used the team’s regular season winning percentage squared as the probability parameter. I squared the winning percentage because the playoff teams have similar winning percentages, and a non-squared parameter seemed to overstate the probability of upsets for the seven and eight seeds. These calculations assume independence of each game and round, discarding factors such as momentum and slumps for the sake of a simpler analysis.

I calculated the odds of four “successes” out of four games. Then I added the odds of four successes out of five, plus four out of six and four out of seven. Let that number be the odds that Team A wins four games. I used the same methodology to get Team B’s odds of winning four games. Let the combined probability of Team W winning four games be TeamA4. The probability of a team advancing is as follows:

Pr(TeamA Wins) = TeamA4 / (TeamA4 + TeamB4)

With that, I calculated the odds of a team advancing in the first round. For the second and third rounds, the math gets trickier. To calculate the probability that a team wins in round two, let’s assume that Team A can only play Team C or Team D in Round 2 (R2). Pr(TeamA Wins in Round2) is equal to, by Bayes Theorm:

Pr(TeamA beats TeamB)*Pr(TeamB makes R2)+Pr(TeamA beats TeamC)Pr(TeamC makes R2)

Take this value and multiply by Team A’s probability of getting to the second round, and this is the probabiliy that a Team A wins in round 2. Do this for all parts of the bracket to get every team’s odds of winning in Round 2. Assume that in Round 3, Team A can only play Team E, F, G or H, so sum the probability that each team reaches Round 3 (winning in R2) times the probability that A beats that team (using the same probability function created for Round 1).  For Team A winning in Round 3: 

Pr(Team A wins R2)*Σ(Team A beats Team X)*(Team X wins R2 )

From that, we get a spreadsheet of of probabilities.


I also graphed the probabilities.


 As you can see, the odds of advancing in each round decrease. By the conjunction fallacy, a team cannot have a higher probability of winning two series than one. This model will always predict the higher seed winning, so it is no surprise that it picks the 1-4 seeds to advance in Round 1, the 1 and 2 seeds to advance in Round 2, and 1 seed Miami to win the conference. Only Miami and New York have over a 50 percent chance of making the conference finals. Miami is the clear favorite to make it to the finals, having a 55% chance of winning the Eastern Conference.



About Divya Parmar

I'm a college graduate and young professional, embarking on the journey of life and learning along the way. Interests include data science, sports statistics, and political economy.
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