It is very unlikely that the NBA draft is rigged in any way. There is just not enough incentive for the NBA to rig the draft (especially for a city like Cleveland) and have it be a viable business strategy. Of course, that doesn’t stop the outcry among NBA fans that the lottery is purposefully rigged. Because of this, I feel that it would be a very interesting approach to use Bayesian inference to express an individual’s personal belief in the draft being rigged.
To start off, let’s look at the basic Bayesian Equation and apply it to our scenario.
P(A|B) = P(B|A)*P(A)/P(B)
Where A represents the probability that the NBA draft is rigged, and B represents the probability of Cleveland receiving the 1st overall pick this year.
Our P(A) is our individual prior belief in whether the draft is a rigged system or not. Prior belief must be assessed without the bias of knowing the current result. This varies based on the individual and can only be estimated.
P(B|A) represents the probability of Cleveland winning the draft given that the lottery is rigged. That is, if we know the lottery is rigged beforehand, what is the chance that Cleveland will get the first pick? Again, this is a belief that you have to build in prior.
P(B) is the percentage chance that Cleveland would win the lottery. This gets complicated because we have to consider both the rigged draft and non-rigged draft scenarios. We know there is a 1.7% chance of Cleveland winning if the lottery is not rigged, and the P(B|A) probability if the lottery is rigged. In order to calculate the total P(B), we multiply the probabilities by our prior belief of the scenarios being true and add them together, ending up with the following equation:
P(B) = P(A)P(B|A) + (1-P(A))(.017)
P(A|B) is the probability that the draft is rigged given that Cleveland won. This is the belief percentage we want to solve for.
Looking at the entire equation as a whole we have:
P(A|B) = P(B|A)*P(A)/ (P(A)P(B|A) + (1-P(A))(.017))
This looks like a mess, but can be simplified by using variables.
P(A|B) = xy/(xy+(1-x)z)
Where x is the prior belief that the draft is rigged, y is the probability that Cleveland wins the draft if it is rigged, and z represents the probability that Cleveland wins the draft given that it is not rigged (which we know to be .017).
Now lets fill in some values in order to get an estimate of an updated P(A|B) through the equation. I will use my personal beliefs to make my calculation. I sincerely believe that the NBA really has no incentive to rig the lottery; therefore, I would assign a 1% chance that the lottery is rigged before Cleveland won the lottery. Additionally, I don’t believe the NBA would have any reason to favor a market like Cleveland, even if the lottery is rigged. However, it is a popular theory that Cleveland would be one of the teams favored by a rigged lottery, so I would put the percent chance that Cleveland would win the lottery at around 10%. We can plug these values into the Bayes equation:
My new belief percentage in the NBA lottery being rigged is about 6.13%. This is certainly up from 1% beforehand, and implies that I should be a bit more skeptical regarding the legitimacy of the lottery. Still, there is a little bit of sample bias here and people only yell about the NBA being rigged when the team that wins the lottery is a good team for the NBA to rig it for. It would be interesting to start from the beginning of the lottery introduction and update probabilities from year to year, to establish a modern probability estimate.
Feel free to use the equation and plug in your own probabilities in order to reach your own belief percentage. At this point, it is hard to do a truly unbiased estimate of belief percentage, but in the future you can use an approach like this. Write down your belief probabilities before hand (including a distribution of possible teams and their percentages for winning the draft if it is rigged), and then update them after the results have been produced.
Image courtesy of Sports Illustrated
Georgetown University Class of 2016