NBA Draft Pick for Worst Team Probability

Question

This is actually a question from "A First Course in Probability" by Sheldon Ross, and I have the solution, but am unclear as to why the solution is the case. Would anyone please clarify?

The question is as follows:

11 teams are entered into the NBA draft lottery. A total of 66 balls are placed in an urn, with the worst team having 11 balls placed, the second worst team having 10 balls placed, etc (with 1 ball having the name of the team with the eleventh worst record). A ball is chosen at random and the team whose name is on the ball is given the first pick in the draft. Another ball is chosen, and if that ball is different from the first pick team, that team becomes the second pick team. If the ball is the same, the ball is discarded and the drawing continues. This goes on for the first three draft pick teams.

After the first three draft pick teams are chosen, the rest of the teams that did not win the lottery invert their win-loss records and that order is awarded for draft picks 4 through 11. For example, if the team with the worst record didn't receive any of the 3 lottery picks, it becomes the 4th pick.

Let X denote the draft pick of the team with the worst record. Find the probability mass function of X.

For P(X = 1), the answer is simply $\frac{11}{16}$. However, for $P(X = 2)$, the solution in the book is $$\sum\limits_{j=2}^{11} \bigl( \frac{12-j}{66} \bigr)\bigl( \frac{11}{54+j} \bigr)$$.

This solution doesn't make sense to me, because it doesn't account for the repeated drawing of the same first draft pick team. For example, if the team with 10 balls gets the first draft pick, the second draft pick team would be decided from another draw. If the first pick team is drawn again, that ball is discarded and the process is repeated. So shouldn't the solution account for when, for example, the first draft pick is the second worst team, the second and up to tenth draws are still the second worst team, and the 11th draw (still choosing the 2nd draft pick) is the worst team?

Sorry for the long post. I didn't know how else to summarize this.

Thank you so much for your help!

Answer 1

To the point of your question: multiple draws of the same number are ignored. They never happened.

Suppose that the draw was complicated by throwing a number of blank balls into the urn. Every time you drew a blank ball, you said, "Oh, ****, another blank!", and threw it away. You didn't count that as a draw, and nothing was recorded.

Now, assume that once a number is drawn for the first time, all the undrawn balls with the same number are magically converted to blank balls. They then can't affect any draw results

The expression you are troubled by has two terms in the sum: the first is the probability that the jth-worst team gets picked first, the second is the probability that, after this draw, the worst team gets drawn from the remaining numbered balls to be the second draft pick, possibly after a number of uncounted duplicate number draws. Add over the ten teams that can beat the worst to first...

Edit: @Andre Nicolas's term "live balls" expressed the concept much better...

Answer 2

Consider the following

P(X=2) = summation( j from 2 to 11 ) {P(X=2|Ij)*P(Ij)} where Ij is the event that jth worst team got to pick first.

P(Ij)=(12-j)/66 is obvious as mentioned in the solution. Now the question arises if P(X=2|Ij) = 11/(54+j)

This can be easily proven by considering the following:

P(X=2|Ij) = summation(k from 0 to 11-j) {P(X=2|A(k,j),Ij)*P(A(k,j)|Ij)}

Where A(k,j) is the event where next k balls drawn are from jth team.

P(X=2|A(k,j),Ij) = 11/(66-(k+1))

P(A(k,j)| Ij) is obvious.

This reduces to P(X=2|Ij) = 11/(54+j).

And hence the solution.

PS: Sorry I don't really know how to insert equations.