The Minnesota Timberwolves are complaining about having the second pick in the NBA Draft. Do they have a point?

Lest this become a sports-draft blog, I feel contractually obligated to post about the NBA Draft Lottery, what with the Cavs winning and the statistical odds and all.

Last night, the Cleveland Cavaliers (my favorite team for you non-twitter followers) won the NBA Draft Lottery, meaning, by chance they were awarded the #1 overall pick in June. This, despite not having the worst record.

For the uninitiated, the NBA Draft is unique among leagues’ player selection drafts in that the top pick does not necessarily go to the worst team. You see, in order to prevent teams from “tanking” (that is, losing intentionally at the end of the season in order to obtain a higher draft pick) the NBA has instituted a lottery system. Teams with worse records have higher odds of having a higher pick, via ping-pong ball selections, but there’s not guarantee that the losingest team in the NBA will have the highest pick. The logic is that by not automatically awarding teams picks based on record, that will potentially prevent teams from throwing games at the end of the season. To be sure, it still happens, but probably not to the extent that it would otherwise.

The end result is that the worse a team does, the better chance they have to pick higher. And in the NBA, which only allows five players on the court at a time, having a single high pick is generally more valuable than in any other sport.

Now, a bit about the mechanics. The Draft Lottery is conducted using an overly complicated method of ping-pong balls and number combinations. Ping pong balls numbered 1-14 are drawn at random, yielding a potential 1001 combinations (order is disregarded, that is 1-2-3-4 = 4-3-2-1). Each of the 14 teams that do not make the playoffs are assigned a certain number of number combinations. The worse a team’s record, the more number combinations they are awarded. (Note: the specific number combination of 11-12-13-14 is disregarded, yielding exactly 1000 assigned number combinations)

The team with the worst record is assigned 250 number combinations, good for a 25% chance at being the first team drawn.

The team with the second-worst record is awarded 199 number combinations (19.9% chance). The team with the third-worst record is awarded 156 (15.6%) and the team with the fourth worst record is assigned 119 (11.9%) combinations. This proceeds all the way down to the team with the highest record of all the non-playoff teams being awarded 5 number combinations, good for an unlikely 0.5% chance of being awarded the first overall pick.

There’s one more catch for the mechanics, then we’ll get to the actual driving question of this: the lottery is only utilized for the top 3 picks. After the top 3 picks are determined, the remaining draft picks are awarded based solely on record. So, in effect, if you have the worst record in the NBA, you will pick no worse than 4th. And that’s if your number combos (totally 25% of the combinations) are somehow not selected the in first three. Similarly, the second-worst team can slip no further than 5th. And so forth.

[Looks at teacher notes: “Check for understanding”]

Does this make sense?

[Sees lots of confused looks]

Here’s the long and short of it:

  • 1: 25%
  • 2: 19.9%
  • 3: 15.6%
  • 4: 11.9%
  • etc.

Now, on to the driving question.


Last night the Cavs were awarded the #1 overall pick. The Minnesota Timberwolves, who had the worst record in the NBA, were awarded the #2 overall pick. David Kahn, the General Manager of the Timberwolves went on record, complaining about how the lottery system is rigged.

“This league has a habit … of producing some pretty incredible story lines,” Kahn said, while smiling, on Tuesday. “Last year it was Abe Pollin’s widow and this year it was a 14-year-old boy… We were done. I told (Utah executive) Kevin (O’Connor): ‘We’re toast.’ This is not happening for us, and I was right.”

(Last year, they were beat out by a widow, representing the Wizards, this year by the 14-year-old son of the Cavs owner, who has a nerve disorder.)

Guiding Questions

  • Does Kahn have a point?
  • Should the Timberwolves be upset with the second pick, or happy that they didn’t slide all the way to fourth?
  • If you have the worst overall record, you have the best odds for the top pick, but that’s only 25%. So what’s your “expected” pick?

Potential Solution

It should be noted that the Timberwolves have had especially terrible “luck” with the NBA Draft Lottery system, at least according to them. They’ve been in the lottery 14 times and never been awarded a higher pick than their record should indicate. In fact they’ve ended up with a lower pick than their record would indicate 8 times. Some Timberwolves fans says this makes them 0-14 in the lottery. I’d say more like 0-8-6. That’s a whole other blog post. There’s also the subtext of Cleveland being awarded the top pick a year after losing LeBron James, but whatever. Our concern is with this particular draft.

Namely, if a team has a 25% chance of being selected (via 250 out of 1000 number combinations), should they expect to get the top pick? Should they be happy they didn’t get the 4th (or 3rd) pick? If not, what pick should they expect to get, and how often? (side note: mean, median, and mode anyone?)

Obviously, the team with the worst record should expect to pick 1st 25% of the time and either 2nd, 3rd, or 4th 75% of the time.

So we need to split up that 75%, right? Depending on who gets the #1 pick, that will affect the remaining odds. Last night, for instance, the Cavs were awarded the #1 pick via a trade with the Los Angeles Clippers, which only had a 2.8% chance of hitting. (Interestingly, the Cavs’ original pick was the “biggest loser”, slipping to 4th, despite having the second-most opportunities.)

The Timberwolves started with a 250/1000 chance of hitting, then after the first pick was awarded, if we eliminate the 28 combinations for the Clippers, they had only a 250/972 chance of being awarded the second pick, good for only 26%. 26% of 75% is 18.75%.

So the T-wolves had a 25% to get #1, a 18.75% chance to get #2, and, by subtracting from 100, a 56.25% chance of ending up with the 3rd or 4th pick.

Had the Cavs won the #1 overall pick with their original pick, with 199 out of 1000 combinations, the T-wolves would have had a 250/801 (31%) chance at the second pick. (for some mental gymnastics, we can say that the T-wolves had a 19.9% chance to have a 31% chance at the second pick)

For general purposes (i.e. no prior knowledge of who the #1 pick is), the NBA Draft Lottery wiki has done the calculations for us:

Seed    Chances    1st    2nd    3rd    4th
1           250         .250  .215   .178    .357

According to this, if you have the worst record in the NBA, the odds of you receiving the 3rd or 4th pick, worse than the Timberwolves landed at #2, total to 54%. The most common (mode) pick the worst team can expect is the worst possible pick at #4.

With odds like that, should the Timberwolves consider themselves lucky to only be picking as low as 2nd?

U-haul Linear Systems problem (updated and improved)

A couple weeks ago I posted this problem. I like the problem, but I wasn’t a huge fan of its solution and frankly, it sort of got away from me. Thankfully, my colleagues are more adept than I am at developing clear problems. So here’s the updated, improved version.


Moving On Up

Artifact / Entry Event

Facilitator : “Great news everybody. Principal ________ gave me a huge raise. It turns out that, contrary to all the media reports, the school district is flush with cash. As such, I’m moving on up. To the East side. I bought a MTV-cribs-like house. Unfortunately, I don’t have my hands on the huge sums of money yet, so I’ll be using a U-Haul to transport my currently meager possessions. Here’s a map of my old house, new house, and U-Haul rental location.”

[Facilitator shows slide / handout of map]

Facilitator: “Now, the local U-Haul rental has different fees for different sized trucks. The larger the moving truck, the more it costs. Here is their pricing scheme.”

[Facilitator shows slide / handout of U-Haul pricing table. Note: alternatively, the facilitator could direct them to the U-Haul website to obtain the specs: 14’ truck and 20’ truck. Note: there is also a 17’ truck for an up front cost of $29.95.]

Truck length U-Haul listed capacity Up front cost (1 day rental) Additional cost per mile
14’ 733 cu. Ft. $19.95 $0.79
20’ 1015 cu. Ft. $39.95 $0.79

Facilitator: “Now, I’m not exactly sure how much stuff I have. It might take 1 trip with these trucks, it might take 10. So I’m going to go calculate my furniture while you guys help me come up with a mathematical model, suggesting which truck I should get depending on the number of trips I have to take.”

Potential solution:

Students should first come up with a basic functional representation of the cost vs. mileage before applying it to the number of trips.

Cost = (up front cost of truck)+$0.79*mileage

, where m=mileage driven. This can be graphed as well.

To make a choice on which truck to get, students will have to determine a distance to and from each location on the map.

From U-Haul rental to Old Home: ~5.5 mi

From Old Home to New Home: ~8.5 mi

From New Home to U-Haul Rental: ~9 mi

Students then will need to set up a model in the form of a linear function, depending on the number of trips between the Old Home and New Home. Each extra trip after the first one will tack on 8.5+8.5=17 miles to the total mileage.

Cost = (Up front cost of truck)+$0.79*(mileage from U-Haul to Old Home + mileage between Old Home and New Home + mileage from New Home to U-Haul) + $0.79*2*(mileage from Old Home to New Home)*(number of trips – 1)

Cost = (Up front cost of truck)+$0.79*(5.5+8.5+9) + $0.79*2*(8.5)*(number of trips -1)

Cost = (Up front cost of truck)+$0.79*(23) + $0.79*2*(8.5)*(number of trips – 1)

Cost = (Up front cost of truck)+$18.17+$13.43*(number of trips – 1)

We can break that down for the three given truck sizes. For the smallest truck (the 14’ truck):

where x = # of trips to and from the New/Old Home

For the other two trucks:

Using a table:

# of trips:











14’ truck











20’ truck











Entry Event #2

Facilitator: “I’m still working on figuring out the number of trips I’ll have to take. I had several moving companies come over – before I decided to rent a U-Haul – and they each gave me a wildly different estimate of the amount of cubic footage of stuff I have.

Estimate 1: 2225 cu. ft
Estimate 2: 2891 cu. Ft.
Estimate 3: 3262 cu ft.

               Which size truck should I rent, based on each estimate?”

Potential solution.

Each estimate would entail a different number of trips based on the size of the truck rented. Recall,

Truck length U-Haul listed capacity
14’ 733 cu. Ft.
20’ 1015 cu. Ft.
Truck capacity

14’ truck (733 cu ft)

20’ truck (1015 cu ft)

Estimate 1: 2025

3 trips

2 trips

Estimate 2: 2891

4 trips

3 trips

Estimate 3: 3462

5 trips

4 trips

We can translate this into cost also using a table. (most cost effective)

Truck capacity

14’ truck (733 cu ft)

20’ truck (1015 cu ft)

Estimate 1: 2025



Estimate 2: 2891



Estimate 3: 3462



So for the three estimates, the first two would be more cost efficient using the 20’ truck while the last estimate would be more cost efficient using the 14’ truck.

Extension questions:

Can we develop a rule of thumb for differentiating between getting a 14’ vs. a 20’ truck (i.e. “If the # of trips using the 14’ truck is two or greater than the 20’ truck then it’s more cost efficient to get the 20’ truck.)?

Can we calculate the actual square footage of the 14’ and 17’ trucks? This incorporates volume standards.


Here’s a downloadable pdf version of the above activity.


I still am not 100% sure this activity is ready to go. I feel like it could still be improved. By all means, if you have any suggestions, let me know in the comments.