Dirk Nowitzki in the high-school diploma questions
May 30, 2008 by Christophe
BallinEurope is not only a basketball blog, we also talk about the stories that are somehow related to the sport we love. So the info I found today on the net made me smile a bit as it shows the importance of Dirk Nowitzki in Germany.
In recent German high school mathematics exams, a question about probability was linked to the German NBA player. For the mathematics freaks among our readers, here is the question:
The German basketball player Dirk Nowitzki plays for the Dallas Mavericks in the American professional league NBA. In the 2006/2007 season, he had a free-throw shooting percentage of 90.4%.
Calculate the probability that he
1. makes exactly 8 shots out of 10 attempts
2. makes a maximum of 8 shots out of 10 attempts
3. is successful four times in a row at most
So personally, I have no clue what the answers are. My math-school days are years away. The chief attraction in the questions, however, is the following. A university mathematics teacher from Bonn has declared that question 3 is “wrongly stated” and “importantly incomplete.”
According to the teacher, without saying in question 3 how many free throws in total Nowitzki must shoot, this question is not solvable.
So it looks like Dirk Nowtzki is not only incapable of leading a NBA team to the championship, but is also reason for lots of German students to have big problems on their math exams.
By the way, if any reader of BallinEurope can solve the three problems, he earns a bag full of Respect… 
Answers:
1. 18%
2. 25%
For the record, this kind of fuzzy thinking is typical of high school “math”. There is an implicit assumption that the probability a free throw is independent of previous free throws. As any basketball fun knows, that a very problematic assumption.
So there wasn’t enough information given to answer any of the questions. No wonder high school graduates have a culture shock when they first encounter logic in the natural sciences.
It’s all based on the binomial distribution, which says that the chances of getting n successes in m tries is [m!/(n!*(m-n)!)]*p^n*(1-p)^m-n
1) .185
2) .2484
3) For Dirk to hit four in a row at most, the first shot has to be a miss, then four makes, then the sixth shot a miss (or three, two, or one make with miss bookends). You don’t need to know how many shots he takes; you just add up the probabilities of a miss, then 4,3,2,1, and 0 makes, then a miss.
.0288. Or that’s my guess at least.
1. Two Putnam exams had problems centered around the free-throw shooting of “Shanille O’Keal”, who is somewhat better at it than her more famous namesake.
2. As Amnon says, the problem from Germany seems to implicitly assume an independence hypothesis. The normal practice when phrasing such problems is of explicitly assuming independence. This is particularly important since I am not aware of any research showing that free throws are independent events.
There’s a well-known (in some circles) paper by Gilovich, Vallone, and Tversky showing that there’s no such thing as the hot hand in basketball, including free throws. No one else has ever found it either. Shooting in general can be described by the mean, just like flipping a coin. So as far as that goes, free throws are independent events.
I suppose question 3 means: “in his immediately upcoming free throws, what is the probability that he does NOT begin a streak of five or more consecutive makes.”
You take the geometric distribution based on the probability of a miss; i.e., (Geo(.096)), which is .096*(1-.096)^n. Add together n=0, n=1, …,n=4.
And the total is… .396, or 39.6%.
Check out the big brain on Alex! Now what do they call a quarter-pounder with cheese in Berlin? Excellent work, stud.
Incidentally, will there be any question’s on next year’s exam about the probability that Dirk Nowitzki will take over a key game in the clutch? Before the question is phrased, i’m going with 0. Or maybe the square root of negative 1.
Cheers.
Dirk has had some clutch performances in international competition…
in the NBA it was all averys fault…
Alex, for #3, you basically said that it is equally likely that Dirk will miss one of his next five shots and that he will never hit five in a row in his career. The number of attempts is essential.
I solved it recursively on number of attempts by calculating the probability of hitting five in a row, then subtracting from 1.
5: .904^5
6: .904^5 + .096*(#5)
7: .904^5 + .904*.096*(#5) + .096*(#6)
8: .904^5 + .904^2*.096*(#5) + .904*.096*(#6) + .096*(#7)
9: .904^5 + .904^3*.096*(#5) + .904^2*.096*(#6) + .904*.096*(#7) + .096*(#8)
10: .904^5 + .904^4*.096*(#5) + .904^3*.096*(#6) + .904^2*.096*(#7) + .904*.096*(#8) + .096*(#9)
11: .904^5 + .904^4*.096*(#6) + .904^3*.096*(#7) + .904^2*.096*(#8) + .904*.096*(#9) + .096*(#10)
12: …
Subtract any of these values from 1, and you have your answer for that value.