Percentages of success in basketball

[Yankel]

Hello all,

I wasn't sure if this question should be placed here or on algebra forum, but it is related to probability, so I put it here.

A basketball player shoots several times during some time period in a game, and scores less than 80% of his shots (let's say after the first quarter, but the time doesn't really matters). At the end of the game, he scored more than 80% of his shots. Is it necessary that at some point during the game, his success was exactly 80% ?

My intuition say that the answer is no, because this is not a continuous function, but every example I tried setting up, showed that 80% was achieved. What do you think ?

 

[squidsk]

Hello all,

I wasn't sure if this question should be placed here or on algebra forum, but it is related to probability, so I put it here.

A basketball player shoots several times during some time period in a game, and scores less than 80% of his shots (let's say after the first quarter, but the time doesn't really matters). At the end of the game, he scored more than 80% of his shots. Is it necessary that at some point during the game, his success was exactly 80% ?

My intuition say that the answer is no, because this is not a continuous function, but every example I tried setting up, showed that 80% was achieved. What do you think ?

Your correct that the answer is no. In terms of finding an example to show this the easiest way is to find some fraction that is greater than and not equal to 80% where subtracting one from the numerator and one from the denominator is less than and not equal to 80%. In other words find a fraction such that making the last shot pushes them over 80% success and that before making the last shot they had less than 80% success.

Find an x,y such that
\(\displaystyle \frac{x}{y} > 80%\) and
\(\displaystyle \frac{x-1}{y-1} < 80%\)

 

[Siron]

Actually, I think the claim is true. First, I tried some elementary examples and they appeared to be true so my intuition was that the claim should be true. Suppose that he shots $n$ times during the game, then each time he shots we can compute his succes rate. The game starts and he shots ... he scores, then his succes rate at that time is $1/1 = 100\%$. Suppose he shots again after a few minutes but misses then his succes rate is now $1/2 = 50\%$. So in fact, you can represent the game as a finite sequence
$$\left(\frac{a_i}{i}\right)_{i=1}^{n},$$
where the $a_i's$ are increasing by one but not necessarily each consecutive term.

Statement
The question can then be reformulated as: given that $a_n/n > (4/5)$ and at some point in the game $a_j/j < (4/5)$ for $j<n$. Does this imply that there exists an index $k$ such that $a_k /k = 0.8$ where $j<k<n$?

Proof
​The answer is yes. We have given $\displaystyle \frac{a_n}{n} > \frac{4}{5} > \frac{a_j}{j}$. Hence, $5a_n>4n$ and $5a_j < 4j$. Let $b_k=5a_k-4k$. As $a_{k+1}$ is either $a_k$ or $a_k+1$, we have $$b_{k+1}=5a_{k+1}-4k-4=\begin{cases}5a_k-4k-4&=b_k-4\quad \text{ or}\\5(a_k+1)-4k-4&=b_k+1\end{cases}$$Thus if the integer $b_k$ increases, it only increases in steps of $1$. From $b_j<0<b_n$ we see that there must be a ​first index $k$ between $j$ and $n$ for which $b_k\ge 0$. Then $b_k\le b_{k-1}+1$ and $b_{k-1}<0$, hence $b_k=0$ as desired.

Please, correct me if I am wrong.