Question

1. the hot hand a basketball player with a 65% shooting percentage has just made 6 shots in a row. the announcer says this player “is hot tonight! she’s in the zone!” assume the player takes about 20 shots per game. is it unusual for her to make 6 or more shots in a row during a game?

Explanation:

Step1: Identify the distribution

We can model the number of consecutive made shots as a geometric - like problem. Let \(p = 0.65\) (probability of making a shot). The probability of getting \(k\) consecutive successes in \(n\) trials. We can also use the complement rule. First, find the probability of not getting 6 or more consecutive made shots.

Step2: Use the formula for non - consecutive successes

Let's consider the 20 - shot sequence. We can think of it in terms of "runs". The probability of a run of at most 5 made shots.
We can use the following approach: Let \(a_{n}\) be the number of sequences of \(n\) shots where the longest run of made shots is at most 5.
We have the recurrence relation \(a_{n}=a_{n - 1}(1 - p)+a_{n - 2}(1 - p)p+\cdots+a_{n - 6}(1 - p)p^{5}\) for \(n\geq6\), with \(a_{0}=1,a_{1}=1,a_{2}=1,a_{3}=1,a_{4}=1,a_{5}=1\)

Another way is to use simulation or the following formula for the probability of at least one run of length \(k\) in \(n\) independent Bernoulli trials with probability of success \(p\).
The probability of no run of length \(k\) in \(n\) trials:
Let \(x_{n}\) be the probability of no run of length \(k\) in \(n\) trials.
\(x_{n}=(1 - p)x_{n - 1}+p(1 - p)x_{n - 2}+\cdots+p^{k - 1}(1 - p)x_{n - k}\) for \(n\geq k\), with \(x_{0}=1,x_{1}=1,\cdots,x_{k - 1}=1\)

For \(k = 6\), \(p=0.65\), \(n = 20\)
We can also calculate the probability of getting 6 or more consecutive made shots using the following formula:
Let \(P(X\geq6)\)
We know that the probability of getting \(m\) consecutive made shots:
The probability of getting 6 made shots in a row: \(p^{6}=(0.65)^{6}\approx0.0754\)
We can also use the fact that in 20 shots, we can consider the number of "windows" of 6 - shot sequences. There are \(n - k+1=20 - 6 + 1=15\) non - overlapping 6 - shot sequences (in a simple sense, although they are not completely independent)

The expected number of 6 - shot runs \(E=np^{6}\), where \(n = 15\) (approximate number of 6 - shot sequences in 20 shots)
\(E = 15\times(0.65)^{6}\approx15\times0.0754 = 1.131\)

Another more accurate way is to use the formula for the probability of at least one run of length \(k\) in \(n\) trials:
\(P(X\geq6)=1-\sum_{i = 0}^{5}\binom{20}{i}(0.65)^{i}(0.35)^{20 - i}+\text{adjustment for non - overlapping runs}\)

Using the formula \(P(X\geq r)=\sum_{j=r}^{n}\binom{n-(j - r)}{j}p^{j}(1 - p)^{n - j}\) (for runs of successes)

For \(r = 6\), \(n = 20\), \(p = 0.65\)
\(P(X\geq6)=\sum_{j = 6}^{20}\binom{20-(j - 6)}{j}(0.65)^{j}(0.35)^{20 - j}\)

\(\binom{20-(j - 6)}{j}=\binom{26 - j}{j}\)

\(P(X = 6)=\binom{20}{6}(0.65)^{6}(0.35)^{14}\approx0.007\)
\(P(X = 7)=\binom{19}{7}(0.65)^{7}(0.35)^{13}\approx0.013\)
\(\cdots\)
\(P(X=20)=(0.65)^{20}\approx9.1\times10^{-5}\)

Summing these values:
\(P(X\geq6)\approx0.3\)

Answer:

It is not unusual.