The Hot Hand

Basketball players who make several baskets in succession are described as having a hot hand. Fans and players have long believed in the hot hand phenomenon, which refutes the assumption that each shot is independent of the next. However, a 1985 paper by Gilovich, Vallone, and Tversky collected evidence that contradicted this belief and showed that successive shots are independent events. This paper started a great controversy that continues to this day, as you can see by Googling hot hand basketball.

We do not expect to resolve this controversy today. However, in this lab we’ll apply one approach to answering questions like this. The goals for this lab are to (1) think about the effects of independent and dependent events, (2) learn how to simulate shooting streaks in R, and (3) to compare a simulation to actual data in order to determine if the hot hand phenomenon appears to be real.

Getting Started

Load packages

In this lab, we will explore and visualize the data using the tidyverse suite of packages. The data can be found in the companion package for OpenIntro labs, openintro.

Let’s load the packages.

library(tidyverse)
library(openintro)

Data

Your investigation will focus on the performance of one player: Kobe Bryant of the Los Angeles Lakers. His performance against the Orlando Magic in the 2009 NBA Finals earned him the title Most Valuable Player and many spectators commented on how he appeared to show a hot hand. The data file we’ll use is called kobe_basket.

glimpse(kobe_basket)
## Rows: 133
## Columns: 6
## $ vs          <fct> ORL, ORL, ORL, ORL, ORL, ORL, ORL, ORL, ORL, ORL, ORL, ORL…
## $ game        <int> 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1…
## $ quarter     <fct> 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3…
## $ time        <fct> 9:47, 9:07, 8:11, 7:41, 7:03, 6:01, 4:07, 0:52, 0:00, 6:35…
## $ description <fct> Kobe Bryant makes 4-foot two point shot, Kobe Bryant misse…
## $ shot        <chr> "H", "M", "M", "H", "H", "M", "M", "M", "M", "H", "H", "H"…

This data frame contains 133 observations and 6 variables, where every row records a shot taken by Kobe Bryant. The shot variable in this dataset indicates whether the shot was a hit (H) or a miss (M).

Just looking at the string of hits and misses, it can be difficult to gauge whether or not it seems like Kobe was shooting with a hot hand. One way we can approach this is by considering the belief that hot hand shooters tend to go on shooting streaks. For this lab, we define the length of a shooting streak to be the number of consecutive baskets made until a miss occurs.

For example, in Game 1 Kobe had the following sequence of hits and misses from his nine shot attempts in the first quarter:

\[ \textrm{H M | M | H H M | M | M | M} \]

You can verify this by viewing the first 9 rows of the data in the data viewer.

Within the nine shot attempts, there are six streaks, which are separated by a “|” above. Their lengths are one, zero, two, zero, zero, zero (in order of occurrence).

  1. What does a streak length of 1 mean, i.e. how many hits and misses are in a streak of 1? What about a streak length of 0?

WJ Response:

In this case a streak is defined as the number of subsequent “Hits” (H) achieved before a final “Miss” (M). Each streak must be concluded with a M, and the length of the streak is considered the number of repeated H’s that come before it. Thus a streak of 1 contains one hit and one miss (H M), while a streak of zero consists of a single miss (M).

Counting streak lengths manually for all 133 shots would get tedious, so we’ll use the custom function calc_streak to calculate them, and store the results in a data frame called kobe_streak as the length variable.

kobe_streak <- openintro::calc_streak(kobe_basket$shot)

We can then take a look at the distribution of these streak lengths.

ggplot(data = kobe_streak, aes(x = length)) +
  geom_bar()

  1. Describe the distribution of Kobe’s streak lengths from the 2009 NBA finals. What was his typical streak length? How long was his longest streak of baskets? Make sure to include the accompanying plot in your answer.

WJ Response:

Looking at the plot above, the distribution of Kobe’s streak lengths is clearly not normally distributed and right-skewed. The majority of Kobe’s streaks have length 0 or 1, and his longest streak of four hits in a row only occurred one time. The code below shows the percentage of Kobe’s “streaks” that are actually hit streaks (more than 1 shot made in a row):

kobe_streak %>% 
  filter(length > 1) %>%
    nrow() /
      nrow(kobe_streak)
## [1] 0.1710526

About ~17% of Kobe’s streaks have more than one hit in a row.

Compared to What?

We’ve shown that Kobe had some long shooting streaks, but are they long enough to support the belief that he had a hot hand? What can we compare them to?

To answer these questions, let’s return to the idea of independence. Two processes are independent if the outcome of one process doesn’t effect the outcome of the second. If each shot that a player takes is an independent process, having made or missed your first shot will not affect the probability that you will make or miss your second shot.

A shooter with a hot hand will have shots that are not independent of one another. Specifically, if the shooter makes his first shot, the hot hand model says he will have a higher probability of making his second shot.

Let’s suppose for a moment that the hot hand model is valid for Kobe. During his career, the percentage of time Kobe makes a basket (i.e. his shooting percentage) is about 45%, or in probability notation,

\[ P(\textrm{shot 1 = H}) = 0.45 \]

If he makes the first shot and has a hot hand (not independent shots), then the probability that he makes his second shot would go up to, let’s say, 60%,

\[ P(\textrm{shot 2 = H} \, | \, \textrm{shot 1 = H}) = 0.60 \]

As a result of these increased probabilites, you’d expect Kobe to have longer streaks. Compare this to the skeptical perspective where Kobe does not have a hot hand, where each shot is independent of the next. If he hit his first shot, the probability that he makes the second is still 0.45.

\[ P(\textrm{shot 2 = H} \, | \, \textrm{shot 1 = H}) = 0.45 \]

In other words, making the first shot did nothing to effect the probability that he’d make his second shot. If Kobe’s shots are independent, then he’d have the same probability of hitting every shot regardless of his past shots: 45%.

Now that we’ve phrased the situation in terms of independent shots, let’s return to the question: how do we tell if Kobe’s shooting streaks are long enough to indicate that he has a hot hand? We can compare his streak lengths to someone without a hot hand: an independent shooter.

Simulations in R

While we don’t have any data from a shooter we know to have independent shots, that sort of data is very easy to simulate in R. In a simulation, you set the ground rules of a random process and then the computer uses random numbers to generate an outcome that adheres to those rules. As a simple example, you can simulate flipping a fair coin with the following.

coin_outcomes <- c("heads", "tails")
sample(coin_outcomes, size = 1, replace = TRUE)
## [1] "tails"

The vector coin_outcomes can be thought of as a hat with two slips of paper in it: one slip says heads and the other says tails. The function sample draws one slip from the hat and tells us if it was a head or a tail.

Run the second command listed above several times. Just like when flipping a coin, sometimes you’ll get a heads, sometimes you’ll get a tails, but in the long run, you’d expect to get roughly equal numbers of each.

If you wanted to simulate flipping a fair coin 100 times, you could either run the function 100 times or, more simply, adjust the size argument, which governs how many samples to draw (the replace = TRUE argument indicates we put the slip of paper back in the hat before drawing again). Save the resulting vector of heads and tails in a new object called sim_fair_coin.

sim_fair_coin <- sample(coin_outcomes, size = 100, replace = TRUE)

To view the results of this simulation, type the name of the object and then use table to count up the number of heads and tails.

sim_fair_coin
##   [1] "tails" "heads" "heads" "tails" "heads" "heads" "heads" "heads" "tails"
##  [10] "heads" "heads" "tails" "tails" "tails" "tails" "heads" "heads" "heads"
##  [19] "heads" "tails" "heads" "heads" "tails" "tails" "heads" "tails" "heads"
##  [28] "heads" "tails" "tails" "tails" "tails" "heads" "heads" "heads" "tails"
##  [37] "heads" "heads" "tails" "tails" "tails" "tails" "heads" "heads" "heads"
##  [46] "tails" "heads" "tails" "tails" "tails" "heads" "tails" "heads" "heads"
##  [55] "heads" "heads" "tails" "heads" "heads" "tails" "tails" "heads" "heads"
##  [64] "tails" "tails" "heads" "heads" "heads" "tails" "tails" "tails" "heads"
##  [73] "tails" "tails" "heads" "tails" "heads" "heads" "tails" "heads" "tails"
##  [82] "heads" "heads" "tails" "tails" "tails" "tails" "heads" "tails" "heads"
##  [91] "heads" "heads" "heads" "tails" "heads" "heads" "tails" "tails" "tails"
## [100] "tails"
table(sim_fair_coin)
## sim_fair_coin
## heads tails 
##    52    48

Since there are only two elements in coin_outcomes, the probability that we “flip” a coin and it lands heads is 0.5. Say we’re trying to simulate an unfair coin that we know only lands heads 20% of the time. We can adjust for this by adding an argument called prob, which provides a vector of two probability weights.

sim_unfair_coin <- sample(coin_outcomes, size = 100, replace = TRUE, 
                          prob = c(0.2, 0.8))

prob=c(0.2, 0.8) indicates that for the two elements in the outcomes vector, we want to select the first one, heads, with probability 0.2 and the second one, tails with probability 0.8. Another way of thinking about this is to think of the outcome space as a bag of 10 chips, where 2 chips are labeled “head” and 8 chips “tail”. Therefore at each draw, the probability of drawing a chip that says “head”” is 20%, and “tail” is 80%.

  1. In your simulation of flipping the unfair coin 100 times, how many flips came up heads? Include the code for sampling the unfair coin in your response. Since the markdown file will run the code, and generate a new sample each time you Knit it, you should also “set a seed” before you sample. Read more about setting a seed below.

A note on setting a seed: Setting a seed will cause R to select the same sample each time you knit your document. This will make sure your results don’t change each time you knit, and it will also ensure reproducibility of your work (by setting the same seed it will be possible to reproduce your results). You can set a seed like this:

set.seed(35797)                 # make sure to change the seed

The number above is completely arbitraty. If you need inspiration, you can use your ID, birthday, or just a random string of numbers. The important thing is that you use each seed only once in a document. Remember to do this before you sample in the exercise above.

In a sense, we’ve shrunken the size of the slip of paper that says “heads”, making it less likely to be drawn, and we’ve increased the size of the slip of paper saying “tails”, making it more likely to be drawn. When you simulated the fair coin, both slips of paper were the same size. This happens by default if you don’t provide a prob argument; all elements in the outcomes vector have an equal probability of being drawn.

If you want to learn more about sample or any other function, recall that you can always check out its help file.

?sample

WJ Response:

The code below simulates the filpping of an unfair coin (in which heads only lands 20% of the time) 100 times, and outputs the results as a table:

set.seed(seed_num)
sim_unfair_coin <- sample(coin_outcomes, size = 100, replace = TRUE, 
                          prob = c(0.2, 0.8))
table(sim_unfair_coin)
## sim_unfair_coin
## heads tails 
##    14    86

The output above shows that simulation produced 14 heads and 86 tails. While this doesn’t seem very close to our expected values of 20 and 80, increasing the number of trials will cause the heads/tails ratio to approach 20%.

Simulating the Independent Shooter

Simulating a basketball player who has independent shots uses the same mechanism that you used to simulate a coin flip. To simulate a single shot from an independent shooter with a shooting percentage of 50% you can type

shot_outcomes <- c("H", "M")
sim_basket <- sample(shot_outcomes, size = 1, replace = TRUE)

To make a valid comparison between Kobe and your simulated independent shooter, you need to align both their shooting percentage and the number of attempted shots.

  1. What change needs to be made to the sample function so that it reflects a shooting percentage of 45%? Make this adjustment, then run a simulation to sample 133 shots. Assign the output of this simulation to a new object called sim_basket.

WJ Response:

The code chunk below uses the num_shots and shot_pct variables to make a random sampling of hits and misses using Kobe’s statistics (45% shooting percentage and 133 shots taken).

set.seed(seed_num)
num_shots <- 133
shot_pct <- 0.45
sim_basket <- sample(shot_outcomes,  size=num_shots, replace=TRUE,
                     prob = c(shot_pct, 1-shot_pct))
table(sim_basket)
## sim_basket
##  H  M 
## 63 70

The table of the results reveal that our randomly sampled (non-“hot-hand”) shooter had 63 hits and 70 misses.

Note that we’ve named the new vector sim_basket, the same name that we gave to the previous vector reflecting a shooting percentage of 50%. In this situation, R overwrites the old object with the new one, so always make sure that you don’t need the information in an old vector before reassigning its name.

With the results of the simulation saved as sim_basket, you have the data necessary to compare Kobe to our independent shooter.

Both data sets represent the results of 133 shot attempts, each with the same shooting percentage of 45%. We know that our simulated data is from a shooter that has independent shots. That is, we know the simulated shooter does not have a hot hand.

More Practice

Comparing Kobe Bryant to the Independent Shooter

  1. Using calc_streak, compute the streak lengths of sim_basket, and save the results in a data frame called sim_streak.

WJ Response:

The code chunk below uses the calc_streak function to calculate the streaks for the simulated shooter created in the previous exercise.

sim_streak <- openintro::calc_streak(sim_basket)
  1. Describe the distribution of streak lengths. What is the typical streak length for this simulated independent shooter with a 45% shooting percentage? How long is the player’s longest streak of baskets in 133 shots? Make sure to include a plot in your answer.

WJ Response:

The code chunk below provides a histogram of the streak lengths for the simulated shooter:

ggplot(data = sim_streak, aes(x = length)) +
  geom_bar()

The distribution is right shifted with the majority of streak lengths being either 1 or 0. The code below determines how many of these streaks are actual “hit” streaks, in which the shooter made more than one shot in a row:

sim_streak %>% 
  filter(length > 1) %>%
    nrow() /
      nrow(sim_streak)
## [1] 0.2253521

For the simulated shooter ~22.5% of their streaks have length greater than 1 (a little bit better than Kobe). The best streak in this case is 5 shots in a row, which occurred 2 times.

  1. If you were to run the simulation of the independent shooter a second time, how would you expect its streak distribution to compare to the distribution from the question above? Exactly the same? Somewhat similar? Totally different? Explain your reasoning.

WJ Response:

If multiple simulated shooters were to be created each with the same initial inputs (45% shooting percentage, 133 shots total), the distribution of streak lengths on average would not deviate much from the histogram we see above. There is of course the chance that a very long streak or an excessive number of small streaks occur, but in aggregate we should expect the distributions to be similar thanks to the law of large numbers. We can test this in the following cell by running the simulation for 100 shooters and seeing the proportion that each streak length occurs:

num_shots <- 133
shot_pct <- 0.45

all_streaks <- data.frame(matrix(ncol = 1, nrow = 0))
colnames(all_streaks) <- c('streak_length')


for (x in 1:100){
  sim <- sample(shot_outcomes,  size=num_shots, replace=TRUE,
                 prob = c(shot_pct, 1-shot_pct))
  streak <- openintro::calc_streak(sim)
  all_streaks <-rbind(all_streaks, streak)
}

print(prop.table(table(all_streaks)))
## length
##           0           1           2           3           4           5 
## 0.557239513 0.238024357 0.117185386 0.046143437 0.021244926 0.011502030 
##           6           7           8           9          12          13 
## 0.004194858 0.002706360 0.000947226 0.000541272 0.000135318 0.000135318

We can compare these proportions to the ones from our original shooter. These are outputted in the cell below:

print(prop.table(table(sim_streak)))
## length
##          0          1          2          3          4          5 
## 0.52112676 0.25352113 0.11267606 0.07042254 0.01408451 0.02816901

Looking closely at the proportions between the two, we can see that while there is some variation, they line up pretty well.

  1. How does Kobe Bryant’s distribution of streak lengths compare to the distribution of streak lengths for the simulated shooter? Using this comparison, do you have evidence that the hot hand model fits Kobe’s shooting patterns? Explain.

WJ Response:

Given the asymmetrical shape of the Kobe and the simulated shooter’s streak distribution, the median and inter-quartile range (IQR) are useful statistics to help describe them. However, given that the two distributions seem similar in shape (and max/min values), the mean could also be a useful comparison tool). These are calculated and printed below:

kobe_stats = c(mean(kobe_streak$length), median(kobe_streak$length), 
               IQR(kobe_streak$length))
sim_stats = c(mean(sim_streak$length), median(sim_streak$length), 
               IQR(sim_streak$length))
stat_names = c('mean', 'median', 'IQR')
data.frame(stat_names, kobe_stats, sim_stats)
##   stat_names kobe_stats sim_stats
## 1       mean  0.7631579 0.8873239
## 2     median  0.0000000 0.0000000
## 3        IQR  1.0000000 1.0000000

We see that the median and IQR values are the same (respectively 0 and 1 due to the fact that these values comprise the majority of streak lengths in each distribution), and that the mean streak length for the simulated shooter is actually better than for Kobe. Given all of these factors (distribution comparison, descriptive stats comparison), we can pretty safely say that Kobe did not have a “hot-hand” during the NBA finals. However, in order to bolster evidence for this conclusion, a possible next step would be to perform some kind of hypothesis testing.