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?

Insert your answer here

# A streak length of 1 means there was 1 hit, followed by a miss, thus the streak ended.
# A streak length of 0 just refers to a miss and no hits, thus a streak of hits didn't begin.

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 <- 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.

Insert your answer here

library(ggplot2)

ggplot(data = kobe_streak, aes(x = length)) +
  geom_bar() +
  geom_text(aes(label = ..count..), stat = "count", vjust = 2, color = "black")

# Of the 133 shots, and the 76 streaks- the most common streak length was length 0 (39 times) 
# His longest streak of baskets was 4, which only occurred once.

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] "heads"

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] "heads" "tails" "tails" "tails" "tails" "heads" "tails" "tails" "heads"
##  [10] "tails" "heads" "tails" "heads" "heads" "tails" "tails" "tails" "tails"
##  [19] "tails" "heads" "heads" "tails" "tails" "tails" "heads" "heads" "tails"
##  [28] "tails" "heads" "tails" "tails" "heads" "tails" "heads" "heads" "tails"
##  [37] "heads" "heads" "heads" "tails" "heads" "heads" "tails" "tails" "heads"
##  [46] "tails" "tails" "tails" "tails" "heads" "heads" "heads" "tails" "tails"
##  [55] "heads" "heads" "heads" "heads" "heads" "heads" "heads" "heads" "heads"
##  [64] "heads" "tails" "tails" "tails" "tails" "tails" "heads" "tails" "tails"
##  [73] "tails" "tails" "heads" "tails" "tails" "tails" "tails" "tails" "heads"
##  [82] "heads" "tails" "heads" "heads" "heads" "tails" "tails" "heads" "heads"
##  [91] "heads" "tails" "heads" "heads" "tails" "heads" "heads" "tails" "heads"
## [100] "heads"
table(sim_fair_coin)
## sim_fair_coin
## heads tails 
##    49    51

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.

Insert your answer here

set.seed(12345)   
sim_unfair_coin <- sample(coin_outcomes, size = 100, replace = TRUE, 
                          prob = c(0.2, 0.8))
sim_unfair_coin
##   [1] "tails" "heads" "tails" "heads" "tails" "tails" "tails" "tails" "tails"
##  [10] "heads" "tails" "tails" "tails" "tails" "tails" "tails" "tails" "tails"
##  [19] "tails" "heads" "tails" "tails" "heads" "tails" "tails" "tails" "tails"
##  [28] "tails" "tails" "tails" "tails" "tails" "tails" "tails" "tails" "tails"
##  [37] "heads" "heads" "tails" "tails" "tails" "tails" "heads" "tails" "tails"
##  [46] "tails" "tails" "tails" "tails" "tails" "heads" "heads" "tails" "tails"
##  [55] "tails" "tails" "tails" "tails" "tails" "tails" "tails" "tails" "heads"
##  [64] "tails" "heads" "tails" "heads" "tails" "tails" "heads" "tails" "tails"
##  [73] "tails" "tails" "tails" "tails" "heads" "tails" "tails" "tails" "heads"
##  [82] "tails" "tails" "tails" "tails" "tails" "heads" "tails" "heads" "tails"
##  [91] "heads" "heads" "tails" "tails" "tails" "tails" "tails" "tails" "tails"
## [100] "tails"
table(sim_unfair_coin)
## sim_unfair_coin
## heads tails 
##    20    80
# With seed 12345, we always receive 20 heads and 80 tails.

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

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.

Insert your answer here

set.seed(112233)  
shot_outcomes <- c("H", "M")
sim_basket <- sample(shot_outcomes, size = 133, replace = TRUE, prob = c(0.45, 0.55))  # Adjust the hit probability to 45%
table(sim_basket)
## sim_basket
##  H  M 
## 59 74
# With seed 112233, and the probability of hitting a shot 45% of the time, a player will get 59 hits and 74 misses from 133 attempts.

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.

Insert your answer here

sim_streak <- 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.

Insert your answer here

ggplot(data = sim_streak, aes(x = length)) +
  geom_bar() +
  geom_text(aes(label = ..count..), stat = "count", vjust = 2, color = "black")

With my randomly generated simstreak at 45% hit probability, the typical and most common streak length is once again 0. Interestingly enough, the seed I chose for this simstreak actually had the longest streak of 6.

  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.

Insert your answer here

# If I were to run the simulation with a different seed, I'd expect to see somewhat similar results, so let's run another example

set.seed(1234)  
shot_outcomes <- c("H", "M")
sim_basket2 <- sample(shot_outcomes, size = 133, replace = TRUE, prob = c(0.45, 0.55))  # Adjust the hit probability to 45%
table(sim_basket2)
## sim_basket2
##  H  M 
## 46 87
# With seed 1234, this has 46 hits and 87 misses, as compared to 59 H and 74 M in the prior example.

sim_streak2 <- calc_streak(sim_basket2)

ggplot(data = sim_streak2, aes(x = length)) +
  geom_bar() +
  geom_text(aes(label = ..count..), stat = "count", vjust = 2, color = "black")

Looking at the distribution, some things remain the same- the largest amount of streaks are of length 0. There is a little over 20 with length 1, and a small handful with length 2 or more. Since this simulation had less hits, there are less streaks with length over 1, but this random simulation actually had 2 occurrences of streaks with length 5. In both simulations of the independent shooter, there was a small handful of streaks with length >= 4. In the first instance, there was 1 streak of length 4 and 1 of length 5. In this second instance, there was 1 streak of length 4, and 2 streaks of length 5. This shows that there will usually be at least 1 streak of longer length.

  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.

Insert your answer here

Looking at Kobe Byrant’s distribution (the first graph), I see that this follows a similar trend to the random simulations above. The majority of streaks are of length 0, with 24 of length 1. Interestingly enough, both random simulations above also had just over 20 streaks with length 1 (23 and 22 respectively). Kobe also had 6 streaks of length 2 and 6 streaks of length 3, which is mathematically similar to the 2 simulations. However, Kobe’s longest streak was of length 4, whereas the 2 simulations had lengths of 6 and 5. Mathematically speaking, I believe this disproves the existence of the hot hand model, as it simply looks to be following the same trends of the ones where we defined P(Hit) = 45%. I believe this shows that the hot-hand model is nothing more than a player’s self-confidence and the enthusiasm of a devoted crowd of fans, looking for a pattern where there is none. Kobe’s performance of the 2009 World Finals was phenomenal, but I believe it was because he is a great player with a supportive and well-coordinated team.