Hot Hands

Our 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. Let’s load some data from those games and look at the first several rows.

download.file("http://www.openintro.org/stat/data/kobe.RData", destfile = "kobe.RData")
load("kobe.RData")
head(kobe)
##    vs game quarter time
## 1 ORL    1       1 9:47
## 2 ORL    1       1 9:07
## 3 ORL    1       1 8:11
## 4 ORL    1       1 7:41
## 5 ORL    1       1 7:03
## 6 ORL    1       1 6:01
##                                               description basket
## 1                 Kobe Bryant makes 4-foot two point shot      H
## 2                               Kobe Bryant misses jumper      M
## 3                        Kobe Bryant misses 7-foot jumper      M
## 4 Kobe Bryant makes 16-foot jumper (Derek Fisher assists)      H
## 5                         Kobe Bryant makes driving layup      H
## 6                               Kobe Bryant misses jumper      M

In this data frame, every row records a shot taken by Kobe Bryant. If he hit the shot (made a basket), a hit, H, is recorded in the column named basket, otherwise a miss, M, is recorded.

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:

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

To verify this use the following command:

kobe$basket[1:9]
## [1] "H" "M" "M" "H" "H" "M" "M" "M" "M"

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?

Answer: A streak length of 1 means hits and then a miss. There is a hit and a miss in a steak 1. A steak length of 0 means a miss.

The custom function calc_streak, which was loaded in with the data, may be used to calculate the lengths of all shooting streaks and then look at the distribution.

kobe_streak <- calc_streak(kobe$basket)
barplot(table(kobe_streak))

Note that instead of making a histogram, we chose to make a bar plot from a table of the streak data. A bar plot is preferable here since our variable is discrete – counts – instead of continuous.

  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?

Answer: The distribution of Kobe’s streak lengths is right skewed. His typical steak length is 0. His longest Steak of Basket is 4.

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 hot hands? 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 hot hands? We can compare his streak lengths to someone without hot hands: 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.

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

The vector 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(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" "heads" "tails" "tails" "heads" "tails"
##   [9] "tails" "tails" "heads" "heads" "heads" "heads" "tails" "tails"
##  [17] "tails" "tails" "heads" "heads" "tails" "heads" "tails" "tails"
##  [25] "heads" "tails" "heads" "heads" "heads" "heads" "tails" "heads"
##  [33] "tails" "tails" "heads" "heads" "heads" "heads" "heads" "tails"
##  [41] "tails" "heads" "heads" "heads" "heads" "heads" "tails" "tails"
##  [49] "tails" "heads" "tails" "heads" "heads" "tails" "tails" "tails"
##  [57] "tails" "tails" "tails" "tails" "heads" "tails" "heads" "tails"
##  [65] "heads" "heads" "tails" "tails" "tails" "heads" "tails" "heads"
##  [73] "heads" "tails" "heads" "tails" "heads" "tails" "tails" "heads"
##  [81] "tails" "heads" "tails" "tails" "tails" "heads" "heads" "heads"
##  [89] "heads" "tails" "tails" "heads" "heads" "tails" "tails" "heads"
##  [97] "heads" "heads" "tails" "tails"
table(sim_fair_coin)
## sim_fair_coin
## heads tails 
##    49    51

Since there are only two elements in 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(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?
table(sim_unfair_coin)
## sim_unfair_coin
## heads tails 
##    25    75

Answer: 22 times flips came up heads.

Simulating the Independent Shooter

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

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

To make a valid comparison between Kobe and our simulated independent shooter, we 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.
outcomes <- c("H", "M")
sim_basket <- sample(outcomes, size = 133, replace = TRUE,prob = c(0.45, 0.55))
sim_basket
##   [1] "M" "M" "M" "M" "H" "H" "M" "H" "M" "M" "H" "H" "H" "H" "M" "H" "H"
##  [18] "M" "M" "M" "H" "M" "H" "H" "H" "H" "M" "H" "H" "M" "H" "H" "H" "M"
##  [35] "M" "H" "H" "H" "H" "M" "H" "M" "M" "M" "H" "M" "M" "M" "H" "M" "M"
##  [52] "H" "M" "H" "H" "M" "M" "H" "M" "H" "M" "M" "M" "H" "H" "M" "M" "M"
##  [69] "M" "M" "H" "M" "M" "M" "M" "M" "H" "H" "H" "H" "H" "M" "M" "H" "M"
##  [86] "M" "M" "M" "M" "M" "M" "H" "H" "M" "M" "H" "H" "M" "M" "M" "M" "M"
## [103] "M" "M" "M" "H" "H" "M" "H" "H" "H" "H" "H" "M" "M" "H" "M" "H" "H"
## [120] "H" "H" "M" "H" "M" "H" "M" "M" "M" "M" "H" "H" "H" "H"
table(sim_basket)
## sim_basket
##  H  M 
## 62 71

Answer: We need to change the size to 133 and add prob of 0.45 vs 0.55 to the function.

On your own

Comparing Kobe Bryant to the Independent Shooter

Using calc_streak, compute the streak lengths of sim_basket.

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?
sim_basket_streak <- calc_streak(sim_basket)
barplot(table(sim_basket_streak))

Answer: The distribution is right skewed. The typical streak length is 0 and the longest streak of baskets is 5.

  • 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.

Answer: If we run the simulation of the independent shooter a second time, the result will be depend on each simulation, but there is no big difference and the distribution will be similar right skewed.

  • 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.

Answer: Kobe Bryant’s distribution of streak is similar to the distribution of streak. They are all right skewed. The hot hand model doesn’t fit Kobe’s shooting patterns because the simulated shooter is similar to Kobe’s patterns. Their typical steak length are 0.