The Hot Hand

library(tidyverse)
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library(openintro)
## Loading required package: airports
## Loading required package: cherryblossom
## Loading required package: usdata

Data

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"…
view(kobe_basket)

Exercise 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?

A streak of length one means that there was one H then a M. A streak of length zero is a M following by another M.

Use the custom function calc_streak to calculate them, and store the results in a data frame called kobe_streak as the length variable.

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()

summary(kobe_streak)
##      length      
##  Min.   :0.0000  
##  1st Qu.:0.0000  
##  Median :0.0000  
##  Mean   :0.7632  
##  3rd Qu.:1.0000  
##  Max.   :4.0000
boxplot(kobe_streak)

Exercise 2

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.

The distribution of Kobe’s streak is right-skewed. He had a max streak length of 4. His median streak length was zero.

Compared to What?

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(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(shot 2 = H|shot 1 = H)=0.60

As a result of these increased probabilities, 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(shot 2 = H|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%.

Simulations in R

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] "tails" "tails" "tails" "tails" "heads" "tails" "tails" "tails" "tails"
##  [10] "heads" "heads" "tails" "heads" "heads" "heads" "heads" "heads" "tails"
##  [19] "tails" "tails" "tails" "tails" "heads" "tails" "tails" "tails" "tails"
##  [28] "tails" "heads" "heads" "tails" "tails" "heads" "tails" "heads" "heads"
##  [37] "tails" "tails" "heads" "tails" "tails" "heads" "tails" "tails" "tails"
##  [46] "heads" "heads" "tails" "heads" "heads" "heads" "tails" "heads" "heads"
##  [55] "heads" "heads" "heads" "tails" "tails" "tails" "heads" "tails" "tails"
##  [64] "heads" "tails" "tails" "heads" "tails" "tails" "heads" "tails" "tails"
##  [73] "heads" "heads" "tails" "heads" "heads" "tails" "heads" "tails" "tails"
##  [82] "tails" "heads" "tails" "tails" "heads" "tails" "tails" "heads" "tails"
##  [91] "tails" "heads" "tails" "heads" "heads" "heads" "heads" "tails" "tails"
## [100] "tails"
table(sim_fair_coin)
## sim_fair_coin
## heads tails 
##    43    57

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%.

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.

Exercise 3

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.

coin_outcomes <- c("heads", "tails")
set.seed(76)
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 
##    23    77

In my simulation of flipping a coin 100 times heads came up 23 times and tails came up 77 times.

?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.

Exercise 4

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.

shot_outcomes <- c("H", "M")
sim_basket <- sample(shot_outcomes, size = 133, replace = TRUE,
                     prob = c(0.45, 0.55))

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

Exercise 5

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

sim_streak <- data.frame(length = calc_streak(sim_basket))

Exercise 6

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.

summary(sim_streak$length)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.0000  0.0000  0.0000  0.7867  1.0000  3.0000

The distribution of streak lengths for the simulated no-hot hands shooter is right-skewed. The median streak length is 0 and maximum streak length is 6. Its spread is best measured by the IQR, which is 1 (Q3-Q1).

Exercise 7

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.

I would expect the distribution to be similar in shape, spread, and center (still right-skewed), but there will be some variability with each new shot.

Exercise 8

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.

Kobe’s distribution of streak lengths very similarly compares to that of the simiulated shooter as the Q1, Q3, and Median measures are the same. The mean is very similar as well and both distributions are right-skewed. I don’t think there is sufficient evidence to say Kobe has a hot hand. Kobe’s shots seem to be independent and making one shot did not increase his probability of making the next shot.