library(usdata)
library(cherryblossom)
library(tidyverse)
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## ✔ tidyr   1.2.1      ✔ stringr 1.4.1 
## ✔ readr   2.1.2      ✔ forcats 0.5.2 
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
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library(openintro)
## Loading required package: airports
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"…

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?

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()

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.

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" "heads" "heads" "heads" "tails" "tails"
##  [10] "heads" "heads" "heads" "heads" "tails" "heads" "heads" "tails" "tails"
##  [19] "tails" "tails" "tails" "heads" "heads" "tails" "heads" "tails" "heads"
##  [28] "heads" "tails" "heads" "heads" "tails" "heads" "tails" "heads" "heads"
##  [37] "tails" "heads" "heads" "heads" "tails" "heads" "heads" "tails" "heads"
##  [46] "tails" "tails" "tails" "heads" "tails" "heads" "heads" "tails" "tails"
##  [55] "heads" "heads" "tails" "heads" "heads" "heads" "heads" "tails" "heads"
##  [64] "heads" "tails" "heads" "tails" "heads" "tails" "tails" "heads" "tails"
##  [73] "heads" "heads" "tails" "tails" "heads" "heads" "tails" "tails" "heads"
##  [82] "heads" "tails" "tails" "heads" "heads" "heads" "tails" "tails" "heads"
##  [91] "tails" "heads" "tails" "heads" "heads" "tails" "tails" "tails" "tails"
## [100] "heads"
table(sim_fair_coin)
## sim_fair_coin
## heads tails 
##    54    46

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%.

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.

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(2022) #make sure to change the seed
coin_flips <- c('heads', 'tails')
sample(coin_flips, size = 1, replace = TRUE)
## [1] "tails"
unfair_coin <- sample(coin_flips, size = 100, replace = TRUE, prob = c(0.2, 0.8))
table(unfair_coin)
## unfair_coin
## heads tails 
##    17    83

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.

Exercise4 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.

set.seed(90)
shot_outcomes <- c("H", "M")
sim_basket <- sample(shot_outcomes, size = 133, replace = TRUE, prob = c(0.55,0.45))

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.

set.seed(202)
shot_outcomes <- c("H", "M")
sim_basket <- sample(shot_outcomes, size = 133, replace = TRUE, prob = c(0.55,0.45))
sim_streak <- 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.

library(ggplot2)
ggplot(data=sim_streak,aes(x=length))+geom_bar()

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.

Answer If I were to run this simulation a second time, I would expect the streak distribution to be somewhat similar to the question above. This is because the probability is still 45% meaning more than half of the ahots will be missed anyway. Therefore, it will still be positively skewed.

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.

Answer Kobe Bryant’s distribution of streak length is very similar to the simulated distribtion we created. It is the same shape, it has the same typical streak length and the same longest streak length.The few other times I ran the simulation - the longest streak length was higher - as high as 9 at one point, but averall the distribution was very similar and the typical streak and ahape remained consistant. I can conclude that hot hand model doesn’t fit Kobe’s shooting pattern. His pattern is similat to our simulated distribution which is independent. I can conclude that hot hand model is incorrect for Kobe.