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.
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.
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.
## 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).
Insert your answer here
strike is initiated by a player when s/he tries to hit the basket, if it is a hit then s/he succeeds and gets one and the sequence of hits defines the length of the strike. In this case, each miss, M, is a strike of 0, Because the player attempt ended with zero hit, the first H after a miss assumed the first strike, if the second is hit now the length of strikes is two, and it continues unless the palyer misses when the strike ends, and a new one begins. Strikes of 1 means; a hit follows with a miss. Strike 0 means, an attempt that leads to a miss.
End of the answer
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.
We can then take a look at the distribution of these streak lengths.
Insert your answer here
Looking into the distribution shows skewness toward 0 and lower strikes which is expected, meaning the majority of the 76 strikes end in missing without any hit. aside from that the second one which can be assumed as a strikes end with 1, meaning one hit follows with a miss.
result <- kobe_streak %>%
filter(length == max(length))|>
summarize(n = n(),
kobe_max = max(length))
result1 <- kobe_streak|>
group_by(length)|>
summarize(n = n())
print(paste("Longest of streaks =", result$kobe_max))## [1] "Longest of streaks = 4"
ggplot(data = kobe_streak, aes(x = length)) +
geom_bar()+ scale_fill_gradient (low="red", high= "yellow" ) +
labs(x = "The streaks", y = "Length of streaks ", title ="Kobe's Streaks")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.
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.
## [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.
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.
## [1] "tails" "tails" "tails" "heads" "heads" "tails" "tails" "heads" "heads"
## [10] "heads" "heads" "tails" "heads" "tails" "tails" "heads" "tails" "heads"
## [19] "heads" "heads" "tails" "tails" "tails" "heads" "heads" "tails" "heads"
## [28] "tails" "tails" "tails" "tails" "heads" "heads" "tails" "heads" "heads"
## [37] "tails" "heads" "tails" "heads" "tails" "tails" "heads" "tails" "tails"
## [46] "heads" "heads" "heads" "tails" "tails" "heads" "heads" "tails" "heads"
## [55] "heads" "tails" "tails" "heads" "heads" "tails" "heads" "tails" "tails"
## [64] "heads" "tails" "tails" "heads" "tails" "tails" "tails" "tails" "heads"
## [73] "heads" "heads" "tails" "heads" "tails" "tails" "heads" "heads" "heads"
## [82] "tails" "heads" "heads" "heads" "tails" "heads" "tails" "heads" "heads"
## [91] "heads" "tails" "tails" "tails" "tails" "tails" "heads" "heads" "heads"
## [100] "heads"
## sim_fair_coin
## heads tails
## 51 49
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.
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%.
Insert your answer here
We will use the code written above and will use seed as sugegste to create a repeatable results.
#defien the cvetro of variable head or tail
coin_outcomes <- c("heads", "tails")
#set seed and use a fair coin
set.seed(2024); sim_fair_coin_kp <- sample(coin_outcomes, size = 100, replace = TRUE)
fair_count <- data.frame(heads_count = sum(sim_fair_coin_kp == coin_outcomes[1]), tails_count = sum(sim_fair_coin_kp == coin_outcomes[2]))
#set seed and set up a unfair coing with 0.2 to 0.8 ratio
set.seed(2024); sim_unfair_coin_kp <- sample(coin_outcomes, size = 100, replace = TRUE, prob = c(0.2, 0.8))
unfair_count <- data.frame(heads_count = sum(sim_unfair_coin_kp == coin_outcomes[1]), tails_count = sum(sim_unfair_coin_kp == coin_outcomes[2]))
print(paste("The number of heads in a unfair coin with 0.2/0.8 is", unfair_count$heads,", and the total counts is", length(sim_unfair_coin_kp)))## [1] "The number of heads in a unfair coin with 0.2/0.8 is 19 , and the total counts is 100"
End of answer
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:
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.
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
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.
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
the way I have approached the problem is by defining the total count and give the “H” the ratio of 45%
shot_outcomes <- c("H", "M")
set.seed(2024); sim_basket <- sample(shot_outcomes, size = 133, replace = TRUE, prob = c(0.45,0.55))
shot_count <- data.frame(hit_count = sum(sim_basket == shot_outcomes[1]), miss_count = sum(sim_basket == shot_outcomes[2]))
print(paste("The hits in a simulation data is", shot_count$hit," with the probqbility of", round(shot_count$hit/length(sim_basket),2), "Vs the required 0.45"))## [1] "The hits in a simulation data is 61 with the probqbility of 0.46 Vs the required 0.45"
End of answer
Note that we’ve named the new vector sim_basket, the
same name that we gave to the previous vector reflecting a shooting
percentage of 50%. In this situation, R overwrites the old object with
the new one, so always make sure that you don’t need the information in
an old vector before reassigning its name.
With the results of the simulation saved as sim_basket,
you have the data necessary to compare Kobe to our independent
shooter.
Both data sets represent the results of 133 shot attempts, each with the same shooting percentage of 45%. We know that our simulated data is from a shooter that has independent shots. That is, we know the simulated shooter does not have a hot hand.
calc_streak, compute the streak lengths of
sim_basket, and save the results in a data frame called
sim_streak.Insert your answer here
sim_streak <- calc_streak(sim_basket)
result_kp <- sim_streak %>%
filter(length == max(length))|>
summarize(n = n(),
sim_max = max(length))
result1_kp <- sim_streak|>
group_by(length)|>
summarize(n = n())
print(paste("Longest of simulated streaks =", result_kp$sim_max))## [1] "Longest of simulated streaks = 4"
ggplot(data = sim_streak, aes(x = length)) +
geom_bar()+ scale_fill_gradient (low="red", high= "yellow" ) +
labs(x = "The streaks", y = "Length of streaks ", title ="Sim's Streaks")End of answer
Insert your answer here
The distribution of streaks differs from both Kobe’s and my simulation; however, the general trends appear quite similar. Maximum streaks occur at 4, and the probabilities for streaks from 1 to 4 decrease. Notably, this simulation exhibits a higher rate of 4-streaks compared to Kobe’s data—where there was only one such streak, this player experiences three. Interestingly, it seems the simulated player might have a better hot hand than Kobe!
Surprisingly enough the longest streak for 45% is 4 same as what it was for Kobe’s.
ggplot(data = sim_streak, aes(x = length)) +
geom_bar()+ scale_fill_gradient (low="red", high= "yellow" ) +
labs(x = "The streaks", y = "Length of streaks ", title ="Sim's Streaks")End of answer
Insert your answer here
I don’t expect it to be identical, but rather different. However, the total number of streaks should remain close to what was observed previously. Although the distribution of streaks may vary, the overall trends are anticipated to be consistent. The maximum streak length might shift to either 5 or 3, and the distribution of the remaining streaks could also change, while still maintaining a total of around 60 or rounding it.
End of answer
Insert your answer here
I printed them side-by-side to compare. Such a comparison reveals that the simulation might outperform Kobe’s results. Specifically, the number of 0 streaks is lower, and the number of 4 streaks is higher in the simulation. However, for the number 3, it is less than Kobe’s. Overall, based on this single point of comparison, it does not appear that Kobe’s results are better than those of the simulated player.
# Create a grouping variable for sim_streak and kobe_streak
sim_streak$group <- "sim_streak"
kobe_streak$group <- "kobe_streak"
# Combine the data frames
combined_data <- rbind(sim_streak, kobe_streak)
# Create a histogram using ggplot2
ggplot(combined_data, aes(x = length, fill = group)) +
geom_histogram(binwidth = 0.5, position = "dodge") +
facet_wrap(~group, ncol = 2) +
labs(title = "Side-by-Side Histograms", x = "The streaks", y = "Length of streaks") +
theme_minimal()End of answer