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 (http://psych.cornell.edu/sites/default/files/Gilo.Vallone.Tversky.pdf). 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.
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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.
## 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:
## [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).
Answer 1:
* A streak length of 1 means 1 hit followed by 1 miss:
\[ \textrm{H M} \]
* A streak length of 0 means 1 miss that is preceded by another miss:
\[ \textrm{M} \]
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.
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.
Answer 2:
* The distribution of streak lengths has a mode at 0 (i.e., no basket), and almost 83% of his streaks had length of 0 or 1. The range of the distribution is 0 to 4, and the median streak length is 0.
* Kobe’s typical streak length, measured by the median, is 0. Measued by the mean, his average streak length is 0.76, i.e., between 0 and 1 but closer to 1.
* His longest streak was of length 4 (i.e., 4 baskets in a row), which occurred only once, or just over 1% of the time.
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0000 0.0000 0.0000 0.7632 1.0000 4.0000
## kobe_streak
## 0 1 2 3 4
## 39 24 6 6 1
## kobe_streak
## 0 1 2 3 4
## 0.51315789 0.31578947 0.07894737 0.07894737 0.01315789
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.
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] "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.
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" "heads" "tails" "tails" "tails" "tails" "tails" "heads"
## [9] "tails" "heads" "heads" "tails" "tails" "heads" "tails" "tails"
## [17] "heads" "tails" "tails" "heads" "heads" "heads" "heads" "heads"
## [25] "heads" "heads" "heads" "heads" "tails" "heads" "tails" "tails"
## [33] "heads" "tails" "tails" "heads" "tails" "tails" "tails" "tails"
## [41] "heads" "heads" "heads" "tails" "heads" "heads" "heads" "heads"
## [49] "heads" "tails" "heads" "heads" "tails" "heads" "tails" "heads"
## [57] "tails" "tails" "tails" "tails" "tails" "tails" "tails" "heads"
## [65] "heads" "tails" "heads" "tails" "tails" "heads" "tails" "heads"
## [73] "heads" "heads" "heads" "tails" "heads" "heads" "tails" "heads"
## [81] "heads" "heads" "heads" "tails" "tails" "tails" "heads" "heads"
## [89] "heads" "tails" "heads" "tails" "heads" "heads" "heads" "heads"
## [97] "tails" "tails" "heads" "tails"
## sim_fair_coin
## heads tails
## 53 47
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.
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%.
Answer 3:
My last simulation produced 21 heads; see frequency table below.
sim_unfair_coin <- sample(outcomes, size = 100, replace = TRUE, prob = c(0.2, 0.8))
table(sim_unfair_coin)## sim_unfair_coin
## heads tails
## 21 79
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 we 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 we use to simulate a coin flip. To simulate a single shot from an independent shooter with a shooting percentage of 50% we type,
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.
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.Answer 4:
We need to adjust the “H” / “M” probability weights to reflect Kobe’s shooting percentage, and then we need to set the sample size to the number of shots that Kobe took during the series (133 shots). See code below.
outcomes <- c("H", "M")
# probability weights
wts <- c(0.45, 0.55)
# sample size
N <- 133
sim_basket <- sample(outcomes, size = N, replace = TRUE, prob = wts)
table(sim_basket)## sim_basket
## H M
## 61 72
## sim_basket
## H M
## 0.4586466 0.5413534
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, we have the data necessary to compare Kobe to our independent shooter. We can look at Kobe’s data alongside our simulated data.
## [1] "H" "M" "M" "H" "H" "M" "M" "M" "M" "H" "H" "H" "M" "H" "H" "M" "M"
## [18] "H" "H" "H" "M" "M" "H" "M" "H" "H" "H" "M" "M" "M" "M" "M" "M" "H"
## [35] "M" "H" "M" "M" "H" "H" "H" "H" "M" "H" "M" "M" "H" "M" "M" "H" "M"
## [52] "M" "H" "M" "H" "H" "M" "M" "H" "M" "H" "H" "M" "H" "M" "M" "M" "H"
## [69] "M" "M" "M" "M" "H" "M" "H" "M" "M" "H" "M" "M" "H" "H" "M" "M" "M"
## [86] "M" "H" "H" "H" "M" "M" "H" "M" "M" "H" "M" "H" "H" "M" "H" "M" "M"
## [103] "H" "M" "M" "M" "H" "M" "H" "H" "H" "M" "H" "H" "H" "M" "H" "M" "H"
## [120] "M" "M" "M" "M" "M" "M" "H" "M" "H" "M" "M" "M" "M" "H"
## [1] "M" "H" "H" "H" "H" "H" "M" "M" "M" "M" "H" "H" "M" "M" "M" "M" "M"
## [18] "M" "M" "M" "M" "H" "M" "H" "H" "M" "H" "M" "M" "H" "H" "M" "H" "M"
## [35] "M" "M" "H" "M" "H" "H" "H" "H" "M" "M" "M" "H" "H" "H" "H" "M" "H"
## [52] "H" "H" "H" "M" "M" "H" "M" "H" "M" "H" "M" "M" "M" "M" "H" "H" "H"
## [69] "H" "M" "M" "H" "M" "M" "M" "H" "M" "H" "M" "H" "H" "M" "M" "H" "M"
## [86] "H" "M" "M" "H" "M" "H" "M" "H" "M" "H" "M" "M" "M" "H" "M" "H" "M"
## [103] "M" "H" "M" "M" "H" "H" "H" "H" "M" "H" "M" "M" "M" "M" "H" "H" "M"
## [120] "M" "H" "M" "H" "M" "M" "M" "M" "M" "H" "H" "H" "H" "M"
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.
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?
Answer:## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0000 0.0000 0.0000 0.8356 1.0000 5.0000
## sim_streak
## 0 1 2 4 5
## 39 22 5 6 1
## sim_streak
## 0 1 2 4 5
## 0.53424658 0.30136986 0.06849315 0.08219178 0.01369863If 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:
I would expect the second simulated distribution to be somewhat similar to the first, e.g., similar mode, median, size of the tail, etc., since the simulations have the same parameters with respect to probability weights and sample size, and since the sample size is reasonably large (133 shots). However, the distribution should not be exactly the same since we are simulating outcomes, and we would expect there to be random fluctuations from one simulation to the next.
For instance, see the results below from running the simulation a second time. Like the first simulation, the distribution for the second simulation has the same mode and median of 0 streak length, and the same mean length of 0.84. However, the second distribution has a longest streak of 7 baskets compared to 5 baskets for the first distribution, and streaks of 4 baskets or more occur roughly 4% of the time compared to almost 10% in the first distribution.
# run the simulation another time (same sample size and probability weights)
sim_basket <- sample(outcomes, size = N, replace = TRUE, prob = wts)
table(sim_basket)## sim_basket
## H M
## 61 72
## sim_basket
## H M
## 0.4586466 0.5413534
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0000 0.0000 0.0000 0.8356 1.0000 7.0000
## sim_streak
## 0 1 2 3 4 7
## 38 21 8 3 2 1
## sim_streak
## 0 1 2 3 4 7
## 0.52054795 0.28767123 0.10958904 0.04109589 0.02739726 0.01369863How 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:
Overall, the simulated shooter who has the same shooting percentage as Kobe – but who doesn’t have a “hot hand” – has a similar distribution of streak lengths as Kobe. Both distributions have the same mode and median of 0 streak length, and the mean streak length is comparable at ~0.8 baskets. However, based on the 2 simulations above (and others not shown here, which admittedly is a small sample size), it appears that the simulated shooter has a consistently higher frequency of longer streaks (4 baskets or more) than Kobe, roughly 4-10% compared to 1%. If anything, repeated distributions from the simulated shooter seem to suggest that Kobe may have had a “cold hand” during the 2009 NBA Finals.
This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel from a lab written by Mark Hansen of UCLA Statistics.