Lab 3: Probability, Independence, and Simulation

Conditional probability, independence, and simulation in R

The question

How can we use observed data and simulation to reason about probability, independence, and conditional probability?

Today we will use historical shot-by-shot data from Kobe Bryant’s 2009 NBA Finals to practice a classic probability question: does making one shot change the probability of making the next?

We are not trying to settle the broader hot-hand debate from one player or one series. The point is to use a familiar sequence of hits and misses to connect the probability ideas from lecture to R.

The lab has four parts:

estimate simple probabilities from observed data;
calculate conditional probabilities;
simulate an independent shooter; and
compare shooting streaks.

For probability problems, keep using the same workflow from lecture:

Name the event. Find the denominator. Calculate. Interpret.

Setup

Load the packages and historical data:

library(dplyr)
library(ggplot2)

load(url("https://raw.githubusercontent.com/GarciaRios/govt_3990/gh-pages/Labs/lab3/resources/hot_hand.RData"))

The file creates two objects:

kobe_basket: one row for each shot attempt;
calc_streak: a function that calculates the length of consecutive made-shot streaks.

Take a quick look:

glimpse(kobe_basket)
count(kobe_basket, shot)

In the shot variable:

H means hit;
M means miss.

1. Estimate probability from observed frequency

A probability model describes what could happen in repeated trials. In observed data, we can estimate a probability using the proportion of times an event occurred.

Start with Kobe’s observed shooting percentage:

kobe_basket %>%
  count(shot) %>%
  mutate(prop = n / sum(n))

You can also calculate the proportion of hits directly:

p_hit <- mean(kobe_basket$shot == "H")
p_hit

Your turn

What is the observed probability of a hit?
What is the observed probability of a miss?
Check that the two probabilities sum to 1.

In probability notation, if \(H\) means a made shot,

\[ P(H) = \frac{\text{number of hits}}{\text{number of shot attempts}}. \]

The complement is

\[ P(H^c)=1-P(H). \]

Here, \(H^c\) is simply a miss.

2. Conditional probability: does the previous shot matter?

The hot-hand question is naturally a conditional-probability question.

We want to compare:

\[ P(H_t \mid H_{t-1}) \]

with

\[ P(H_t \mid M_{t-1}), \]

where \(H_t\) means the current shot is a hit, \(H_{t-1}\) means the previous shot was a hit, and \(M_{t-1}\) means the previous shot was a miss.

First create a variable containing the previous shot.

kobe_transitions <- kobe_basket %>%
  group_by(game) %>%
  mutate(previous_shot = lag(shot)) %>%
  ungroup() %>%
  filter(!is.na(previous_shot))

We group by game so that the first shot of a new game is not treated as if it immediately followed the final shot of the previous game.

Now build a two-way table and calculate proportions within the previous-shot category:

kobe_transitions %>%
  count(previous_shot, shot) %>%
  group_by(previous_shot) %>%
  mutate(prob = n / sum(n))

Now visualize the same conditional comparison using a proportional bar chart:

ggplot(kobe_transitions, aes(x = previous_shot, fill = shot)) +
  geom_bar(position = "fill") +
  labs(
    x = "Previous shot",
    y = "Proportion of next shots",
    fill = "Current shot",
    title = "What happens after a hit or miss?"
  ) +
  theme_minimal()

Because position = "fill" makes each bar total 1, the plot compares the conditional distribution of the current shot within each previous-shot category.

Your turn: watch the denominator

Answer the following:

What is the observed value of \(P(H_t \mid H_{t-1})\)?
What is the observed value of \(P(H_t \mid M_{t-1})\)?
Are the two conditional probabilities exactly the same?
If shots were independent, what relationship would we expect between them and the overall probability of a hit?
What does the proportional bar chart show about the chance of a hit after a previous hit versus after a previous miss?
Does a difference in this one sample automatically prove dependence? Why not?

Recall the definition of independence:

\[ P(H_t \mid H_{t-1}) = P(H_t) \]

if the current shot is independent of the previous shot.

Observed sample proportions will almost never be exactly equal, even when the underlying process is independent. That is one reason simulation is useful.

3. Simulate an independent shooter

A simulation lets us create data from a process whose rules we control.

For an independent shooter, every shot has the same probability of being made, regardless of what happened on the previous shot.

We will use Kobe’s observed shooting percentage as the fixed probability of a hit.

First, set a seed so that your simulation is reproducible:

set.seed(311)

Now simulate the same number of shots:

shot_outcomes <- c("H", "M")

sim_basket <- sample(
  shot_outcomes,
  size = nrow(kobe_basket),
  replace = TRUE,
  prob = c(p_hit, 1 - p_hit)
)

Check the simulated shooting percentage:

mean(sim_basket == "H")
table(sim_basket)

Why this simulation is independent

Every simulated shot uses the same fixed probability p_hit. The computer does not increase or decrease the probability after a hit or miss.

Now calculate conditional probabilities in the simulated sequence:

sim_transitions <- tibble(shot = sim_basket) %>%
  mutate(previous_shot = lag(shot)) %>%
  filter(!is.na(previous_shot))

sim_transitions %>%
  count(previous_shot, shot) %>%
  group_by(previous_shot) %>%
  mutate(prob = n / sum(n))

Your turn

Compare \(P(H_t \mid H_{t-1})\) and \(P(H_t \mid M_{t-1})\) in the simulation.
Are they exactly equal?
We know the simulated shots are independent. Why can the sample conditional probabilities still differ?

This is an important idea that will return later in the course:

Random samples vary, even when the underlying probability model stays fixed.

4. Streaks: what does independence look like?

People often interpret a long run of hits as evidence that the probability of success has changed.

But an independent random process can also produce streaks.

Calculate Kobe’s streak lengths:

kobe_streak <- calc_streak(kobe_basket$shot)

ggplot(kobe_streak, aes(x = length)) +
  geom_histogram(binwidth = 1, boundary = -0.5) +
  labs(
    x = "Streak length",
    y = "Count",
    title = "Observed shooting streaks"
  ) +
  theme_minimal()

Summarize the streaks:

kobe_streak %>%
  summarise(
    mean_streak = mean(length),
    median_streak = median(length),
    max_streak = max(length)
  )

Now do the same for the independent shooter:

sim_streak <- calc_streak(sim_basket)

ggplot(sim_streak, aes(x = length)) +
  geom_histogram(binwidth = 1, boundary = -0.5) +
  labs(
    x = "Streak length",
    y = "Count",
    title = "Streaks from an independent-shooter simulation"
  ) +
  theme_minimal()

sim_streak %>%
  summarise(
    mean_streak = mean(length),
    median_streak = median(length),
    max_streak = max(length)
  )

Your turn

Compare the observed and simulated distributions.

Are long streaks possible under independence?
How do the typical and maximum streak lengths compare?
Why would one simulated sequence not be enough to make a strong conclusion about the hot-hand question?

The key point is not that the two distributions must look identical. They will not. The simulation gives us a comparison process in which we know shots are independent.

On Your Own

Complete the following in your student template.

A second independent-shooter simulation

Choose a new seed and simulate another independent shooter with:

the same number of shots as Kobe;
the same overall hit probability, p_hit.

Then:

calculate the simulated shooting percentage;
calculate \(P(H_t \mid H_{t-1})\) and \(P(H_t \mid M_{t-1})\);
calculate the maximum streak length;
compare your second simulation to the first simulation and to the observed Kobe data in 3–4 sentences.

Your interpretation should explain why differences between simulations do not imply that the probability model changed.