Project 1 - Kobe Basketball Data

Part 2.a – Conditional probabilities

1. With the data, a logical statement, and the function mean, first calculate the total percentage of shots that resulted in a basket in the 2009 NBA finals, as []

num_hits = 0
i = 1 # counter
total_shots = nrow(kobe) # the number of total shots
# In the loop, get the number of hits
for(i in 1:nrow(kobe_streak)){
    if(!(mean(kobe_streak[[i, "streak"]]) == 0)){
        num_hits = num_hits + kobe_streak[[i, "streak"]] 
    }
}
prob = num_hits/total_shots # calculate the probability
pct <- round(prob * 100, 2) # get the percentage with rounded
print(paste(pct, "%")) 

## [1] "43.61 %"

2. We need to filter out the streaks that had at least the first shot resulting in a Hit – by doing this we are conditioning the data to make the conditional statement. Since those streaks in which Kobe made the first shot have two shots or more, use the variable “shot.num” in the dataset to calculate \[P(\text{shot 2 = H} \, | \, \text{shot 1 = H})=\frac{\# (\text{shot 2 = H} \cap \text{shot 1 = H)}}{ \# (\text{shot 2 = H} \cup \text{shot 2 = M})}=\frac{\# \text{shot 2 = H}}{ \# \text{ shot 2}}\] by identifying those observations corresponding to the second shots (i.e., those with “shot.num==2”).

# We should figure out first what does P(shot 2 = H | shot 1 = H) means. It means what is the probability that Kobe hits in his second shot in a streak given that the first shot was hit. We should aware that when second shot is hit(H), the shot.num should be bigger than 2 (so, 3, 4, or 5). Similarly, first shot will be hit if shot.num is 2,3,4, or 5. Thus the conditional probability we need to get is the same as n(shot.num is 3,4,5)/n(shot.num is 2,3,4,5). 

# Initialize the values
shot.2=0
shot.3=0
shot.4=0
shot.5=0

for(i in 1:nrow(kobe)){
    if(kobe[i,]$shot.num == 1){
        next
    } else{
        if(kobe[i,]$shot.num == 2){
            shot.2 = shot.2 + 1
        }else if(kobe[i,]$shot.num == 3){
            shot.3 = shot.3 + 1
        }else if(kobe[i,]$shot.num == 4){
            shot.4 = shot.4 + 1
        }else{
            shot.5 = shot.5 + 1
        }
    }
}
cond_prob <- round(((shot.3+shot.4+shot.5)/(shot.2+shot.3+shot.4+shot.5)) * 100, 2)
print(paste(cond_prob, "%"))

## [1] "36.84 %"

3. Is there evidence to think that Kobe has hot hands? How reliable is this conclusion? Provide an objective argument to justify your answer.

The calculation we get in the above question, P(shot 2 = H|shot 1 = H) means what is the probability that shot 2 is hit when given shot 1 is hit. In other words, it means tat what is the probability that Kobe success the second hit when we know his first shot was hit in a streak. We got approximately 0.36(36%) for the conditional probability, and we got 43.61% for the probability that Kobe shots hit. We could know that the conditional probability we get is less than the probability that Kobe shots hit(Part 2.a.1). This means Kobe less likely gets a hit(H) in his second shot. Thus this gives reliable conclusion that the conditional probability above supports that hot hand theory is not real.

Part 2.b – Simulating independent shooters

The second alternative is to compare Kobe’s streak lengths to the streak lengths of shooters without hot hands, or in other words to independent shooters. We don’t have any data from shooters we know to have independent shots, but this type 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. To simulate a single shot from an independent shooter with a shooting percentage of 50% we can use the code below (switch the chunk option eval=FALSE to eval=TRUE so that this chunk is evaluated).

outcomes <- c("H", "M")
sample(outcomes, size = 1, replace = TRUE, prob = c(0.5,0.5))

## [1] "H"

Keep in mind that 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.

If you want to learn more about sample or any other function, recall that you can always check out its help file.

?sample

Work on the following problems:

1. Simulate 133 shots from an independent shooter comparable to Kobe and using calc_streak, compute the streak lengths of this independent shooter.

sample_streak <- sample(outcomes, size = 133, replace = TRUE, prob = c(0.5,0.5)) # simulate 133 shots from an independent shooter
a_shooter_streak <- get_streak(sample_streak)

2. Use the R functions table and quantile to compare the streak length distribution of Kobe and that for this independent shooter

# Used the function 'table'
table(kobe_streak)

## kobe_streak
##  0  1  2  3  4 
## 39 24  6  6  1

table(a_shooter_streak)

## a_shooter_streak
##  0  1  2  3  4  6 
## 40 18 10  1  2  2

# Used the function 'quantile'
quantile(kobe_streak$streak)

##   0%  25%  50%  75% 100% 
##    0    0    0    1    4

quantile(a_shooter_streak$streak)

##   0%  25%  50%  75% 100% 
##    0    0    0    1    6

3. Build the R function sim_generation, which takes the number of independent shooters (\(N\)) to simulate, the number of shots taken by each shooter (\(m\)), and the shooting percentage (\(perc\)); and returns a data frame containing the outcomes for all shot taken by a single shooter in each row.

sim_generation <- function(N, m, perc){
    df <- data.frame("Shooters" = c(), "Shots" = c()) # create a data frame
    i = 1 # a counter
    for(i in 1:N){
        shots <- c()
        for(j in 1:m){
            shots <- c(shots, sample(outcomes,size=1, replace=TRUE, prob = c(perc, 1-perc)))
        }
        temp_df <- data.frame("Shooters" = i, "Shots" = shots) # temporary data frame
        df <- rbind(df, temp_df) # update the data frame
    }
    return(df)
}

4. Use the function sim_generation to simulate \(N=500\) shooters taking \(m=133\) shots and \(perc=0.45\).

simulation <- sim_generation(500, 133, 0.45)

5. Get creative, use the results from the previous exercise to evaluate if Kobe has in fact hot hands. For this you may use any R function of your choice, some options are quantile, mean, median, histogram, barplot.

summary(simulation)

##     Shooters        Shots          
##  Min.   :  1.0   Length:66500      
##  1st Qu.:125.8   Class :character  
##  Median :250.5   Mode  :character  
##  Mean   :250.5                     
##  3rd Qu.:375.2                     
##  Max.   :500.0

quantile(a_shooter_streak$streak)

##   0%  25%  50%  75% 100% 
##    0    0    0    1    6

a_shooter_list <- table(a_shooter_streak)
barplot(a_shooter_list)

barplot(a_shooter_streak$streak)

In this problem, our goal was comparing Kobe’s streak lengths and the samples(random shooters)’ streak lengths. We assumed that the samples(random shooters) do NOT have hot hands. As we could see through the graphs above, the distributions of the samples appear similar in shape. Kobe’s biggest streak length was four, and the sample’s biggest streak length is five. Kobe had one streak of length 4 whereas a random shooter had more than one streak of length 4. So this is contradiction, we cannot conclude that Kobe has hot hands.