(Sections 6.5 - 6.7)
What is the chance that they meet?
This app will help you understand the situation.
The random process is:
From 10 to 11am, watch for Anna and Raj enter/leave the coffeeshop.
Modeling involves getting Anna and Raj’s arrival times (in minutes after 10am):
Anna and Raj will connect if their arrival times are no more than 10 minutes apart.
That is, they connect if and only if
\[-10 < anna - raj < 10,\]
which is the same as
\[\vert\, anna - raj \,\vert < 10.\]
Try this a few times:
runif() can generate many values. Let’s use this fact to repeat eight times.
meetupSim <- function(reps = 10000, table = FALSE,
seed = NULL) {
if (!is.null(seed)) {
set.seed(seed)
}
anna <- runif(reps, 0, 60)
raj <- runif(reps, 0, 60)
connect <- (abs(anna - raj) < 10)
if ( table ) {
cat("Here is a table of the results:\n\n")
print(table(connect))
cat("\n")
}
cat(
"The proportion of times they met was ",
mean(connect), ".\n", sep = ""
)
}meetupSim2 <- function(reps = 10000, table = FALSE,
seed = NULL) {
if (!is.null(seed)) {
set.seed(seed)
}
# set up vector to hold results
connect <- logical(reps)
# loop to repeat random process
for (i in 1:reps) {
# get one arrival time for anna
anna <- runif(1, 0, 60)
# and a time for raj
raj <- runif(1, 0, 60)
they_meet <- abs(anna - raj) < 10
# store in results vector:
connect[i] <- they_meet
}
if ( table ) {
cat("Here is a table of the results:\n\n")
print(table(connect))
cat("\n")
}
cat(
"The proportion of times they met was ",
mean(connect), ".\n", sep = ""
)
}system.time() records time to perform a task.
Note: If you ran this code in R Studio, it would be able to compute the “user” and “system” times, which add up to the total “elapsed” time”.
Avoid looping if you can repeat with vectorization.
(But it doesn’t matter much, if your job is “small”.)
What’s the chance they will vote the correct way?
Try this app to understand the situation.
Set up the scenario in R:
Now imagine that they hear a case.
Add to find the number of correct votes in each case:
Now we can find whether they got each case right or wrong:
Try it for yourself a few times, from scratch:
courtSim <- function(reps,
probs = c(0.95, 0.94, 0.90, 0.90, 0.80),
seed = NULL) {
if (!is.null(seed)) {
set.seed(seed)
}
# get the probabilities
aProb <- probs[1]
bProb <- probs[2]
cProb <- probs[3]
dProb <- probs[4]
eProb <- probs[5]
# simulate decisions of each judge:
a <- sample(
c(0, 1),
size = reps, prob = c(1 - aProb, aProb),
replace = TRUE
) # Judge A
b <- sample(
c(0, 1),
size = reps, prob = c(1 - bProb, bProb),
replace = TRUE
) # Judge B
c <- sample(
c(0, 1),
size = reps, prob = c(1 - cProb, cProb),
replace = TRUE
) # Judge C
d <- sample(
c(0, 1),
size = reps, prob = c(1 - dProb, dProb),
replace = TRUE
) # Judge D
e <- sample(
c(0, 1),
size = reps, prob = c(1 - eProb, eProb),
replace = TRUE
) # Judge E
# count the number of correct votes in each case:
correctVotes <- a + b + c + d + e
# determine whether court decided correctly, in each case:
courtCorrect <- (correctVotes >= 3)
# report estimate of probability
mean(courtCorrect)
}Suppose the judges convince Judge E (the weakest one) to always vote the same as Judge A (the best one) does.
Will this increase the court’s chance of making the correct decision?
courtSim2 <- function(reps = 10000,
probs = c(0.95, 0.94, 0.90, 0.90),
seed = NULL) {
if ( !is.null(seed) ) {
set.seed(seed)
}
# get the probabilities
aProb <- probs[1]
bProb <- probs[2]
cProb <- probs[3]
dProb <- probs[4]
# simulate decisions:
a <- sample(
c(0, 1),
size = reps, prob = c(1 - aProb, aProb),
replace = TRUE
) # Judge A
b <- sample(
c(0, 1),
size = reps, prob = c(1 - bProb, bProb),
replace = TRUE
) # Judge B
c <- sample(
c(0, 1),
size = reps, prob = c(1 - cProb, cProb),
replace = TRUE
) # Judge C
d <- sample(
c(0, 1),
size = reps, prob = c(1 - dProb, dProb),
replace = TRUE
) # Judge D
# count the number of correct votes in each case:
correctVotes <- 2*a + b + c + d
# determine whether court decided correctly, in each case:
courtCorrect <- (correctVotes >= 3)
# report estimate of probability
mean(courtCorrect)
}Compare with using all five judges:
Kicking out the worst judge made the court not as good!
Does this make sense?
Consider this chance process:
Let \(X\) be the number of numbers you have to pick. Let’s estimate its expected value.
This app will help you understand the situation.
Here is a function to simulate one “try” of \(X\):
numberNeeded <- function(verbose = FALSE) {
# set up initial sum:
my_sum <- 0
# set up initial count of numbers selected:
count <- 0
# repeatedly select numbers until sum exceeds 1:
while (my_sum < 1) {
# select a random real number between 0 and 1:
random_real_number <- runif(1)
# add it to the current sum:
my_sum <- my_sum + random_real_number
# increase the count of numbers by 1:
count <- count + 1
# report (if user wants a report):
if (verbose) {
cat("Just now picked ", random_real_number, ".\n", sep = "")
cat(count, " number(s) picked so far.\n", sep = "")
cat("Their sum is ", my_sum, ".\n\n", sep = "")
}
}
count
}Try this:
This estimates the distribution of \(X\), the number needed.
Now we can make a good simulation function:
numberNeededSim <- function(reps = 1000, seed = NULL,
report = FALSE) {
if (!is.null(seed)) {
set.seed(seed)
}
needed <- numeric(reps)
for (i in 1:reps) {
how_many_needed <- numberNeeded()
needed[i] <- how_many_needed
}
# a table of proportions (if the user wants one):
if (report) {
cat("Proportions table of simulated values:\n")
print(prop.table(table(needed)))
}
# make sure to return the estimate of the expected value:
mean(needed)
}It is known that the expected number of numbers needed is:
\[\mathrm{e} \approx 2.71828 \ldots\]