The Set-up

Hoss has a 1-in-3 chance of making a hole-in-one on six holes, a 1-in-6 chance on another six holes, and a 1-in-20 chance on the remaining (and most difficult) six holes.

Each day, he plays an 18 hole round and is looking at breaking Doc’s record of 6 hole-in-ones on a single round.

Problem 1

What is the probability that Hoss ends the round without making a single hole-in-one?

Let \(X_1\), \(X_2\), and \(X_3\), be the number of hole-in-ones he gets on the easiest, medium-difficulty, and hardest 6 holes. Then \(X_i\) (for \(i = 1,2,3\)) is a binomial random variable with \(n=6\) and \(p_1 = 1/3\), \(p_2 = 1/6\), and \(p_3 = 1/20\). To answer problem 1, we would like to solve \(P(X_1+X_2+X_3 = 0)\), This can be thought of as \(P((X_1=0)\cap (X_2=0)\cap (X_3=0))\). and since these events are independent, we can multiply the three probabilities together.

prod(dbinom(0,6,c(1/3,1/6,1/20)))
## [1] 0.02161258

The answer to this is about 2.16%.

Problem 2

What is the probability that Hoss completes the challenge during his first round and gets seven or more holes-in-one on the 18-hole course? Also, what is the mean number of days it will take Hoss to break Doc’s record?

To answer the first part of this problem, a function will be used to create all groupings of r items into k groups, as we will be interested in how many ways we can get r hole-in-ones in the 3 groups of 6 holes. An additional argument will be added so that it is not possible to get more than 6 hole-in-ones on each group of 6 holes.

Then, a for loop will be used to go through the complement of getting at least 7, which is 0 through 6 holes in one.

all_groups <- function(r, k) {
  grids <- expand.grid(rep(list(0:r), k))
  out <- as.matrix(grids[rowSums(grids) == r & apply(grids, 1, max)<=6, ])
  rownames(out) <- NULL
  out
}

probNot <- 0
for(i in 0:6){
  outcomes <- all_groups(i,3)
  numOutcomes <- nrow(outcomes)
  for(j in 1:numOutcomes){
    x <- outcomes[j,]
    probNot <- probNot + (dbinom(x[1],6,1/3) *
                            dbinom(x[2],6,1/6) *
                            dbinom(x[3],6,1/20))
  }
}

1-probNot
##       Var1 
## 0.02644432

The probability of getting seven or more holes-in-one is about 2.64%. Thinking of the entire round as an bernoulli trial with a successful trial being that of getting at least 7 holes-in-one, we can answer the second part by thinking of this as a negative binomial distribution. The mean number of trials (days, in this case) it will take for Hoss to break Doc’s record is \(1/(.02644432) \approx 37.8\) or about 38 days.

Problem 3

Bonus: What is the probability that the first day Hoss completes the challenge, he will do so in epic fashion, demolishing the record with at least 8 holes-in-one during the round?

Let \(S = X_1+X_2+X_3\). Then, we are after \(P(S\geq 8 | S\geq 7)\), which is \(1 - P(S=7 | S\geq 7)\). We already know that \(P(S\geq 7) = 0.02644432\), but we need \(P(S=7)\).

prob7 <- 0 
outcome7 <- all_groups(7,3)
for(i in 1:nrow(outcome7)){
  x <- outcome7[i,]
  prob7 <- prob7 + (dbinom(x[1],6,1/3) *
                            dbinom(x[2],6,1/6) *
                            dbinom(x[3],6,1/20))
}

1 - prob7/(1-probNot)
##      Var1 
## 0.2446118

As a sanity check, it can be computed directly with a for loop.

probMore8 <- 0
for(i in 8:18){
  outcomeMore8 <- all_groups(i,3)
  for(j in 1:nrow(outcomeMore8)){
    x <- outcomeMore8[j,]
    probMore8 <- probMore8 + (dbinom(x[1],6,1/3) *
                            dbinom(x[2],6,1/6) *
                            dbinom(x[3],6,1/20))
  }
}
probMore8/(1-probNot)
##      Var1 
## 0.2446118

So, for the bonus question, the probability that Hoss completes the challenge in epic fashion is about 24.46%.