Baanee’s Version

Problem 6.

It asks for the probability that strategy 1 results in goat and strategy 2 results in car when \(N=5\), for which I wrote the following code:

require(magrittr)
require(dplyr)

# provided game function
game <- function(N){
  if(N<3) stop("Must have at least 3 doors")
  prize <- sample(c(rep("goat",N-1),"car"), N)
  guess <- sample(1:N,1)
  game <- data.frame(door = 1:N, prize = prize, stringsAsFactors = FALSE) %>% 
    mutate(first_guess = case_when(
      door == guess ~ 1
      , TRUE ~ 0
    )) %>% 
    mutate(potential_reveal = case_when(
      first_guess == 1 ~ 0
      , prize == "car" ~ 0
      , TRUE ~ 1
    )) %>% 
    mutate(reveal = 1*(rank(potential_reveal, ties.method = "random") == 3)) %>% 
    mutate(potential_switch = case_when(
      first_guess == 1 ~ 0
      , reveal == 1 ~ 0
      , TRUE ~ 1
    )) %>% 
    mutate(switch = 1*(rank(potential_switch, ties.method = "random") == 3))
  c(game$prize[game$first_guess == 1], game$prize[game$switch == 1])
}

# setting parameters
num_simulations <- 10000
doors <- 5

# initial count of how many results are have both strategies yield "goat"
strat_1_goat_count <- 0
strat_2_car_count <- 0

# loop to run simulation and add onto the goat_and_car_count counter
for (k in 1:num_simulations) {
  result <- game(doors)
  if(result[1] == "goat") {
    if(result[2] == "goat") {
      strat_1_goat_count <- strat_1_goat_count + 1 
    }
  }
  
  if(result[2] == "car") {
    strat_2_car_count <- strat_2_car_count + 1
  }

}

# finding the probability
probability <- goat_and_car_count / num_simulations
strat_1_probability <- strat_1_goat_count / num_simulations
strat_2_probability <- strat_2_car_count / num_simulations

# print probability
cat('The probability of strategy 1 yielding "goat" is:', strat_1_probability)
cat('The probability of strategy 2 yielding "car" is:', strat_2_probability)

The probabilities are \(0.5343\) and \(0.2677\) respectively

Problem 7

We are given the following,

Strat 1/Strat 2 \((N = n)\)	Car	Goat
Car	0	\(\frac{1}n\)
Goat	\(\frac{n-1}{n(n-2)}\)	\(\frac{(n-1)(n-3)}{n(n-2)}\)

Strat 1/Strat 2 \((N = 5)\)	Car	Goat
Car	0	\(\frac{1}4 = 0.20\)
Goat	\(\frac{5-1}{5(5-2)} = 0.267\)	\(\frac{(n-1)(n-3)}{n(n-2)} = 0.533\)

| Strat 1/Strat 2 $(N = n)$ | Car                  | Goat                        |
|---------------------------|----------------------|-----------------------------|
| Car                       | 0                    | $\frac{1}n$                 |
| Goat                      | $\frac{n-1}{n(n-2)}$ | $\frac{(n-1)(n-3)}{n(n-2)}$ |

| Strat 1/Strat 2 $(N = 5)$ | Car                          | Goat                                |
|------------------------|------------------------|------------------------|
| Car                       | 0                            | $\frac{1}4 = 0.20$                  |
| Goat                      | $\frac{5-1}{5(5-2)} = 0.267$ | $\frac{(n-1)(n-3)}{n(n-2)} = 0.533$ |

And using the following code,

goat_goat_count <- 0
goat_car_count <- 0
car_goat_count <- 0
car_car_count <- 0
sim_count = 10000

for (k in 1:sim_count) {
  result <- game(doors)
  if(result[1] == "goat") {
    if(result[2] == "goat") {
      goat_goat_count <- goat_goat_count + 1
    }
    else {
      goat_car_count <- goat_car_count + 1
    }
  }
  else {
    if(result[2] == "goat") {
      car_goat_count <- car_goat_count + 1
    }
    else {
      car_car_count <- car_car_count + 1
    }
  }
}
cat(car_car_count, car_goat_count, goat_car_count, goat_goat_count)
cat(car_car_count/sim_count, car_goat_count/sim_count, goat_car_count/sim_count, goat_goat_count/sim_count)

I get this contingency table from running,

Strat 1/Strat 2 \((N = n)\)	Car	Goat
Car	0	\(0.1972\)
Goat	\(0.2604\)	\(0.5424\)

| Strat 1/Strat 2 $(N = n)$ | Car                  | Goat                        |
|---------------------------|----------------------|-----------------------------|
| Car                       | 0                    | $0.2039$                 |
| Goat

Given that the relative error is \(0\), \(0.014\), \(0.023\), and \(0.018\) respectively, I’d say it fits the proposed solution well.

Problem 8.

More specifically, from Problem 6,

Strategy 1 Absolute Error: \(|0.5343 - 0.533| = 0.0013\) Relative Error \(0.0013/0.533 = 0.002\)

Strategy 2 Absolute Error: \(|0.2677 - 0.267| = 0.0007\) Relative Error \(|0.0007/0.267| = 0.002\)

Mrunal’s Version

Problem 6.

It asks for the probability that strategy 1 results in goat and strategy 2 results in car when \(N=5\), for which I wrote the following code:

require(magrittr)
require(dplyr)

# provided game function
game <- function(N){
  if(N<3) stop("Must have at least 3 doors")
  prize <- sample(c(rep("goat",N-1),"car"), N)
  guess <- sample(1:N,1)
  game <- data.frame(door = 1:N, prize = prize, stringsAsFactors = FALSE) %>% 
    mutate(first_guess = case_when(
      door == guess ~ 1
      , TRUE ~ 0
    )) %>% 
    mutate(potential_reveal = case_when(
      first_guess == 1 ~ 0
      , prize == "car" ~ 0
      , TRUE ~ 1
    )) %>% 
    mutate(reveal = 1*(rank(potential_reveal, ties.method = "random") == 3)) %>% 
    mutate(potential_switch = case_when(
      first_guess == 1 ~ 0
      , reveal == 1 ~ 0
      , TRUE ~ 1
    )) %>% 
    mutate(switch = 1*(rank(potential_switch, ties.method = "random") == 3))
  c(game$prize[game$first_guess == 1], game$prize[game$switch == 1])
}

# setting parameters
num_simulations <- 10000
doors <- 5

# initial count of how many results are have both strategies yield "goat"
strat_1_goat_count <- 0
strat_2_car_count <- 0

# loop to run simulation and add onto the goat_and_car_count counter
for (k in 1:num_simulations) {
  result <- game(doors)
  if(result[1] == "goat") {
    if(result[2] == "goat") {
      strat_1_goat_count <- strat_1_goat_count + 1 
    }
  }
  
  if(result[2] == "car") {
    strat_2_car_count <- strat_2_car_count + 1
  }

}

# finding the probability
probability <- goat_and_car_count / num_simulations
strat_1_probability <- strat_1_goat_count / num_simulations
strat_2_probability <- strat_2_car_count / num_simulations

# print probability
cat('The probability of strategy 1 yielding "goat" is:', strat_1_probability)
cat('The probability of strategy 2 yielding "car" is:', strat_2_probability)

The probabilities are \(0.5281\) and \(0.2683\) respectively

Problem 7

We are given the following,

Strat 1/Strat 2 \((N = n)\)	Car	Goat
Car	0	\(\frac{1}n\)
Goat	\(\frac{n-1}{n(n-2)}\)	\(\frac{(n-1)(n-3)}{n(n-2)}\)

Strat 1/Strat 2 \((N = 5)\)	Car	Goat
Car	0	\(\frac{1}4 = 0.20\)
Goat	\(\frac{5-1}{5(5-2)} = 0.267\)	\(\frac{(n-1)(n-3)}{n(n-2)} = 0.533\)

| Strat 1/Strat 2 $(N = n)$ | Car                  | Goat                        |
|---------------------------|----------------------|-----------------------------|
| Car                       | 0                    | $\frac{1}n$                 |
| Goat                      | $\frac{n-1}{n(n-2)}$ | $\frac{(n-1)(n-3)}{n(n-2)}$ |

| Strat 1/Strat 2 $(N = 5)$ | Car                          | Goat                                |
|------------------------|------------------------|------------------------|
| Car                       | 0                            | $\frac{1}4 = 0.20$                  |
| Goat                      | $\frac{5-1}{5(5-2)} = 0.267$ | $\frac{(n-1)(n-3)}{n(n-2)} = 0.533$ |

And using the following code,

goat_goat_count <- 0
goat_car_count <- 0
car_goat_count <- 0
car_car_count <- 0
sim_count = 10000

for (k in 1:sim_count) {
  result <- game(doors)
  if(result[1] == "goat") {
    if(result[2] == "goat") {
      goat_goat_count <- goat_goat_count + 1
    }
    else {
      goat_car_count <- goat_car_count + 1
    }
  }
  else {
    if(result[2] == "goat") {
      car_goat_count <- car_goat_count + 1
    }
    else {
      car_car_count <- car_car_count + 1
    }
  }
}
cat(car_car_count, car_goat_count, goat_car_count, goat_goat_count)
cat(car_car_count/sim_count, car_goat_count/sim_count, goat_car_count/sim_count, goat_goat_count/sim_count)

I get this contingency table from running,

Strat 1/Strat 2 \((N = n)\)	Car	Goat
Car	0	\(0.2039\)
Goat	\(0.2588\)	\(0.5373\)

| Strat 1/Strat 2 $(N = n)$ | Car                  | Goat                        |
|---------------------------|----------------------|-----------------------------|
| Car                       | 0                    | $0.2039$                 |
| Goat                      | $0.2588$ | $0.5373$ |

Given that the relative error is \(0\), \(0.0195\), \(0.031\), and \(0.008\) respectively, I’d say it fits the proposed solution well.

Problem 8.

More specifically, from Problem 6,

Strategy 1 Absolute Error: \(|0.5281 - 0.533| = 0.005\) Relative Error \(0.005/0.533 = 0.009\)

Strategy 2 Absolute Error: \(|0.2683 - 0.267| = 0.0013\) Relative Error \(|0.0013/0.267| = 0.005\)