Step by Step solution about How to resolve the algorithm Monty Hall problem step by step in the Ruby programming language

This Ruby program simulates the classic Monty Hall game show problem thousands of times to determine the probability of winning with each strategy: staying with your original guess or switching to the other unopened door. . Here's a step-by-step breakdown:

Initializer:

n = 10_000: This sets the number of times the game will be played to 10,000. stay = switch = 0: These variables will store the number of wins for each strategy: staying with the original guess and switching to the other door. Game Loop:

The n.times block runs the game n times. For each iteration:

doors = [:goat, :goat, :car].shuffle: Creates an array of three door labels (goat, goat, car) and shuffles them randomly. guess = rand(3): Randomly selects a door (0, 1, or 2) as the initial guess. begin shown = rand(3) end while shown == guess || doors[shown] == :car: Finds a random unopened door (neither the guess nor containing the car) to be shown as a goat. If the guess is the car (doors[guess] == :car), the stay strategy wins, so stay is incremented by 1. Otherwise, the switch strategy wins because the other unopened door must contain the car, so switch is incremented by 1. Output:

Finally, the program prints the winning percentages for both strategies as a percentage of the total games played. Example Output:

Staying wins 33.33% of the time. Switching wins 66.67% of the time. This output demonstrates the well-known result that switching is the better strategy in the Monty Hall problem, winning two-thirds of the time, while staying (not switching) wins only one-third of the time.

n = 10_000                  #number of times to play

stay = switch = 0           #sum of each strategy's wins

n.times do                  #play the game n times
  
  #the doors reveal 2 goats and a car
  doors = [ :goat, :goat, :car ].shuffle
  
  #random guess
  guess = rand(3)
  
  #random door shown, but it is neither the guess nor the car
  begin shown = rand(3) end while shown == guess || doors[shown] == :car
  
  if doors[guess] == :car
    #staying with the initial guess wins if the initial guess is the car
    stay += 1
  else
    #switching guesses wins if the unshown door is the car
    switch += 1
  end
  
end

puts "Staying wins %.2f%% of the time."   % (100.0 * stay   / n)
puts "Switching wins %.2f%% of the time." % (100.0 * switch / n)