Q:

"Bug" in if statement parameters Learn Ruby The Hard Way

On line 7 of http://learnrubythehardway.org/book/ex35.html, there is a comment stating there is a bug.
When this program runs and you choose a number which contains, neither a "0" nor a "1" it will say it's not a number.  How do I get it to detect all numbers?
Here is the code:
def gold_room
  puts "This room is full of gold.  How much do you take?"

  print "> "
  choice = $stdin.gets.chomp

  # this line has a bug, so fix it
  if choice.include?("0") || choice.include?("1")
    how_much = choice.to_i
  else
    dead("Man, learn to type a number.")
  end

  if how_much < 50
    puts "Nice, you're not greedy, you win!"
    exit(0)
  else
    dead("You greedy bastard!")
  end
end

A:

You can use regular expression here. \d means any digit, + means one or more times, =~ is a pattern match operator, so:
if choice =~ /\d+/
  how_much = choice.to_i
else
  dead("Man, learn to type a number.")
end

