[jayvanguard]

Yes, _this time_. But how on earth are you evaluating probabilities without considering all the priors and/or doing multiple trials? Is this some new kind of mathematics you've invented?

The hidden assumption is that the odds that he opens a door containing a car is zero. That is what changes the odds in the second step. Without that they don't change.

Seriously. Write the simulation.

[skizm]

Read your second question again. It says that the host shows a goat. Then asks if you would switch. Yes, you would switch.

I understand what you were trying to ask but you worded the second question wrong. You can't tell the listener that the host picks a goat. That defeats the purpose of the "randomness" which is meaningless since we know the host picks a door with a goat.

[jayvanguard]

I found this on the internet and modified it so you can just change the commented out options to run it under the different scenarios.

  car = wins = 0
  many = 100000
  actual = 0

  many.times do
    choice1 = rand(3)
    car = rand(3)
    #host_opts = [0, 1, 2] - [choice1, car]  # host knows what is behind
    host_opts = [0, 1, 2] - [choice1]  # host doesn't know what is behind
    #choice2 = [choice1]  # don't switch
    choice2 = [0, 1, 2] - [choice1, host_opts.first]  # switch
    if host_opts.first == car then
        # Discard? Automatic win? It doesn't actually matter!
        # It only changes the denominator.
    else 
        # According to the puzzle, it is decision time!
        wins += 1 if choice2.first == car 
        actual += 1
    end
  end

puts "#{(wins * 100) / actual}%"

[dllthomas]

I believe it doesn't change the result, but this is technically Monty always picking "the lower (leftmost?) door you didn't pick" not "a random door you didn't pick", right?

Also, "it only changes the denominator" may be misleading, as it does not change the only denominator actually visible in the code. It changes the total numbers of wins and losses, but not the total numbers of wins and losses given that you saw a goat.

[jayvanguard]

Re: first pick, right it doesn't affect things, but here is the code anyways using Ruby Array.sample().

  open_door = host_opts.sample
  choice2 = [0, 1, 2] - [choice1, open_door]  # switch
  if open_door == car then

As for the denominator, good point.

[skizm]

> if host_opts.first == car then

That is false 100% of the time given the parameters that the outcome of the host's choice is a goat (as the question states)

[dllthomas]

Yes, and we don't bump the wins or the number of tries for that case. The question is your odds if you switch in the other branch of the if, and that's what we are calculating.

Edited to add:

I think there may be a different misunderstanding here...

Set aside, for the moment, whether the above implements the originally posed question or the variant. Imagine a gameshow that runs the above code, but the switch/stay code is moved below the "decision time!" line and the decision is made based on a prompt rather than hard coded. When you hit that prompt, "Monty" has already picked his door, and we've checked that it's not a car.

Do you expect your chance to win, on answering "switch" at that prompt, to depend on which line above is commented out?

[dllthomas]

WRITE THE SIMULATION

[jayvanguard]

No doubt. See seriuslyguys post above. He did it and now agrees.