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Comment by Sandman | Hacker News Reader

Hmm.. I don't think there's a difference. In two out of three cases, one of the remaining two doors contains a car. If the host opens the one that does, ok, bummer, you lost. But if not, there's a good chance that the other door contains a car. Remember that you always had only one in three chances of choosing the right door, this doesn't change.

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There is a difference. Write a simulation in your favorite language. You'll be forced to decide whether the door with the car ever gets revealed to the contestant. In those cases does the contestant get to switch and win the car? Or did you just throw those cases out in your code? If so, you made an assumption that isn't aways stated clearly in the problem.

But your question was if the host chooses a door at random and reveals a goat. Which means it doesn't matter if he had knowledge or not. You told us the outcome. If you were writing a computer simulation there would be no rgn for the host picking his door. The host would always open the door with the goat, same as if he knew since you already told us he opens the door with the goat.

He does in this single instance, but in order to calculate the probably you need to consider all the possible cases. Does the case where opens the door with the car calculate in? Why or why not? Seriously, write the simulation to calculate the probabilities. It makes the assumptions very obvious.

You told us he opens the door with the goat. All cases consist of the host opening the door with a goat after you have picked.

But it matters whether you have the guarantee or not.

If the host picked at random and happened to get a goat, it is 1/2 vs 1/2. If the host picked a door he knew had a goat, it's 2/3 vs 1/3.

Again, write the simulation. It shouldn't take you 5 minutes.

You do have a guarantee. The premises of #2 above is that the host shows a goat.

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.

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.

WRITE THE SIMULATION

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

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}%"

> 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)

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?

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.

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.

You don't need to because you're asking what is the probability of winning by switching once the goat has been revealed. We want to know:

(# of possible worlds in which we win after switching) / (all possible worlds in which we are shown a goat after the initial pick)

(# of possible worlds in which we win without switching) / (all possible worlds in which we are shown a goat after the initial pick)

in this case 2/3 vs 1/3. We are only considering the cases where a goat has been shown so what happens when a goat isn't shown or what the host's intentions are when a goat is show is irrelevant.

But 1 in 3 times the host opens a door with a car. That changes the probabilities. Or are you assuming this never happens? Why?

Let's suppose we are doing a computer simulation.

1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat

2. We simulate the host randomly choosing one of the remaining doors to open.

3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total'

4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'.

5. The probability of winning after switching is 'wins' / 'total'

It doesn't matter how many times the host shows a car because those trials are discarded.

Did you actually write that simulation? Because once up on a time (many years ago) I followed the same line of reasoning, wrote a simulation to show that I was right, and that simulation showed me I was wrong.

If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.

@dllthomas

I've written the simulation, and you're right. It is 50% when the host chooses randomly!

You didn't write the simulation.

See my cousin comment[0] about this, I think you're getting tripped up with the idea of the host's "knowledge" mattering. That's a distraction here, what's important is the probabilities.

Two game shows, two hosts, both reveal a goat. The random host revealed a goat with 1/3 probability; the canonical host revealed a goat with 1 probability.

We know that both hosts revealed a goat. We also know (for the purposes of this problem) that they had different likelihoods of doing so before the fact.

Does it make intuitive sense to you that the prior probability of an event could affect how you react to it? That's all that's going on here.

[0] https://news.ycombinator.com/edit?id=8664771

The premises of the riddle in both cases is that the host reveals a goat. There is no randomness. The fact that there was a chance the host might show a car doesn't matter. He showed a goat. The probability of the host showing a goat in both cases is 1.

I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.

> The probability of the host showing a goat in both cases is 1.

Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat.

Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?

There are two separate things here. First, the premise of the riddle. Second, the procedure of the game show (as fictionalized within the riddle).

If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.

The question is phrased ad-hoc. We know the host picked a goat in both scenarios. Decision time. Do we switch or no? It didn't ask: "On an infinite timeline where the host might not always pick the goat, should we switch and if so when?" Which I think is what op was trying to say.

No. Even in the case of a single trial, it matters.

If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point.

If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch.

Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.

"If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point."

1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.

Right, to be clear, that's your total chance of walking away from the show with a car. Switching doesn't change it.

Right.

It matters why the host picked the goat. (Or, more precisely, what the odds were of him doing otherwise when he picked). That we're looking at a goat doesn't tell us everything.

Again, write the simulation. Either the switch strategy will win 2/3 of the time or 1/2 of the time, depending on how you write Monty. Only one of those approaches matches the results we saw IRL - Monty never revealed the car (so far as I'm aware) - but either is consistent with many phrasings of the question, and OP correctly specified the other case above.

you keep telling me to write a simulation. It won't matter. the code is the same for both questions since they are asking the same thing.

That's not true, but we'll pick it up under the code jayvanguard supplied.

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