The Monty Hall Problem : Why does the answer to the Monty Hall problem differ from the "Monty Fall" problem?

This happens because, when you receive new information, you have to consider not just what you've been told but also which circumstances lead you to being told it. It's kind of like the difference (hopefully no flame wars start here) between seeing political analysis on PolitiFact versus on CNN or Fox News. (This is an oversimplification, but the point is that CNN and Fox are more likely to give you certain pieces of information than others, but PolitiFact will [hopefully] not discriminate. Sorry if that doesn't float your boat.) Put yourself in the Monty Hall problem for a second. You pick door #1, and the host opens door #3, with a goat behind it. If the car is behind door #2, then Monty's only choice was to open door #3. But if you had originally picked the car, he could've opened either door #2 or door #3, each with probability 1/2.[1] That is, you are more likely to see door #3 get opened and reveal a goat in worlds where the car is behind door #2 than in worlds where the car is behind door #1. This is the same logic that goes behind the http://en.wikipedia.org/wiki/Principle_of_restricted_choice in contract bridge; you're more likely to see someone make a given choice (in this case, about which card to play) if they have fewer choices to make. This is the real reason switching in Monty Hall provides an advantage; the events you see are more likely to occur if switching nets you the car. This is not the case in Monty Fall. In effect, there are two binary choices that the host makes: open your door or not open your door, and open the door with the car or not open the door with the car, each of which is split into one choice with probabilitity 2/3 and one with probability 1/3. Call these events Opens Yours (OY), Opens Other (OO), Opens Car (OC), Opens Goat (OG). OY/OC has probability 1/9, OY/OG has probability 2/9, OO/OC has probability 2/9, OO/OG has probability 4/9. You happen to have landed in OO/OG. But the same reasoning as above doesn't apply, because Monty wasn't more likely to open door #3 when door #2 contained the car; before, it was his only choice, but now it was one of three equally likely ones. You just happen not to have seen him pick the other two. Since Monty Fall's opening of door #3 was independent of which door the car was initially behind, you can't use it to reason about which door the car was initially behind. You've only got an even chance of winning now. To continue with the bridge analogy, it's as if you know the Jack and Queen of Spades are out and one of your opponents shuffles his hand and randomly happens to throw out the Jack. Seeing the Jack isn't less likely to occur in worlds where your opponent also has the Queen, unlike before. Where did the "missing" probability go? It went into the other possible, but unrealized, situations! There was probability 1/3 that Monty would open the door with the car and you would win with probability 1 by just switching to that door, for a total of a 1/3 chance of winning. There was probability 2/9 Monty would open your door and reveal a goat, and you would have had a 1/2 chance of guessing the right door, for a total of a 1/9 chance of winning. And there was a 4/9 chance Monty would open a different door with a goat and then a 1/2 chance that you would guess the correct remaining door, for a total of a 2/9 chance of winning. All these winning chances sum to 2/3, so if given the choice to play Monty Hall or Monty Fall you shouldn't have any preference for one or the other. It just so happens that you win with probability 2/3 in all cases for Monty Hall, but different cases yield different probabilities of winning for Monty Fall. This result is, in fact, more general. It's a simple application of Bayes' theorem to show that, as long as Monty Hall always opens a door (other than the one you picked) with a goat and always offers you the choice to switch, switching wins with probability 2/3. I haven't shown it yet myself, but I strongly suspect your chances can't be worse than 2/3 if Monty also sometimes opens the door with the car behind it (though given that we're talking about intuitions about probability...perhaps!), though the fully general case (Monty is allowed to assign probabilities to opening any door at all, including the one you originally chose, and possibly with different probabilities depending on which goat you chose) is more laborious to work out. [1] I'm assuming the standard Monty Hall formulation, in which Monty chooses which door to open uniformly at random whenever he has the choice.

Let's say you run a simulation of the Monty Fall problem 3 million times, each run independent of the last.   You'll expect to pick car first 1 million times. Monty will always reveal a goat during these one million runs, so clearly you shouldn't switch.   Conversely, you expect to pick goat 2 million times. The difference now is that Monty reveals the last two doors at random. So during the (expected) 2 million times you picked goat first, Monty will reveal car (an expected) 1 million times, and goat (an expected) 1 million times.   In total, we expect Monty to reveal goat 2 million times. However, in one million of those two million simulations, an expectation of 1/2, you'll have picked car first. Therefore there is no advantage to switching.

The Monty Fall scenario doesn't give better odds for switching because the host doesn't know where it http://is.So when there are two doors left, they are indeed both equally likely of having the car.

Probabilities are expected outcome when same experiment is repeated man times. If Monty Fall is one such instance when accidentally opened door happens to be empty, you should still switch since incremental information is still the same as Monty Hall and just reason and source of information changed which has no bearing on optimum strategy. If accidentally opened door (not including chosen door) happens to be with prize, you should of course switch to opened one if you can, if not you are losing anyway. If falling Monty accidentally opens your chosen door which happens to be empty then you can randomly switch one of two unopened doors for 50:50 chances of winning.

In the Monty Hall problem, you choose a door, and then the host (who knows what's behind the doors) opens one you didn't choose. By switching, you win if you originally chose a goat (probability 2/3) and lose is you chose a car (probability 1/3). You can't use this strategy in Monty Fall. You can still pretend like you made an initial choice, but the host need not respect that; he may very well open that same door you chose, making the situation incomparable to Monty Hall. Once you've made your initial choice, you may think that its probability must remain 1/3. While that's true in Monty Hall, that is not necessarily the case when survivor bias comes into play. That is, future events can imply past probabilities. Here's an example: You flip a fair coin and it comes up heads. Conditional on that observation, the probability that it came up heads that specific time is not 1/2, but 1. The coin could have equally likely landed tails, but that would not be consistent with our observations. That would put us in a different universe. But in this universe, the probability of heads was 1. In Monty Fall, the fact that the host opened a different door from the one we "chose" puts us in a different universe from the one we would be in if he opened the chosen door. In the latter case (which happens with probability 1/3), the probability our door contained the car is zero, since we observed a goat in our door. In the former case (which happens with probability 2/3), let p be the probability our door contained the car. Since 1/3 is the overall probability our door contained the car before the host's slip, we have \frac{1}{3} = \frac{1}{3} (0) + \frac{2}{3} p, so p = 1/2.