Monty Hall problem explained visually

This is, indeed, the most effective way to explain it, imo.

Without that information, the interlocutor naturally assumes that the offeror opens doors randomly - in which case the fact that he knows what's behind each door is an irrelevant "red herring" and the correct conclusion is different.

In the random-opening case the intuitive conclusion is right - that the agent (offeree)'s decision does not affect the odds. But if the question-poser revealed that the offeror is trying to maneuver the offeree into the result that's less valuable for the offeree, then the supposed mystery of the whole thing evaporates and the correct conclsion is obvious.

Of course it matters.

The host uses his knowledge to AVOID opening the door with the car (and thus ending the game prematurely).

So he essentially gives out this information: "the door with the car is one that I, the host, haven't opened".

The host is not opening just one door.

EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.

EDIT: Leaving my wrong answer, but bvk is correct.

My stab at it: If he doesn't know, then he's essentially another contestant, and his odds are the same 1 in 3 as yours. You've made selections one after the other, but since your selection has yet to be revealed as either right or wrong, you are essentially just picking different options from the same scenario. There are exactly equal chances that you won the car, he flubs and reveals the car, or he reveals a goat.

But if you guess wrong, there's a 50% chance that the host reveals a car by accident.

So the probablities look like:

33%: you guess right, the host reveals a goat 33%: you guess wrong, the host reveals a goat 33%: you guess wrong, the host reveals a car

So, of the situations where the host reveals a goat, the car will be behind your door 50% of the time.

If the host knows and avoids the goat, the probabilities look like this:

33% you were right; the host reveals a goat 33% you were wrong; the host reveals a goat behind door B 33% you were wrong; the host reveals a goat behind door C

So in the situations where the host knows about and avoids the goat, 66% of them have the car behind the other door.

In the case where the host doesn't know about the car, picking the car on your first choice makes it more likely for the host to reveal a goat. That's the difference.

I guess that the odd player can always choose 1. In that case he can either win 3 or lose 2. Making that the favorable choice for the odd player.

Even player could always choose 2 in which case he can either win 4 or lose 3. Making this the favorable choice for the even player. However, if both players play their favorable choice, odd player always wins: 2 + 1 = 3.

I'm not sure if I got this right, but that's my thinking so far...

I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse.

If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%).

If I flip tails, I answer 2Y (X) or 2N (1-X).

So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X)

And combined, Y = 0.25+0.5X

What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)

Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/

  var runs = 1000,
    honest_admissions = 0,
    automatic_admissions = 0;

  function flip() {
    return Math.random() < 0.5 ? 'h' : 't';
  }

  for (var i = 0; i < runs; i++) {
    if (flip() == 't') {
        honest_admissions++;
    } else if (flip() == 'h') {
        automatic_admissions++;
    }
  }
  document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.')

Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.

With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.

Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.

IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.

In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.

So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.

In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.