There are 3 doors. Behind 2 doors are apples. Behind the final door is a yummy pastry. Only the vending machine knows where they are, and the doors are black.

You make your selection. The snack machine pleasantly asks you if you might want to change your mind, and shows you an apple behind one
of the other doors. (The machine always reveals an apple, and NEVER shows the pastry. It knows where they are, remember.)

Your friends behind you are screaming for you to stick, now convinced that you've made the right choice.

But, if you really want that pastry (and I certainly do), you should switch to the other unknown door, improving your odds of getting it from .33 to .66. (Now, don’t argue with me, I’ve run the math. But if you persist, see the link first.)

While the in-crowd (mainly halfbakery veterans) will get fat, most will not appreciate the non-intuitive odds, somtimes switching, sometimes not, getting one apple with every pastry on average. But, whenever you score that pastry, girl, it is ever soooo sweet!

You can beat the sniffer dog by representing the doors and the edible goods as an electronic display and dispensing the actual good as chosen using the display.

Whether or not one is better off "switching doors" depends upon the probability that one will be allowed to switch if one has chosen the right door.

To clarify, the contestant chooses a door, then the host opens a door. If the prize is behind the door the host opens, the contestant loses. Otherwise the contestant is given the option to switch. The question is how does the host choose which door to open? Consider three cases.

First of all, imagine that the host selects one of the two remaining doors at random. In this case, there's a 1 in 3 chance that the door the host reveals will have the prize, in which case the player loses. Otherwise, the player is given the chance to switch; at that point, the player's odds are 50:50 whether he switches or not.

Secondly, imagine that the host will always show the prize door if the player hasn't already chosen it. In this case, the probability of the host showing the prize door is 2/3, and the player will only be allowed to switch doors if he started with the right one. In this case, the probability of winning after the switch is zero so the player should stand pat.

Thirdly, imagine that the host will always show an empty door (THE empty door if the player's initial choice was wrong; an arbitrarily-selected door if the player's initial choice was right). In this case, the player will have a 2/3 chance of winning by switching doors, and should do so.

Perhaps if I had Game Show Network, I would analyze the old reruns of "Let's Make a Deal" to see which mode of operation most closely matches what Monty Hall actually did on his show, since such knowledge would be vital to knowing how to proceed.

[supercat]
You make a selection.
The machine then always makes the offer to let you switch.
It always shows you what is behind one of the other 2 doors.
And what it shows you is always an apple.
Once you are shown were that apple is, you have the option of keeping your original (still unopened) door, or switching.
Your odds of eating pastry are 1/3 if you stick, 2/3 if you switch.
It’s strange, but true.

So in other words, you're saying the machine acts like the third kind of host. Fair enough. My point was that on the real show, Monty does not always give the opportunity to switch (unlike your machine), so the question of whether to switch or not is far more complicated.

The problem is always stated as it is here. Whether it was actually a game on the show at all isn't important; it's just an illustration used to demonstrate that human intuition is terrible at statistics. The other variations aren't really interesting.

Building a vending machine around it seems like the idlest of all possible whimsies. Isn't this just a pointless restatement of a widely-discussed problem?

[supercat] Actually I believe that Monty always made the offer before the contestant knew what was behind his curtain. That was necessary, because if he opened the curtain to reveal that the contestant won, that would clue in the next contestant that he didn’t have a winner behind the curtain if Monty didn't open it, but instead made the offer. That was all part of the Let's Make a Deal concept.

The 'Monty Hall' link is inaccurate; it makes one big mistake in that it assumes that probabilities are fixed throughout the game. I've been fighting that one for years. They are not, they are a function of the inputs and outputs; the knowns and unknowns, which change during the show.

Let's say a contestant picks a door. He's got a 1/3 shot of picking the right door.

Now Monty offers a switch, showing another door with an apple.

Now that the apple door is known, the odds drop to 50-50 that the consumer has chosen the right or wrong door.

[RayfordSteele] Think about it this way. You have two sets. The first set is the first door that you picked. The odds that it will have the pastry are 1/3. The second set is the other two doors. The odds for the second set are 2/3. The machine is really offering to let you switch between the first set and the second set. That it has shown you what was behind one door is immaterial, as it can always do that, since it knows where the apples are. That doesn’t change the odds.

pluterday: Let me offer another explanation (assuming the "third style" of host): assume the contestant has chosen door #3. At that point there are three possibilities:

-1- Prize is behind door #1. Host shows door #2.
-2- Prize is behind door #2. Host shows door #1.
-3- Prize is behind door #3. Host shoows door #1 or #2 at random.

The three main possibilities each have probability 1/3. The two sub-outcomes of #3 each have probability 1/6.

Suppose the host opens door #1, showing (surprise surprise) that it's not the winner. At this point, we eliminate possibility #1 and one of the sub-possibilities from #3. We are left with possibility #2 and the other sub-possibility from #3. Since the latter was half as probable as the former, there's at this point a 2/3 chance that possibility #2 was the correct one.

To look at it another way, the host could have chosen #1 because he was forced (by the combination of initial guess and prize placement), or he could have selected it randomly (if the contestant initially guessed the correct prize). In the former case he'd always do door #1; in the latter case he'd only do it half the time.

I haven't seen the game in ages but I thought the door that was shown wasn't dependent upon which door was chosen. If Monty is creating an informational link by always showing a non-prize door, then yeah, the probabilities become skewed.

So now what I want to know is, if I wear a chicken costume, will the odds of the machine picking me out of the crowd of contestants increase?

I was thinking about this last night and independantly reached the same conclusions as ray (nods). Being given the choice to switch after the door is opened effectively resets the counter. You now have a choice between two doors, one of which has a prize. 50/50. The open door is now irrelevant. The flaw in the logic is to allow that the previous choice affects the outcome now. That is like saying that a coin *must* come up heads because last time it came up tails.

Unless supercat's premise is true, and the host actually does allow the previous choice to affect the outcome. In which case it is not a simple statistical problem.