[iamjdg]

What I don’t get is in the first table, game 3 and game 5 no longer exist after he opens doors 3. The auto can’t be behind door 3, because he revealed there is a goat there. So yes, relative to the original choices, she is correct. But we have been presented with new information, so if you reset the problem based on the new information, don’t we now have a choice between 2 doors, one with a car and one with a goat?

[ColinWright]

The information you have gained is not about the door you've chosen. It doesn't matter what door you choose, the host can always open a door to reveal a goat. So there is no information given about the door you've chosen, so the chances of that door containing the prize remain at 1/3.

But you have been given information about the door he didn't open, because he didn't open it. That's why it's possible for the odds on that door of holding the prize can change.

And yes, we do now have a choice between two doors, one with a goat and one without. The error is in believing that these are equal choices.

I roll a die, and you can choose "1" or "not 1". You have two choices, but they have unequal chances of being correct. Similarly with the doors. Just because there are two choices, they may have different odds.

In the Monty Hall problem, they do.

[indigodaddy]

Odds aren't changing at any point. It's as simple as you were initially sitting in a 1/3 "probability block" (the way I like to think of it)... ; and then you shifted over to a 2/3 "probability block."

When or where are any odds changing?

[ColinWright]

The probability on the two you didn't open is 2/3. At that point you have no information as to which of those two doors holds the prize, so they each have probability of 1/3.

Then the host opens one. The probability of the door opened holding the prize goes to 0. But the probability on the set of two is still 2/3, and so the probability of the door that is both unchosen and unopened goes to 2/3.

[aw3c2]

So this is a psychology puzzle rather than pure math?

[ModernMech]

No. The host could be a robot, and the contestant could be a robot, and the could play the game a million times. The robot contestant who chose to switch would win 2/3 of the time, while the robot contestant who chose to stay would win 1/3 of the time. It's math, not psychology.

[ColinWright]

I'd be interested to know what it is about my reply that makes you think this is a psychology puzzle. In particular, it seems to be people's psychology that prevents them from understanding the mathematics underneath. People seem to assume that if there are two choices then they must be equally likely. That's a psychological thing, although to be honest, I don't understand it.

But the Monty Hall Problem as stated is about the probabilities, not on the psychology. Computing the probabilities is simple math, once you understand the situation. My explanation was to help the reader understand why the two choices given don't have equal probability.

[aw3c2]

Hm, then I do not understand it. "But you have been given information about the door he didn't open" was what made me comment.

If I chose a door before, then something happens that leads to only two doors being left, both those doors have the same probability so I could just choose the same door again.

[ColinWright]

aw3c2> Hm, then I do not understand it. "But you have been given information about the door he didn't open" was what made me comment.

aw3c2> If I chose a door before, then something happens that leads to only two doors being left, both those doors have the same probability so I could just choose the same door again.

The original said this:

CW> The information you have gained is not about the door you've chosen. It doesn't matter what door you choose, the host can always open a door to reveal a goat. So there is no information given about the door you've chosen, so the chances of that door containing the prize remain at 1/3.

CW> But you have been given information about the door he didn't open, because he didn't open it. That's why it's possible for the odds on that door of holding the prize can change.

So let's recap what's going on. There are three doors. For the sake of concreteness let's call them A, B, and C. You choose one of them. For the sake of concreteness let's suppose you choose A.

So now there are two doors, B and C, remaining unchosen by you. Currently those two doors, B and C, each have probability 1/3 of having the prize. The door you chose, door A, has probability 1/3 of holding the prize.

Now the host opens a door, taking care to open a door that does not hold the prize. So the pair {B,C} still has total probability of holding the prize, but you are being shown that one of them certainly does not. This doesn't affect the probability that your chosen door, door A, holds the prize -- the probability that the prize is behind door A is still 1/3.

The pair {B,C} still has total probability 2/3 of holding the prize. You're now given the choice of staying with A, or switching.

Quoting you again, you said:

aw3c2> something happens that leads to only two doors being left, both those doors have the same probability ...

That turns out not to be the case. Just because there are two doors they don't have to have equal probability of holding the prize, and in this case they don't. The probability that your door holds the prize has not changed and is still 1/3. The probability that the door neither chosen by you nor opened by the host holds the prize is now 2/3.

Does that help?

[aw3c2]

Thanks!

[sametmax]

I think the easiest way to grasp this without getting into the math is to make such a counter intuitive problem intuitive again.

Let's change the number of doors !

First, I'll rephrase the original problem ================================================

You have 3 doors, 1 hides a car, all others hide goats.

You pick one door, then __ALL doors you have NOT selected AND that hide goats are revealed, except one__

Here it's only one door because we have a total of 3. But it's one door that is indeed, ALL except one, for this particular situation.

Do the same, but with 100 doors ====================================================

You have 100 doors, 1 has a car, all others have goats.

You pick one, then __ALL doors you have NOT selected AND hide goats are revealed, except one__

Now you get:

- 1 selected, 99 not selected => the car is most probably among the 99 ones.

- Monty reveals 98 goats

- 1 selected, 98 goats, 1 not selected => the car is most probably in the later one.

[iamjdg]

This is a very clear explanation. Thank you. Now I understand.

[alangpierce]

I think the second table is more clear than the first, since it shows that Monty Hall's choice depends on your choice. The main nuance to this problem is that Monty Hall is not picking randomly. There's always some unpicked door with a goat, and he'll pick that one to reveal. So if your original choice is door #1, then when deciding to stay or switch, you're really deciding whether go with door #1 or with the better of doors #2 and #3 (since the worse of the two doors has already been shown to you).

[speakeron]

That's how I got to understand it when I first heard about the problem in the early 90s. (Of course, I also wrote a simulation.)

The problem itself can be rephrased in an even simpler manner where Monty doesn't open any doors at all. Once you've picked a door, he then gives you the option of either staying with that door or changing to both of the other doors (where you get to keep the best prize behind them). Given that formulation, almost everybody would switch.

[erpellan]

Yes, if you selected one of the remaining doors at random you have a 50% chance of winning.

If you always switch to the door you didn't initially choose your odds of winning rise to 2/3.