OT - 4 Door Monty Hall Problem

[Ganchrow]

From another forum (I've edited slightly for clarity):

There are four doors. Two are empty. Two have prizes. A host is present who knows what is behind each door. You do not.

First, you choose a door.

Two of the remaining doors must be now opened - one by you, the other by the host. It can be in either order, so long as an empty door and a prize are both revealed. This will leave two doors unrevealed - an empty door and a prize. Your chosen door is one of them.

To win the prize should you stick or swap? Or are the odds 50/50?

Rules:

You can choose who opens the first door.

If the host is to open first door, you can choose to have him reveal either an empty door or a prize - he will follow your instruction. The game proceeds if the door you open contains the opposite of what the host revealed. If the door you open shows the same as what the host has picked, everything is known and the game restarts from the beginning.

If you open the first door, the host will always show the opposite of whatever you have revealed.

After you've made your initial pick, there are four possible scenarios that would require a decision to be made:

1. You reveal an empty door, host reveals a prize. Stick? Swap? Doesn't matter?
2. You reveal a prize, host reveals an empty door. Stick? Swap? Doesn't matter?
3. Host reveals an empty door, you reveal a prize. Stick? Swap? Doesn't matter?
4. Host reveals a prize, you reveal an empty door. Stick? Swap? Doesn't matter?

What's the generalized optimal strategy?

[maxpower79]

These confuse the hell out of me, but I'll try.

If you choose to open a door before the host, and it is empty, the door you selected to begin with now has a 2/3 chance of containing a prize. You should stick with it. If you choose to open a door before the host, and it contains a prize, the door you selected to begin with now has a 1/3 chance of containing a prize. You should switch doors after the host opens an empty door, since the remaining door has a 2/3 chance of containing a prize.
If you command the host to open a door first, regardless of what type of door you order him to open, and then you open a door of a different type, the remaining two doors each have a 1/2 chance of containing a prize. It doesn't matter which door you take. If you open a door of the same type, the game resets.
It seems like the paradox is that when the host opens a door, you don't gain any information that you can use. If you pick first, the host's door is "predetermined" and doesn't affect your knowledge of the probabilities about the remaining doors. If the host picks first, you have a chance at gaining useful information, but if you pick a door that tells you something new, the game ends before you can make use of it.
The optimal strategy is to open a door first, keep your original door if the door you open is empty, and swap doors if the door you open contains a prize.

I think that's right. Anyone have a better explanation?

[Wheell]

You should win two thirds of the time no matter how you play it. If you want to open a door first then remember there is a 2/3 chance that whatever you open is the opposite of what you picked in your original choice. If you reveal a prize, swap, if you reveal an empty room, stick.

If you would like the host to open first and you want it to be empty then you are planning to swap later. If your original choice was an empty room (50% shot) then you will reveal a prize and then swap for a prize. If your first choice was an empty room (50%) then you have a 50% shot of forcing a redo and a 50% shot of choosing a prize and swapping into an empty room. Thus, 50% of the time you win, 25% you lose, and 25% you waste time.

If you want the host to reveal a prize, it is the mirror opposite of the options above, and thus just remember to stick.

I would rather just save the time, choose first, and let that guide me on whether to stick or swap without worrying about a replay.

[Ganchrow]

You should win two thirds of the time no matter how you play it. If you want to open a door first then remember there is a 2/3 chance that whatever you open is the opposite of what you picked in your original choice. If you reveal a prize, swap, if you reveal an empty room, stick.

If you would like the host to open first and you want it to be empty then you are planning to swap later. If your original choice was an empty room (50% shot) then you will reveal a prize and then swap for a prize. If your first choice was an empty room (50%) then you have a 50% shot of forcing a redo and a 50% shot of choosing a prize and swapping into an empty room. Thus, 50% of the time you win, 25% you lose, and 25% you waste time.

If you want the host to reveal a prize, it is the mirror opposite of the options above, and thus just remember to stick.

I would rather just save the time, choose first, and let that guide me on whether to stick or swap without worrying about a replay.

Yep. You got it.

So in general, regardless of who picks first, then (assuming you don't redo):

if host reveals an empty then switch;

if host reveals a prize then stick.

In both cases you'd win 2/3 of the time that you don't redo.

[Wheell]

A long time ago I had to solve a similar problem in High School. A lot of game shows are really public IQ tests.

[Cyclone]

This has been covered at another website http://www.straightdope.com/classics/a3_189.html
They use three doors in their example; I don't know if that changes things.

[Ganchrow]

This has been covered at another website http://www.straightdope.com/classics/a3_189.html
They use three doors in their example; I don't know if that changes things.

The straight dope column demonstrates the traditional (3-door) "Monty Hall problem."

The 4-door problem just adds an extra complication.