In this post I won't go into that detail, I'm going to assume that you're now familiar with both versions and you know that the Monty version wins 2/3 of the time if you switch, while the Opponent version is still 50/50, even if you don't understand why. In the comments of that original post I described an alternate way to view the problem that, for me at least, makes this whole thing really obvious. I haven't seen it described that way elsewhere, so I want to elaborate in the hopes that this will help some people intuitively see why Monty Hall works, as well as why that isn't initially obvious.

I want to show you the smoke and mirrors at work.

I'm going to describe a number of scenarios. It will not be obvious at first, but every one of these scenarios will be equivalent to the standard Monty Hall problem. I will begin with a version in which it is (probably) super-obvious to you that switching is the right thing to do, and then move gently towards the standard Monty Hall description where things are less obvious but still identical when you look behind the curtain. In all versions, the problem is the same as the standard Monty Hall except in the ways described.

It is my hope that you will see that these problems are all the same, and that they only differ in how well they hide the advantage you receive by switching.

__Version 1. The super-obvious version:__

You pick Door 1, but it is not opened.

Monty gives you the option to instead choose to open both Doors 2 and 3, and if the prize is behind either one of them, you win.

Should you switch?

This one is obvious, or should be. You get to win if the prize is behind either Door 2 or Door 3, giving you a 2/3 chance to win if you switch. Critical concepts that will be repeated in all other versions are:

- This is a 2-for-1 guess, that is why you get a 2/3 chance to win if you switch.

- Another way to look at that is that you've been given the opportunity to make 1 choice that actually reveals a pair of doors.

All future versions will contain these same features; that is why they work.

__Version 2:__

You pick Door 1, but it is not opened.

Monty tells you that, if the prize is behind Door 2, he is going to move it to Door 3. If the prize is behind Door 3, he is going to leave it there.

You are then given the chance to open Door 3 instead of Door 1.

Should you switch?

Yes, absolutely. Door 3 now wins if it had the prize originally, or if the prize was originally behind Door 2. Only if the prize was originally behind Door 1 does switching lose. You have a 2/3 chance, and again we have been given a single choice that essentially reveals a pair of doors.

__Version 3__:

You pick Door 1, but it is not opened.

Monty tells you that you can pick Doors 2 and 3 as a pair, and if the prize is behind either one of them, he will open that door, and you will win.

Should you switch?

Yes. If the prize is behind either door you win, giving you a 2/3 chance by switching. Again the doors are treated as a pair.

__Version 4:__

You pick Door 1, but it is not opened.

Monty tells you that you can pick Doors 2 and 3 as a pair. If you do so and the prize is behind one of them, he will open the other one (the loser) and then let you actually open the winner.

Should you switch?

Yes. If the prize is behind either door you win, giving you a 2/3 chance by switching. Again the doors are treated as a pair. Notice how close we are to the description of the original problem.

__Version 5:__

You pick Door 1, but it is not opened.

Monty tells you that he is going to open a losing door from the pair of doors made by Doors 2 and 3. He will then give you the option to switch to the unopened door.

Should you switch?

Yes. Same as before, these doors are treated as a pair. 2/3 chance to win by switching. Notice how this is identical to the previous version.

__Version 6, the original problem re-framed to not specify which door Monty opened__:

You pick Door 1, but it is not opened.

Monty tells you that he is going open a losing door, but he will never open the door you've already picked. He will then give you the option to switch to the remaining door.

Should you switch?

Yes. 2/3 chance to win by switching. Though he hasn't said so explicitly, Monty is treating Doors 2 and 3 as a pair because he isn't going to open the door you already picked. His choice therefore is out of Doors 2 & 3, and by using his knowledge to eliminate a sure loser from that pair, he makes it so that you have 1 choice that wins if either of these doors has the prize. It is more obvious that they are a pair here, than in the standard Monty version, because we have not specified which of the two is a loser, and which remains.

__Version 7, the original problem:__

You pick Door 1, but it is not opened.

Monty tells you that he is going to open a losing door for you.

Monty opens Door 3, revealing a goat.

You are given the option to switch to Door 2.

Should you switch?

Yes. You have a 2/3 chance to win by switching.

**What makes this trick your mind is that you are likely to frame the problem as "Should I pick Door 1 or Door 2?" when the correct framing is "Should I pick Door 1 or the door that remains from the Door 2/Door 3 pair?"**(which in this particular case happens to be Door 2) As in all other versions, Doors 2 & 3 are treated as a pair. You get to win if the prize was behind either of them. You get a 2-for-1 guess, but if you're thinking of Door 2 as itself and not as representative of the pair, then your brain is likely to experience the disbelief that makes this a veridical paradox. Your choice is not between two doors, but between a single door and a pair of doors. When given the opportunity to switch to a pair of doors, your brain gets it. When given the opportunity to switch to a door that

*represents*a pair of doors, your brain can easily miss the hidden value there.

The Monty version (knowledge based choice) isn't the same as the Opponent version (random choice) because in the Opponent version the opponent is opening a door that might be a winner. If the prize is revealed, you lose. Monty never reveals the prize, he reveals information that gives you the option to make a 2-for-1 guess. In this way, Monty's action is guaranteed to help you, whereas the opponent will cause you to immediately lose 1/3 of the time (the times when he guesses correctly.) Nothing that the opponent does creates a situation where you can win if either of 2 doors has the prize. The opponent does not create a 2-for-1 guess for you.

Does this help make the problem intuitive for anyone else?

As I have said as before in common sense everything should be put in place and all the elements of it has to be used

ReplyDeleteMy thesis

The psychological factor of any contest.

It is known that the host will put ALL hindrances for the contestant to even win that is a good assumption. If the contestant wins the host looses money :) Isnt that logically obvious EXCEPT if the host is benevolent enough that he will give the prize straight away :) As I said in the previous post :)

Ex 1.

The assumption is wrong if you look closely the contestant will not win any prize at all a no win situation exists :)

1. [Monty gives you the option to instead choose to open both Doors 2 and 3, and if the prize is behind either one of them, you win.]

So Door number 1 is eliminated already as a prize winner coz the host will give anything on door no 2 and 3 only. Ok this is the situation if I do not switch

There is no car in number 2 so you hope on number 3 and still there is no car. Even if you chose no 1 and opened it you will still not win the car and why???? Monty promised you the prizes on door 2 and 3 and not in door no. 1.

Now if I choose to switch you do not win any prizes at all except for the goat :)

Ex. 2

Here I will give a different tack I will not switch even with 2/3 advantage coz of my thesis above. But this time I really have a chance to win coz unlike Ex 1 he never promises anything except opening of the doors.

Ex. 3

No again the explanation on number 1 and my thesis will provide the answer almost the same promises are made in number 1

Ex. 4

No again my thesis counts :) Again Monty will not let you win any prize in this one :) See also no. 1

Ex. 5

This is the only example where you can win any prize :) But I will still not switch a 50/50 chance of winning since the loosing door is already opened not 2/3 as alleged. I will STILL not switch unless the benevolent host shows itself and why? Why do he want me to choose door no 3 and persuade me to do so? As in my thesis he do not want me to win any prize :)

Ex. 6

Again the same as no. 5

Ex. 7

Again the same as no 5

All in all Common sense wins all factors has to be put into its proper perspective and should not be used :) Door no 1 is the only viable alternative on winning :)

There is a good alternative to this being used in Hollywood films of increasing the chance of winning

Let 3 cards be on the deck the winner should be the King of hearts. The host will choose one of the loosing card and you can pick any winning or loosing card of the two depending on your pre-picked condition. You will win if you chose the winner card

I will choose this way I will choose the loosing card and tell that to my host and let the unopened card be the winner card. The onus here is this the host does not want you to win any so if the king of hearts card is there or not (mostly it is not there making it another looser card) you have a 50/50 chance of winning :)

In serious contest if the card is not there at all you have 100% chance of winning if you choose the looser card :)

Errata

DeleteAll in all Common sense wins all factors has to be put into its proper perspective and all factors in regards to common sense should be used

All of this has been published in facebook using your blog I hope you would understand my first statement :)

DeleteAs I've explained to you before, this is the standard version of the Monty Hall problem. There is no psychology at work. Monty has no hidden motives. He does the same thing every time - open one losing door. In the various versions, he does what I describe.

DeleteThis is a logic problem, nothing more. The purpose of it is to demonstrate that there are times when "common sense" will lead you astray, and more thorough analysis is required to lead you to truth. In all cases of the Monty version, switching your guess is the right thing to do. If you've come up with a different result than that, then you're mistaken and need to look at the problem more closely.