One day you decide to go out to buy a
new puzzle book at the massive Honty Mall.
When you enter, however, you are confronted by three doors and a rather
dishevelled amateur philosopher who swiftly talks you into trying out a
variation of an old game show puzzle (you obviously like puzzles, so it was an easy
task).

The puzzle is put to you as briefly yet
comprehensively as possible:

- There are three doors, there is a goat behind two of the doors and behind the third is a car.
- If, at the end of the game, you open the door with the car behind it, you win the car.
- First, you select
doors (not the one door of the Classic Monty Hall Problem).*two* - The philosopher will then open one of the doors you selected, revealing a goat.
- You then have the option to switch from your remaining selected door or stay.
- Before being allowed to open a door, you
provide the likelihood that a switch will win you the car (even if you choose to stay).*must* - The placement of the goats and car is randomised.

Do you switch or stay, and what is the
likelihood of winning from a switch?

Note that we are assuming that you want
a nice new car, rather than a goat!

Original Monty Hall Problem: "You select the one door you want to pick".

ReplyDeleteRephrased, but exactly the same: "You select the two doors you don't want to pick"

Original Monty Hall Problem: "If you change selection (of the one you want to pick), you will have a 2/3 chance of winning"

Rephrased, but exactly the same: "If you change your selection (of the one you don't want to pick), you will have a 2/3 chance of losing."

Addition: The two rephrased sentenced can be re-rephrased to fit your version:

Delete"You select the two doors you want to pick"

"If you change your selection (of the ones you don't want to pick), you will have a 2/3 chance of winning."

Thanks, yes, I did go through a sequence of thinking about "selected doors" and the "unselected door", and saw that it mirrored the classic Monty Hall problem with it single "selected door" and the two "unselected doors". We could imagine that the host tells you to pick a door, any door, as the door that doesn't become yours.

DeleteI fully understand the logic of this and it's very difficult to see just why it is wrong. While I worked the problem out (over a couple of hours), I experienced cognitive dissonance, as mentioned at reddit. Part of that was cycling furiously between this obviously correct answer and the other obviously correct answer (obvious to me, because of where I was coming from).

For some reason, a key feature of this scenario remains hidden to most people (or I am simply wrong and ridiculously confident that I have a valid solution, when it's total bollocks - I do accept that there is a likelihood of this due to good old human psychology).

I'd choose to stay with my door. The likelihood of winning this way is 2/3.

ReplyDeleteYou stay, with 2/3 odds of winning.

ReplyDeleteThere are 3 pairs of doors you can start with, AB, BC, and AC. In any configuration, 2/3 of these options will have one goat and one car in the two doors you selected, and 1/3 option will have 2 goats. In either of the 2/3 options with one goat and one car, if the goat is eliminated and the car remains. In the 3rd of 3 options, a goat is remaining. Therefore, always stay after the goat has been eliminated, and your chance of winning by staying is 2/3.

Can I suggest thinking about it this way ... you are offered two randomly selected doors by the philosopher (this is functionally the same as choosing two doors at random yourself). The philosopher says "Here are two doors, one of the doors will definitely have a goat behind it, the other may or may not have goat behind it. The other door, the unselected door, may or may not have a car behind it."

ReplyDeleteDoes this in anyway make you doubt the 2/3 probability with staying solution?

No, because the probabilities don't depend on the presentation of the problem. If they do, then you're not presenting the same problem.

DeleteThe interesting thing about the Monty Hall problem is that it shows how bad humans are at intuitive reasoning about probability, as evidenced by the experts who got it wrong. However, it's no match for correctly applied high-school techniques. The problem you present, as with the original Monty Hall problem, is trivially solvable with a tree diagram.

I suspect that I am NOT presenting the same problem, because I changed something important (you select two doors rather than one, and the host opens one of your selected doors, rather than the one you didn't select - I am uncertain as to which of these is most critical, I suspect the former, but it could be both in combination). Does it help, or change anything, when you consider that the goats are separate entities and not interchangeable (this is a feature of goats, so I don't think I need to make it explicit - thus I don't think it is a "cheat". I may be wrong on this)? Does it help to consider what the host knows before he or she opens the first door?

Delete> I suspect that I am NOT presenting the same problem

DeleteBut, this is exactly the same problem. Here is a neutral presentation :

* There are three unopened boxes.

* Two contain miniature goats, and one contains miniature car.

* You are asked to put two boxes on the right side of the room and one box on the left side of the room.

* The host then opens one of the two boxes that are on the right side, revealing a miniature goat. (and he will always reveal a goat)

* You can now choose to pick the box on the right side or the box on the left side, at your convenience.

This formulation works perfectly for both situations (Monty Hall and reverse Monty Hall). Which means that the two problems are completely equivalent

New Anonymous (it would help if people at least left an initial I could respond to, or a couple of them, or a nick of some kind),

DeleteThis is a good presentation. I agree with you that the Monty Hall and the Reverse Monty Hall are the same, but then I've been suspecting that quite a few comments ago :)

We could possibly make this scenario even better, in such a way as to eliminate host all together. Say we have two white spheres about the size of billiard balls, inside of which are miniature models of goats (nicely padded so they don't move around). We also have a black sphere of the same size and total weight, inside of which is a miniature model of a car (nicely padded so it doesn't move around). We put the spheres into a barrel like the ones they use for PowerBall and then we put the barrel into a pitch black room and turn it on for as long as we need to to totally mix up the balls (all three of them) and set it to spit out the balls into three hidden slots.

Then we walk into the pitch black room and we select two of the hidden slots at random, and we take out the balls. One of the balls we put into an urn, which we then seal so that we cannot see the ball, and the other we hold in our hand. We then walk out into the light. If the ball in our hand is black, meaning that a car was selected, then we void the scenario and repeat and we keep repeating until we are holding a white ball in our hand.

What is the likelihood that the ball in the urn is also white - and thus contains a goat, making a switch, if it were offered to you a winning option?

This scenario, while apparently more complex, is also exactly the same. It just highlights that some draws simply aren't possible (the ones that were voided because you held a black ball in your hand).

The logic which applies to this applies to the Two Balls, One Urn puzzle which I posted a short while back. I encourage you to read that one, then see what answer you come up when there are 2 million balls in the barrel. If you come up with a likelihood of the other ball in the urn being white of 1/(N-1), then you must also conclude that the likelihood of you holding two white balls (one in your hand, one in the urn) is 1(N-1) = 1/2. And this flows directly on to the Monty Hall Problem.

> What is the likelihood that the ball in the urn is also white - and thus contains a goat, making a switch, if it were offered to you a winning option?

ReplyDeleteIt's 1/2. I have absolutely no problem with that.

> This scenario, while apparently more complex, is also exactly the same.

It's not. This is a common variant of the problem. It has been adressed by so many people over the years that I won't repeat the same thing over and over again. Just acknowledge that you are not the first person to have thought about this.

> The logic which applies to this applies to the Two Balls, One Urn puzzle which I posted a short while back.

Yes absolutely. You are perfectly correct on this puzzle, the logic perfectly applies.

> And this flows directly on to the Monty Hall Problem.

No it doesn't. I don't know what to say. We can make the experiment if you want, and you will see by yourself. But I have the feeling that even an experiment will not convince you ...

I am more than happy to do the experiment, I've encouraged people to do the experiment (via a role play). I have written what amounts to a script to assist the reader to experience a role play (someone is still checking it over, or maybe it's just languishing in her inbox - she might be taking my "sporadically updated" far too seriously).

ReplyDeleteI've actually done the experiment with four individual people now. I might even be forced to video the experiment to show people what I mean and how it works.

Could you do me the great favour of explaining how you would do your experiment to test my scenario? Note that if you are testing something that is not my scenario, I will tell you and try to get you to see the difference.