Monty Hall Problem Intuition

I was thinking about the Monty Hall problem and I thought of a possible intuitive explanation:

• You choose a door.
• Monty gives you the option of sticking with your original choice or instead choosing both of the other two doors.
• If you decided to switch (which now becomes an obvious choice), Monty first opens the door with the goat behind it (say, to add to the excitement), and then opens the other door.

My question then is, is this reasoning flawed? Is this even the same problem as before? Because now, choosing to switch from one door to two doors becomes quite obvious, and so does the $2/3^{rd}$ chance of winning the car on switching.

• I think this is one reasonable way of explaining the problem, yes. Never thought of it that way, but I like it! Good job! – 5xum Jun 9 '16 at 8:13
• This is how I first figured out why the odds favor switching, and from this line of reasoning you can recognize a VERY important subtlety: "Monty knows". What is meant by that, is that the reason it is favorable to switch is because you knew monty would open a door with a goat behind it. If Monty didn't know, then in the event that you found yourself with one door open which had a goat, you would actually have 50/50 odds whether or not you switch. (!) – Justin Benfield Jun 9 '16 at 9:29
• This is nice. The only problem is that is a bit difficult to vote an answer since the answer is: 'no, your reasoning is correct' – MiKo Jul 22 '16 at 8:56

I cannot find a flaw in your reasoning.

My own reasoning (if you are interested).

Someone who sticks to his original choice will win if his original choice was correct.

Probability on that: $\frac13$.

Someone who switches will win if his original choice was wrong.

Probability on that: $\frac23$.

• So simple it made me smile – Dleep Jun 10 '16 at 8:18

I believe the best way to intuitively understand the Monty Hall problem is by playing the game with a $100$ doors, $99$ goats and one supercar.

I can choose a door, doing so will give me a probability of $1\%$ of choosing the car. The host then opens $98$ doors, showing $98$ goats. At this point I know that the door I chose either contains the supercar (with a probability of $1\%$) or more likely a goat ($99\%$).

Now I'm given the oppurtinity to switch doors, it's clear that doing so will increase my chance of getting the supercar to $99\%$.

• This explanation never really did it for me. Why would this be equivalent? Wouldn't the equivalent situation be one where you have $99$ goats and $1$ car, then you pick one door, the host opens one other door, and asks me if I want to switch? – 5xum Jun 9 '16 at 8:39
• I mean, I understand the original problem, I just never considered this a good "intuitive" explanation. – 5xum Jun 9 '16 at 8:39
• Well actually, it doesn't matter how many doors he opens, the chance will always increase if you pick another door. I chose to open $98$ doors to get an extreme situation were it becomes clear that you should switch. If he opens one other door, switching to some other door is still better, but the increase in chance isn't as dramatic. – Mathematician 42 Jun 9 '16 at 9:05
• I understand that, but what I am trying to say is that for me, the intuitive situation is not "he opens all other dors", but "he opens one other door", and then, switching being the better option is no more obvious than in the base case. – 5xum Jun 9 '16 at 9:06
• Work backwards then, opening $98$ doors clearly is awesome, $97$ is still awesome, $\dots$ , $50$ better, but not that great, $\dots$ , $1$ still better than nothing. You can work your way down from one extreme situation that is very intuitive to the case you want. For the first so many cases, the intuition is very clear, so why wouldn't this extend down the sequence. It's intuition after all. – Mathematician 42 Jun 9 '16 at 9:12

Yes this is the same problem. Since in the original solution Monty opens one door, and thus that one is bad, hence choosing that one too does not make a difference.

My favourite way of convincing people is to, instead of considering 3 doors, consider 1000 doors. If, after you have choosen 1 door, Monty opens all other doors except one, then obviously you will switch.

Here is the way I convinced myself of the truth that: "the odds of winning the car by switching the original choice is 2/3," given that the host knows the locations of the car and the two goats in advance (stated fact in the problem.)

I changed my thinking to consider that the host is not choosing a DOOR, but rather he is choosing a PRIZE...i.e. he will always pick a goat-door in every case, regardless of its number (assumption that must be inferred because we must assume he won't give the car away by opening its door.) I think much confusion is generated by the wording in the problem (and much of the debates) that the contestant chooses "Door 1" and the host chooses "Door 3."

Then when we look at the three possible arrangements of prizes and doors, and all possible door-choice and prize-choice sequences (see matrices elsewhere), the 2/3 probability becomes clearer.

I think there's a slight difference from the original problem here.

In your version, neither the elimination nor the switching has any effect. In the original version, these processes matter. Here's how:

Let's say you choose Door 1. There's a 1/3rd chance you're right, and a 2/3rd chance you're wrong.

At this point, if you switch Doors (without Monty eliminating a choice for you) you're just moving to another door with the same 1/3rd chance of being right.

Now, Monty opens Door 2. You're then asked to switch or not.

At this point, if you switch, you will win if you were wrong at first (probability 2/3).

If you were right at first (probability 1/3) and you switch, you lose!

Essentially, when Monty removed that empty door, the question "What are the chances that you will win if you switch" became equivalent to "What are the chances that you were wrong at first", which is 2/3rd.