The Monty Hall Problem

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Moderator note: This was split off from another thread so as not to derail the original. Plus it's pretty interesting and deserving of it's own place.

In the Monty Hall problem, there is a 1/3 chance that the car is behind your curtain and a 2/3 chance that it is behind one of the other two. When he opens a curtain with a goat behind it, there is still a 2/3 chance that it is behind one of the other two curtains, it's just that now you know that that it is not behind one specific curtain of the other two.

You have a 1/3 chance that it is behind your curtain and there is no chance that it is behind the curtain that Monty Hall showed you. Thus, there is a 2/3 chance that it is behind the third curtain.

Logical fallacy. You've fallen for one of the oldest tricks in the book. There is an explanation on YouTube about the Monty Hall Problem that purports to explain it, but it's all smoke and mirrors. It intentionally confuses with erroneous information.

Follow this in your head closely:

Remember that Monty already knows where the goats are. He's going to open one door with a goat behind it. We already know that right? Now, just remember that and I'll get back to it.

Now picture the three doors in your mind. You face the choice of picking one door out of three, right? That's a one in three chance that you may have picked the car. Good so far.

Now Monty opens one of the other two doors, and he ALWAYS opens the door with the goat.

But here's where people get lost. Now he asks if you want to keep the door you chose or choose the remaining closed door. Now think about this very carefully. Are your chances still 1 in 3?

NO! they aren't, because you're about to make a SECOND choice, and now you're dealing with a completely NEW problem, NOT the original problem!

In your mind now, envision the door Monty just opened, erased from the picture. (You're not going to pick THAT door anymore are you? Of course not! So it's no longer part of the problem!).

Now go back to the point I originally made. We already know, before we've picked ANY doors, that Monty will open one of them and there WILL be a goat behind it. Catching on yet?

Your chances are 50/50 and have been all along!

The only reason they throw the first choice in there is for the added drama.

Wrong.

Your chances were 1 in 3 at the beginning. After seeing what was behind the one curtain, they are still 1 in 3 that the curtain you originally chose was correct. That Monty Hall shows you another curtain does not make any difference in those odds. But if you change to the other curtain that was not opened, you improve your odds to 2 in 3.

It would seem so, yes, but the point is that what you're chances are at the beginning is totally irrelevant as to the final outcome. It doesn't matter that they are 1 in 3 at the beginning, because he WILL always eliminate one of the doors. Your chances of winning would be the same if he just opened one of the doors before you made any choices, so the door he opens is really never part of the problem at all.

Remember, it is only the second choice that matters, not the first. In the end, no matter what, there will ALWAYS be only two doors to choose from. That is 50/50 no matter now you cut it.

The third door (the one he always opens), is a red herring.

Hey guess what? I've got to go clean up the egg on my face!

It appears eric76 is correct. This is a really interesting paradox for sure.

Go figure I take the time to write this and then when I go to post, you've just locked the thread and already realized your mistake.

I'll post it here anyway...

This is where you go wrong slightly. It is NOT a completely new problem. A new problem, yes, but it still relates to the original problem. This is because your choice in the first problem directly affects the situation in the second problem. Not for a simple mathematical reason, but because you have the knowledge of which door you picked in the first problem.

There is only a 1/3 chance that you will pick the right door in the beginning. In the case that this happens, you should pick the same door in the second round as you did in the first round.

However, there is a 2/3 chance that, in the first round, you will pick one of the wrong doors. In this case, you should pick the door in the second round that you didn't pick in the first round in order to win.

So, 1/3 of the time you should pick the same door, and 2/3 of the time you should change doors...not 50/50.

What you are saying would be correct if the person choosing in the 2nd round had no knowledge of the choice made in the first round. Then the chance of winning for them is 50/50, as you say, because the two doors are identical to them. But because you have knowledge of the choice made in the first round, it cannot be regarded as a completely separate problem.

In this case, the red herring is not the first choice as you claim. In fact, the red herring is in the 2nd choice being presented as a separate choice between two doors. It's presented as a 50/50 scenario, which distracts you from considering the effects of your first choice on the given situation.

You know? I don't feel so bad knowing this guy was convinced of exactly the same thing for quite a while too.

http://en.wikipedia.org/wiki/Paul_Erd%C5%91s

Well at least it's still here, just in a different place now.

Great explanation, but still having a little trouble following some of the logical points. I think there are some gaps in there. Not lack of logic, but gaps in the explanation.

(You didn't really think you were going to get off that easy did you? )

BTW, I read that computer simulations proved this, so I'm convinced it's true now. The statistics produced by a simulation can't lie.

The logic is escaping me though and that bugs me so I'm going to give you a hard time until it stops bugging me!




Okay. Following so far...



First part, fine. Lost me with the second sentence. Why should I switch just because the first choice had a 2/3 chance of being wrong?

Let's say, for example, that I had chosen door one, and he opens door two. Now there are only two doors open. Explain (please) how the 2/3 chance of being wrong the first time relates to the odds now.





What knowledge do I have about the first round? I know my chances of being wrong were 2/3 but I'm not yet seeing how that knowledge actually changes the odds now that there are only two doors.

If you have the means to run it, here is a small C++ program that will simulate the problem. Feel free to try it out:

For what it's worth, I have personally known a couple of very capable mathematicians who had to think about it for quite a while before they saw it correct. One of them, my best friend for more than 30 years, with an Erdős number of 2, loves all kinds of mathematical puzzles but this one really threw him for a loop.

Hmm, I don't know much about probability to be honest. Here's my perspective, you choose a door first and it could be the prize or the gag. This defines what door the game show host opens. He will always open the one that isn't the car assuming that you didn't pick the car in the first place 33.32% chance. So there is a 66.67% chance that he will end up narrowing it down to one choice for you, and a 33.32% chance that he won't narrow it down to one choice for you. So you might as well pick the 66.67% chance that you will pick the correct choice. Your original choice sorta defines what the game show host will choose, well, you know, it influences it. So it is just one problem.

I get it, and it sort of makes sense, except for one thing. If you chose the car, he could open either of the other doors.

Waaaaaait a minute....

I think you've got something here. And by "something," I mean something that might actually jump my brain into finally "getting" this dumb thing.

lol. Well the point is that if you chose the car and switched, then you will lose, and thus have a 33.32% chance of losing. If you chose a goat and switched, then you win, thus you have a 66.67% chance of winning. If you choose a goat, he will choose the other goat.
If you don't switch, you will have a 33.32% chance either way.

No, that would be 100%. Don't forget, we're talking about knowledge changing the odds. Here you've revealed that I chose the car then switched. I have that knowledge, therefore I know with 100% certainty that I am going to lose.

Sorry, late at night, probably could have worded that better. In this case, we're assuming you picked one of the wrong doors. To make this simpler, let's assume that in the first round you pick door number 1. And we can assume the prize is behind door 2. In this event, the host has no choice. He HAS to open door number 3 and reveal that goat. This leaves you with two doors. The door you picked, door number one, has a goat. The door you didn't pick, door number two, has the prize. So, in this case, you should switch doors.

Since you pick the wrong door originally in 2 out of 3 cases, this exact scenario plays out 2/3 of the time (assigning numbers like I did can be done arbitrarily, and does not affect the actual outcomes no matter which number you assign your choice to be, or which number you assign the prize to be). So, 2/3 of the time, you should switch doors.


The knowledge you have is that you know which door was chosen in the first round. The confusing part is you do not know anything else about the door you chose. However, you know more about the other two doors - specifically, you are told what is behind one of the other doors.


This might be an easier way to look at it. At the beginning, you are presented with 3 doors. Clearly, there is a 1 in 3 chance that the door you pick will have the prize, right?

Nothing that follows changes those odds. The chance of making the right choice at the beginning can not be affected by anything that comes afterwards. Since there is only a 1 in 3 chance that you made the right choice in the beginning, if you do not ever make another choice, the chance of winning remains 1 in 3.

Since the total probability has to be 1 the chance of winning by switching doors has to be 1 - 1/3 = 2/3.

(Don't worry, I never expect to get off easy. Besides, I enjoy these types of problems )

Well, the player didn't know that... you have a 33.32% of picking the car first. If you plan to switch, you will lose. You have a 66.67% chance of picking a gag. If you plan to switch you will win. So if you switch, you have a 66.67% chance of winning. If you don't switch, you have a 33.32% chance of winning. My sister's psychology teacher said that it's literally impossible for the human mind to intuitively grasp the concept. It took her a couple hours to convince me when she told me about the Monty's Hall problem.