Monty Hall Problem... For Newbies

[UKBoy1977]

You sneakily added the highlighted part.

Nothing sneaky at all. You indicated that you found my previous wording unclear so I used your wording instead. It was to make our communication clearer so there is no talking at cross purposes.

And the rest of your post is not relevant. If you disagree, look at the whole scenario I described and don't try to narrow the focus to a subset of possible outcomes. You set up a strawman there. And I resent your claim I wouldn't know about conditional probabilities. But they're not relevant here at all.

No strawman, you have to narrow the focus. Once he has randomly chosen a door with a goat, the scenarios where he randomly got the car are no longer relevant.

My comment about conditional probabilities is an inference based on the fact you don't seem to think conditional probability is relevant when it is entirely what this problem is about.

 

[AbleSugar]

I saw that Mythbusters show (November 23, 2011) I thought I totally understood the MHP.
Now my head hurts.

 

[ddt]

Nothing sneaky at all. You indicated that you found my previous wording unclear so I used your wording instead. It was to make our communication clearer so there is no talking at cross purposes.

Are you deliberately being obtuse? In my post #99, which you now reacted to I referred to, and quoted, the start of our exchange, being my post #43 and your post #77.

No strawman, you have to narrow the focus. Once he has randomly chosen a door with a goat, the scenarios where he randomly got the car are no longer relevant.

My comment about conditional probabilities is an inference based on the fact you don't seem to think conditional probability is relevant when it is entirely what this problem is about.

Yes, it is a strawman. My post #43 lays out the obvious switching strategy where Monty randomly opens a door, which includes cases where he happens to open the door with the car. And my claim there is that your probability of success by switching is still 2/3.

That is what you disagreed with.

And then sneakily narrowed the focus.

Reread those posts. You're treading close to wilfully lying simply about what was written.

 

[azzthom]

I think it might be a little easier to understand from the opposite POV. The contestant is trying to avoid a goat rather than win a car.

After the initial pick, there is a 2/3 chance the contestant picked a goat. When Monty opens the door to reveal a goat, the contestant gets a chance to switch. If the contestant switches, there is only a 1/3 chance that the 'new' door has a goat behind it because the position of one goat is known, no new goats have entered the game and the contestant only had a 1/3 chance of avoiding the goats with pick 1. The chances were that the non-goat door was one of the two, so by revealing one of the goats Monty is telling the contestant which door to switch to.

Yes. I know it's untidy. It's 1:50 am, I've been working all day and I've had a few beers. If someone could tidy my explanation up a bit, I'd be very grateful. Thanks.

 

[bruto]

How does this follow? If you choose one of three doors, the odds are 1 in 3.

If there are only two doors (since one door has been eliminated before you even choose), as per my scenario, it is no different than only having two doors in the first place. 50:50.

Norm

so it would seem, but this is not what really happens. It really does amount to this: If you picked right (33 percent) to begin with, you end up wrong if you change. If you picked wrong (67 percent) to begin with, you end up right if you change. You cannot end up wrong, because your initial wrong choice is one, and the opened door is the other wrong choice. The only result you can get if you change is the right one. The odds for the result are the odds for the initial pick.

The trick is that the host always opens all the wrong doors that are left, and always knows they're not the right one.

Imagine if there were a hundred doors instead of three. You pick one (one to 99 chance). The host then opens all the 98 remaining wrong doors. Your scenario would mean that doing this suddenly makes your initial choice one of 50/50. Of course not. Your initial choice was one in a hundred. If you abandon that choice, you abandon only that one in a hundred and choose, in essence, ANY of the remaining 99. It just happens that only one is hidden.

 

[jt512]

Clearly, many votes are from people who have previously seen the solution.

 

[Ray Brady]

Are the prizes really a major cost component of a game show production? I would GUESS not. Anybody know?

Games shows that offer merchandise as prizes don't pay for them at all. Those prizes are donated by the manufacturers as a kind of product placement. That's why you never see cases where there's two "zonk" prizes on the actual "Let's Make a Deal".

On the actual game show, the three doors game isn't played the way it is in the puzzle. Instead, you have two contestants, each of whom gets to pick a door. They know that behind one of the doors is a jackpot prize, usually a car. The host then opens one of the doors to reveal a mid-range prize, something like a washer/dryer. He then attempts to persuade the contestants to trade the sure thing of a mid-level prize for a chance at the jackpot prize. Only occasionally would the end result be them trading down to a zonk.

But the real conundrum is, why isn't this the Wayne Brady problem? Get out of the 20th century, people!

 

[fromdownunder]

so it would seem, but this is not what really happens. It really does amount to this: If you picked right (33 percent) to begin with, you end up wrong if you change. If you picked wrong (67 percent) to begin with, you end up right if you change. You cannot end up wrong, because your initial wrong choice is one, and the opened door is the other wrong choice. The only result you can get if you change is the right one. The odds for the result are the odds for the initial pick.

The trick is that the host always opens all the wrong doors that are left, and always knows they're not the right one.

Imagine if there were a hundred doors instead of three. You pick one (one to 99 chance). The host then opens all the 98 remaining wrong doors. Your scenario would mean that doing this suddenly makes your initial choice one of 50/50. Of course not. Your initial choice was one in a hundred. If you abandon that choice, you abandon only that one in a hundred and choose, in essence, ANY of the remaining 99. It just happens that only one is hidden.

Once again somebody who did not read my post(s), which said that one goat door had already been opened before you get to choose. You only have two doors to choose from, from the beginning. In these circumstances it is 50/50.

Norm

 

[ehcks]

Let's not forget, though, that the show also had The Mystery Box.

A goat's a goat, but a mystery box could be anything. It could even be a goat!

 

[bruto]

Once again somebody who did not read my post(s), which said that one goat door had already been opened before you get to choose. You only have two doors to choose from, from the beginning. In these circumstances it is 50/50.

Norm

Sorry about that. Quick reading. Of course if there are only two doors to choose from then the choice is one in two. If a goat door is opened before you choose, then it is not part of the game.

 

[Manopolus]

As I understand it (explained by Derren Brown) You start with a car and two goats (1:3 car - 2:3 goat) you pick one, let's say door 1 and are left holding your choice (1:3) while MH opens door 3 and shows a goat. If you maintain your original choice, door 1 (1:3) your odds are still as they were from the outset. However, if you swap, you are now picking from a 1:2 or fifty - fifty option, thereby improving your chances of winning the car.

I think!

LOL, it takes a lot of silliness to believe this. As soon as you know what is behind one door, it becomes a 50/50 chance, regardless of whether you switch your choice or not. There are only two doors remaining, a goat behind one and a car behind the other. The door that was opened ceases to be a factor as soon as it is eliminated (opened), and has nothing to do with your choices.

Yes, your chances are improved, but through no action of your own. It doesn't matter what you do, the statistics are independent of your choice.

 

[Drs_Res]

This is the explanation that gelled it in my mind.


http://youtu.be/cphYs1bCeDs

 

[Manopolus]

This is the explanation that gelled it in my mind.


http://youtu.be/cphYs1bCeDs

The problem is that the first choice (1/3) ceases to matter once you are given a second opportunity with a 1/2 chance. This analysis ignores the time/revelation factor. Just to insure I'm correct, I'm actually going to devise a field test, but I'm fairly sure that it's a perfect 50/50 at the time of the second choice, because the opportunity for a second choice completely negates the first in reality.

Your chances are a coin flip (50/50) from the beginning, as far as I can tell... not 1/3, not 2/3, but 1/2. The mistake is in thinking your first choice really makes any difference whatsoever to the final outcome. In any case, a field test should be rather simple... I'll post the results of one later. Admittedly, I am not a statistician, but I am quite capable of simulating the basic operation involved 100 or so times and checking the results.

Another thought: Lets take away the generic "goat" and name one "Fred" and the other "Nancy." Now use the same method to try to find "Nancy" instead of "car" and you'll find that your chances of getting "Nancy" are exactly the same as getting the car using the method described (assuming "Fred" is eliminated). The same goes for Fred if Nancy is eliminated. Thus, it appears to be 50/50 the moment you are given a second choice.

...unless you'd suggest the opposite strategy in order to try and find Fred than you would the car? I'm pretty sure it wouldn't make a difference, anyway. Each goat would have a 25% chance of getting picked (at the beginning, not after one is eliminated) no matter what you do, if I'm right.

So, the moral is that not all goats are alike, no?

 

[jt512]

The problem is that the first choice (1/3) ceases to matter once you are given a second opportunity with a 1/2 chance. This analysis ignores the time/revelation factor. Just to insure I'm correct, I'm actually going to devise a field test, but I'm fairly sure that it's a perfect 50/50 at the time of the second choice, because the opportunity for a second choice completely negates the first in reality.

I was you, once. Be prepared to be humbled.

 

[Modified]

Your chances are a coin flip (50/50) from the beginning, as far as I can tell... not 1/3, not 2/3, but 1/2. The mistake is in thinking your first choice really makes any difference whatsoever to the final outcome. In any case, a field test should be rather simple... I'll post the results of one later. Admittedly, I am not a statistician, but I am quite capable of simulating the basic operation involved 100 or so times and checking the results.

Suppose you decide ahead of time to always stick with your original guess. According to you, you can choose one of three doors and somehow have a 1/2 chance of getting the car. Does that make sense?

 

[Manopolus]

I was you, once. Be prepared to be humbled.

Yep, I can see the mechanics of it now. I don't even need to play it out to see the results, now that I've drawn up the method.

Here's what I did:

Instead of using hidden objects, I decided to randomize the choice... 3 boxes: one with car, one with "Fred" and the last with "Nancy" (my names for the goats). A 6-sided die cast the first choice (1/4="car" 2/5="Fred" 3/6="Nancy"). As for the choice of the door to open, I decided to do what I think it is likely they did for the show: Always open door #3 unless the choice is #3, in which case open #2.

Results:

If 1or4 then "Fred"
If 2or5 then "car"
If 3or6 then "car"

Nancy actually has a 0% chance of getting selected by this method, oddly enough (dependent upon my specific door opening procedure).

Sticking with your first choice would lead to essentially the same thing as not having a second choice to begin with (1/4=car, 2/5=Fred, 3/6=Nancy).

Well, at least I learned something.

 

[Soapy Sam]

Having just made a very similar error in this thread- http://www.internationalskeptics.com/forums/showthread.php?postid=9806911#post9806911 -my thanks to Modified for patiently pointing out where I was going wrong- I'm of the opinion that all the talk of probabilities is (probably) adding to the confusion which hits most of us until we reframe the MH problem in our heads.

First , lets clear any unexplained confusion.
Monty always knows which door the car is behind.
So he is careful never to open that door .

He always opens a door on a goat. Always.

From the participant's POV for her first guess there are not 4 possible scenarios, or five or six. There are three, two of which are functionally identical.

These are

1. She picks the car .
2. She picks goat 1
3. She picks goat 2
Monty's response to all three is to open a door on a goat. The only difference from his POV is that in case 1 he can pick from 2 goats and in cases 2 and 3, he can pick only one goat- the one the participant did not pick.

In case 1, she picked the car, so if she sticks with her first choice, she wins the car.
In cases 2 and 3, she picked a goat, so if she sticks, she gets a goat.

At no point is there a need for probabilistic analysis. The situation is astonishingly simple. Yet most of us fail to see this and start hunting complications based on prior conclusions about chance, very much as I did in the other thread.


ETA.
We do the same over complication with the wine and water mixing problem.
I have seen a high school science teacher tie himself in knots over this. His son, aged about 12, solved it instantly by simply realising that anything not in one container has to be in the other.
http://en.wikipedia.org/wiki/Wine/water_mixing_problem

 

[jt512]

From the participant's POV for her first guess there are not 4 possible scenarios, or five or six. There are three, two of which are functionally identical.

These are

1. She picks the car .
2. She picks goat 1
3. She picks goat 2
Monty's response to all three is to open a door on a goat. The only difference from his POV is that in case 1 he can pick from 2 goats and in cases 2 and 3, he can pick only one goat- the one the participant did not pick.

In case 1, she picked the car, so if she sticks with her first choice, she wins the car.
In cases 2 and 3, she picked a goat, so if she sticks, she gets a goat.

At no point is there a need for probabilistic analysis.

Easy for you to say, since you don't provide a solution, much less a method for arriving at one.

 

[Soapy Sam]

Easy for you to say, since you don't provide a solution, much less a method for arriving at one.

A solution to what?

I am trying to make it clear that there is no problem here.

You can choose a car or one of two goats. You can then choose to stick or swap. If you chose a goat and swap you win, if you chose a car and swap you lose.
2 out of every 3 swap choices gets you a car. (Dammit, I just typed "goat").
What further analysis do you feel is required?

 

[fromdownunder]

2 out of every 3 swap choices gets you a car.

That sums up the whole MH "problem" with beautiful simplicity.

Norm