Are you as smart as the smartest woman on earth? (A Probability Problem)

[Maddy]

In memory of Monty Hall, who died yesterday, I give you the following:

Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats.

“Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?”

Statistically, which choice gets you the car: keeping your original door, or switching?

Formulate your answer, then click on the link below for the correct response together with a very interesting article about the so-called Monty Hall problem, the smartest woman on earth (who, incidentally, thought IQ tests were unreliable bunk), and the scores of mathematicians who argued (wrongly) that she was a flaming idiot.

http://www.zerohedge.com/news/2017-10-0 ... test-woman

[sweetbthescrivener]

I read about this years ago, got it wrong, and now know the answer, so I will, as important types say, recuse myself.

As a consolation, here is a video of MVS being interviewed by David Letterman and also proving that brains don't necessarily translate into social intelligence.

Because that dummy is running rings around her.


https://www.youtube.com/watch?v=EsGc3jC9yas

[barrett]

Once you understand how this works (or I guess even if you don't) it's easily demonstrated with three playing cards...one card for each door.

[Maddy]

The article didn't provide an explanation that made any sense to me. Nor, apparently, did it make sense to the majority of readers who commented on the article and who continue to insist that when Door 3 is eliminated, the probabilities become 50/50 that the car is behind either of the two remaining doors. I lay awake last night thinking about it, not able to see how the answer could possibly be other than 50/50, and the answer finally came to me (albeit in a way that I've not heard it explained before):

The key to the problem is that Monty alters the probabilities when he chooses which door to open. Look at it this way:

When Monty chooses which door to open, he must follow two rules:
(1) He cannot open Door 1 (because that’s the door the contestant has chosen).
(2) He must choose a door with a goat.

Because there is a 1/3 probability that Door 1 has a goat, there must be a 2/3 probability that there is a goat is behind either Door 2 or Door 3.

When Door 3 is revealed to have a goat behind it, the 2/3 probability that attached to the sum of both Door 2 and Door 3 shifts entirely to Door 2, while the probability that Door 1 is the winner remains at 1/3.

So while it appears that with the elimination of Door 3 the two remaining choices each have a 50 percent probability of being right, that approach fails to account for the fact that Monty’s choice to open Door 3 was not random; he could not open Door 1, and he had to open a door with a goat. Thus his choice to open Door 3, rather than Door 2, altered the remaining probabilities, making it substantially more likely that Door 2 would not reveal a goat.

[dualstow]

I love revisiting the Monty Hall problem, even though it hurts my head to think about it.

[barrett]

dualstow, I am impressed! I can only remember the explanation by going back and reading through it again. There's a little book called Paradox by Jim Al-Khalili that starts off with The Monty Hall Paradox. I finally understood it by using three cards on a cousin of mine who insisted the remaining odds had to be one in two. The pattern quickly becomes pretty convincing. But, again, kudos to you for figuring it out.

[dualstow]

Heh,I don't know if any praise is deserved. But, my wife is into that Jim al-Khalili for his quantum physics stuff if you want to praise her instead. Over my head.

Maddy, your explanation is good.

[Maddy]

If you like this kind of stuff, there's a very neat book out there called "The Drunkard's Walk." I began reading it probably two years ago but got sidetracked with other things, always intending to pick it up again. It examined a large number of similarly counter-intuitive probability-related phenomena and explored the ways in which our minds unconsciously distort the probabilities of certain things. I'll bet it addresses the Monty Hall problem, but I never got far enough to find out.

[barrett]

Dang! I read Maddy's explanation and thought it was yours dualstow. I have to take those kudos away and pass them Maddy's way instead. Just a little rebalance. Selling a little SCV and buying a bit of IAU. Seriously, Maddy, you figure this stuff out when you can't sleep at night?

[eufo]

Brilliant explanation! I've never understood this problem until now.

[Xan]

A way to make it easier to think about is to imagine 100 doors.

You pick one door, leaving 99 hidden. Monty opens 98 of them, revealing a goat in each one, leaving one left unopened. In this case, the correct answer (switching) "feels" more obvious. Of course there's a 99% chance that the prize is in the other door, and only a 1% chance that the prize is in the door you originally selected.

Same thing with only 3 doors, except not as extreme.

[Maddy]

Seriously, Maddy, you figure this stuff out when you can't sleep at night?

Not only that, I got out of bed and went to the computer to write it down while it was still fresh in my head. That's more than a little OCD, wouldn't you say?

I really need some sleep tonight, so I shouldn't even be thinking about this, but the following is bothering me:

Suppose the very same scenario, with three doors, two goats, and one car. The goats and the car are hidden behind the same doors as in the original game. Only this time Monty doesn't open a door. He simply eliminates Door 3 from the game without revealing what was behind it. What is the probability that Door 2 is the winner?

My take: This time the probability that the car is behind Door 2 is 1/3. But wait-- It's the very same set of choices, and the very same prizes are behind the very same doors.

So what is the difference? Only the fact that in the original game we knew more about Door 3 than we know in the present one. In the original game, it was revealed that Door 3 had a goat. Now we don't know what is behind Door 3.

So how objective is this thing called probability if the odds of a particular event occurring wax and wane depending upon the knowledge of the observer? Does this have anything to do with Schrodenger's cat?

Now I'm going to bed.

[Mountaineer]

Maddy, I too really liked your explanation - made much more sense than most I've read. Kudos!!!!

Next time you are thinking of Schrödinger's cat and can't sleep, try this book.
https://www.goodreads.com/book/show/129 ... s-religion

... Mountaineer

[buddtholomew]

I understand the explanation but can't process the following:
Why does removing door 3 from the scenario transfer the probability to door 2 only?
The car could exist in door 1 just as it could exist in door 2.
He didn't tell you a goat exists in door 1, just couldn't open that door.

[Maddy]

That's right. It's totally counter-intuitive. The correct result still doesn't set well with me, and the only way I was able to reason my way through the problem was by knowing in advance what the correct answer was and working backwards. It's interesting that this problem tripped up even the mathematicians and that a bunch of them are STILL holding fast to the 50/50 idea. You could raise a whole helluva lot of money at a charity event by playing this game!

[buddtholomew]

I see it now.
Of the 2/3's, 1/3 has been removed so naturally one should pick door 2. Door 1 is irrelevant since it only represents 1/3 of the original equation.

Fine

[dualstow]

@barrett, actually it's a relief to learn what was behind the unfounded praise. For a while there, I felt like Obama receiving the peace prize.

[dualstow]

I understand the explanation but can't process the following:
Why does removing door 3 from the scenario transfer the probability to door 2 only?
The car could exist in door 1 just as it could exist in door 2.
He didn't tell you a goat exists in door 1, just couldn't open that door.

Think of it this way (I see that Budd got it later, but for those of you still feeling the pull of common sense) -

What were the odds on door 1 being right when it was selected, because that's what counts. Opening door 3 and finding a goat does not take us back in time, thus it cannot alter the odds on door 1. Those odds are locked.

The next choice, on the other hand, is a truly fresh choice between two doors. For me, it's 1/2, 50/50 b/c it's like door 3 doesn't exist. Maddy said 2/3, and that does add up to 1 (100%). Either way, the odds are better than the frozen 1/3 with door 1.

[Libertarian666]

The article didn't provide an explanation that made any sense to me. Nor, apparently, did it make sense to the majority of readers who commented on the article and who continue to insist that when Door 3 is eliminated, the probabilities become 50/50 that the car is behind either of the two remaining doors. I lay awake last night thinking about it, not able to see how the answer could possibly be other than 50/50, and the answer finally came to me (albeit in a way that I've not heard it explained before):

The key to the problem is that Monty alters the probabilities when he chooses which door to open. Look at it this way:

When Monty chooses which door to open, he must follow two rules:
(1) He cannot open Door 1 (because that’s the door the contestant has chosen).
(2) He must choose a door with a goat.

Because there is a 1/3 probability that Door 1 has a goat, there must be a 2/3 probability that there is a goat is behind either Door 2 or Door 3.

When Door 3 is revealed to have a goat behind it, the 2/3 probability that attached to the sum of both Door 2 and Door 3 shifts entirely to Door 2, while the probability that Door 1 is the winner remains at 1/3.

So while it appears that with the elimination of Door 3 the two remaining choices each have a 50 percent probability of being right, that approach fails to account for the fact that Monty’s choice to open Door 3 was not random; he could not open Door 1, and he had to open a door with a goat. Thus his choice to open Door 3, rather than Door 2, altered the remaining probabilities, making it substantially more likely that Door 2 would not reveal a goat.

What if Monty Hall opens the door with the goat only if the contestant has already chosen the correct door?
In that case, the chance of winning by switching is 0%.

[WiseOne]

Ha, I love this puzzle. My Dad used to have us solve conundrums when we were kids and this was a favorite. Second only to the one about the 3 guys in a hotel with the thieving bellhop and the missing dollar.

This is actually a quite simple one, and it's a useful example of conditional probabilities. Tenn got it right:

If you don't switch, your chances of picking the car remains at 1/3 (one car out of three doors).
If you switch, your chances of getting the car are 2/3. Whether you picked a goat or a car, you will get the other item if you switch. You had a 2/3 chance of picking a goat initially, so if you switch you have that same chance (2/3) of getting the car.

If only such logic could apply to airlines...finally coming back to the US tomorrow and there's some ticket number issues with my reservation that I'll have to battle through at the airport. Logic has nothing to do with THAT situation!

[dualstow]

That's right. It's totally counter-intuitive... You could raise a whole helluva lot of money at a charity event by playing this game!

I was thinking about the money raising angle. Maybe I'm being dense, but I can't think of an game that takes advantage of the Monty Hall Paradox and that people would want to play.

How about a game show? We could call it something snappy like Let's Make a Deal.

[Libertarian666]

What if Monty Hall opens the door with the goat only if the contestant has already chosen the correct door?
In that case, the chance of winning by switching is 0%.

The rules of the game are (i) Monty always opens a door and (ii) a goat is behind the door that Monty opens.

I didn't know that it was mandatory for him to open the door.

In fact, I'm pretty sure that was not specified in the original version of the problem.

[Xan]

What if Monty Hall opens the door with the goat only if the contestant has already chosen the correct door?
In that case, the chance of winning by switching is 0%.

The rules of the game are (i) Monty always opens a door and (ii) a goat is behind the door that Monty opens.

I didn't know that it was mandatory for him to open the door.

In fact, I'm pretty sure that was not specified in the original version of the problem.

I believe you're right for the actual game show, but for the intellectual exercise of "The Monty Hall Problem", it is definitely always mandatory.

[Maddy]

That's right. It's totally counter-intuitive... You could raise a whole helluva lot of money at a charity event by playing this game!

I was thinking about the money raising angle. Maybe I'm being dense, but I can't think of an game that takes advantage of the Monty Hall Paradox and that people would want to play.

Well, some of us are easily entertained.

[barrett]

This is actually a quite simple one, and it's a useful example of conditional probabilities. Tenn got it right:

Well this is weird. First, barrett credits dualstow. Then WiseOne credits me.

Again, it was Maddy.

I feel the need to point out that neither dualstow nor I were involved in writing the Beach Boys song "Good Vibrations".

Just to clarify, Tenn, I never claimed you and dualstow wrote "Good Vibrations." My point was merely that you two recorded the best version. Now where is that friggin' goat I was looking for?