Please helpe me understand variable change.

October 13, 2008 1:14 PM

Please help me understand the concept of variable change as it relates to statistics. (Provoked by the movie 21.)

I watched the movie "21" last night and am having a tough time agreeing with what was said at the beginning of the movie. The professor (Kevin Spacey) calls it the game show host problem. He says that there are three doors - behind one is a car and behind the other two are goats. The contestant picks door A. At this point, the contestant has a 0.33 chance of winning a car (and a 0.66 chance of getting stuck with a goat). The host reveals that there is a goat behind door C. He then gives the contestant a chance to change his decision. Now, in the movie, they say that the contestant should switch to door B because there is a 0.66 chance that there is a car there (as opposed to the unchanged 0.33 chance in sticking with door A). I understand how they reached that conclusion, but I disagree with it. When the host reveals that there is a goat behind door C, wouldn't that 'extra' 0.33 chance be divided between A and B equally rather than just all going to door B?

Sorry if I've confused you. Basically I think that, after the host reveals door C, there is a 50/50 chance of being right so there is no benefit to switching to door B. Please prove me wrong so I can stop wondering about this.

posted by siclik to education (21 comments total)

This is called the monty hall problem, clear explanation can be had all over
posted by a robot made out of meat at 1:21 PM on October 13, 2008

It's about the number of choices you originally had, not the remaining number of choices. Essentially boils down to choosing 1 out of 3, or 1 out of 2.
posted by SaintCynr at 1:31 PM on October 13, 2008

We have discussed this extensively.
posted by mr_roboto at 1:33 PM on October 13, 2008

Here's a You Tube video which I found very helpful in explaining the problem.
posted by Rewind at 1:34 PM on October 13, 2008 [1 favorite]

Yeah, this is one of the all-time classics, on a par with the airplane conveyor for its ability to stump otherwise rational people.
posted by ghost of a past number at 1:36 PM on October 13, 2008

Easy way to visualize it:

Imagine that instead of three doors, there are a million doors, with 1 car and 999,999 goats.

You pick a door at random. I now open 999,998 doors and show you that they're all have goats. Which one will you choose -- the random, one-in-a-million door you initially chose? Or the other door, which was NOT one of the 999,998 that definitely had goats?
posted by Cool Papa Bell at 1:37 PM on October 13, 2008 [1 favorite]

@ a robot made out of meat

Thanks for that - the section titled "Why the probability is not 1/2" was actually helpful ;)

My problem with the answer was (as stated on wiki) that the past should be ignored when determining probability, but then I forgot that the contestant's first choice DOES influence which door the host reveals (so the past is relevant).

I'll read up on some of these links and see if it clears it up.

@ Cool Papa Bell

Thank you! That does help a lot. I guess I am still stuck on the fact that, when it comes down to it, you have to ignore past events and look at it as a new problem. There are two doors left, one has a car and one has a goat. I guess I don't understand why past events are even considered.
posted by siclik at 1:43 PM on October 13, 2008

Ex-Mefite Ethereal Bligh created www.montyhallproblem.com/ precisely to answer this question.
posted by matthewr at 1:43 PM on October 13, 2008

Basically I think that, after the host reveals door C, there is a 50/50 chance of being right so there is no benefit to switching to door B. Please prove me wrong so I can stop wondering about this.

Here is a different way to think about it. Before anything happens with anything, there are 9 scenarios. All of them are equally likely at the start.

Scenario 1) You pick A, and the car is in A
Scenario 2) You pick A, and the car is in B
Scenario 3) You pick A, and the car is in C
Scenario 4) You pick B, and the car is in A
Scenario 5) You pick B, and the car is in B
Scenario 6) You pick B, and the car is in C
Scenario 7) You pick C, and the car is in A
Scenario 8) You pick C, and the car is in B
Scenario 9) You pick C, and the car is in C

In 3 of the scenarios (1, 5, and 9) you have the car from the start. In 6 of the scenarios (2, 3, 4, 6, 7, and 8) you have the goat at the start. No matter which scenario you are in, the host can open a door to reveal a goat. Opening the door provides you with no new information because the host can ALWAYS show you a goat. It only makes sense to hold onto you original door in 3 out of the 9 scenarios, 1/3 of the time.
posted by 23skidoo at 1:49 PM on October 13, 2008 [1 favorite]

There are two doors left, one has a car and one has a goat. I guess I don't understand why past events are even considered.

I think your trouble is not the past, but that you're not recognizing the true nature of the current situation.

It's not "two doors left, one with car, one with goat." That's not a true observation.

It's "you chose one of three doors, and of the remaining two, I showed you which one doesn't have the prize."

Again ... make it a million doors. Which would you rather have?

* One in a million chance? You almost certainly chose a goat there, right? Otherwise, that'd be like winning the lottery, right?

* The door that's not 999,998 goats? That'd be like knowing nearly all the numbers in the lottery before you picked the final one. Better than that, actually.
posted by Cool Papa Bell at 1:56 PM on October 13, 2008 [1 favorite]

@ Cool Papa Bell

I think you're right - I'm not making a true observation. But here's my problem - when you're eliminating those other 999,998 doors, I'm confident that the last door is a goat too (because I think behind my door there's a car). You make it seem like there is a 1/999,999, but I think it's 0/999,999 (because the car is behind my door).
posted by siclik at 2:05 PM on October 13, 2008

There are two doors left, one has a car and one has a goat. I guess I don't understand why past events are even considered.

Cool Papa Bell has it right. The past matters because you had a known probability of getting it right to begin with, and Monty revealing a goat gives you more information about the one that he didn't reveal. The information and original probability don't seem to matter with 3 doors, but it becomes more obvious with more doors.

Another similar situation: Let's say there are 10 cards labeled 1-10, and you and a friend are randomly drawing for highest card. You draw a 6, and your friend draws one and keeps it a secret. You you have 5 / 9 chance of winning, because he can have any card other than 6 and 1-5 would give him a losing card. If you take a peak at some of the remaining cards and they turn out to be 1, 3, 4 and 5, you now only have a 1 / 5 chance of winning, even though you haven't changed you or your friend's cards. In fact, if you look at all of the remaining cards, your probability of winning will jump to 100% or fall to 0%.

I'm confident that the last door is a goat too (because I think behind my door there's a car)

Are you disputing the fact that you have a 1 in 1,000,000 of picking the car in your first guess? The Monty Hall problem is not about where you "think" the car is, it's about objectively measuring how likely the car is to be behind a given door. You don't have any objective reason to think you have more than a 1 in 1,000,000 chance of having the door with the goat, unless you think you had better odds than that of picking it in the first place.
posted by burnmp3s at 2:17 PM on October 13, 2008

I have been curious about this since I saw 21, thanks for clearing it up
posted by DJWeezy at 2:22 PM on October 13, 2008

Thanks everyone. Reading montyhallproblem.com really helped.

I think burnmp3s hit it home when he suggested I wasn't "objectively measuring how likely the car was to be behind a given door."

In my mind, my first choice was right and I let my emotions (ego) get in the way of rationality. It was my opinion that there was no car in the other 999,999 doors but, rationally, there most likely is (and that's what this discussion is all about). Thank you, burnmp3s, for helping me realize that.

That website and all of your comments have helped tremendously.

Thanks again!
posted by siclik at 2:25 PM on October 13, 2008

because I think behind my door there's a car

Why do you think that? You have no evidence. You made a random choice.

With the other door, you have some evidence -- you know what it is not.

OK, let's look at it another way ... forget about it being a gameshow. There's just a bunch of doors.

Let's say you're going to pick one door at random, and I'm going to automatically get the other 999,999 doors. My odds are far better than yours, right?

* You choice = 1 in a million
* Me = 999,999 in a million

I mean, the odds are astronomically in my favor, right? You'd much rather be me, right?

All right, let's say I start opening doors one by one. And wonder of wonders, miracle of miracles -- every door I'm opening has a goat! No car in sight!

So, I get to the very last one of my doors -- No. 999,998 -- and I stop. "Fuck it," I say. "I'm tired of opening all of my goddam doors! Maybe I should just stop right here and walk away?"

Should I stop? Or am I about to find the car? What are my odds?

They're actually the same as they were before I started.

* You = 1 in a million
* Me = 999,999 in a million

That's why the guy in the movie clip says, "And I'll thank you for giving me the other 33 percent." Because the game show host has done extra work for him.
posted by Cool Papa Bell at 2:26 PM on October 13, 2008

The tricky thing about probability is what you know (knowledge) matters, just as the physical description of the problem matters. burnmp3s example is a good display of this.
posted by chairface at 2:35 PM on October 13, 2008

@ Cool Papa Bell, thanks for explaining that. It was entertaining and actually helped a lot - thanks for being patient and finding such a creative way to explain this to me. You're absolutely right when you say "They're actually the same as they were before I started".

Thanks for proving me wrong and enlightening me.
posted by siclik at 2:42 PM on October 13, 2008

The trick of this problem isn't that he opens 1 door, it's really that he offers you two of his doors, one of which is opened for you. Opening one door is just a ruse to blind you to that fact. Imagine instead that he says "hey, you can have both of my doors in exchange for your door. Done. Ok, now I'll go ahead and open one of your doors for you as well". There really is no difference between him opening the door before or after you make the switch except in presentation.
posted by Green With You at 2:48 PM on October 13, 2008 [2 favorites]

But here's my problem - when you're eliminating those other 999,998 doors, I'm confident that the last door is a goat too (because I think behind my door there's a car). You make it seem like there is a 1/999,999, but I think it's 0/999,999 (because the car is behind my door).

Here's another thought experiment. Imagine a metagameshow with 100 contestants. In front of each contestant are 100 doors: that's 100 contestants each with 100 doors in front of them. Each contestant picks his door. At this point, nothing is revealed.

Now, you are playing a different game. Your game is this: For each contestant, you have to decide whether you think the car is behind the door they picked, or whether the car is behind one of the other 99 doors. Of the 100 contestants, how many of them do you think picked a goat door from the very beginning? 99? 50? 1? Some other number? Remember your answer, and read on.

Now the host of the show reveals 98 goats for each contestant. But each contestant still has whatever they initially chose at this point. How many people out of 100 did you think picked the correct door from the beginning? 99? 50? 1? Some other number? I'd be willing to bet (ha) that you thought that most of the people picked a goat door from the beginning. So what we have now is 100 contestants, most of whom you think are holding onto a goat, being given a chance to switch to the only door left. What do you think most contestants should do: keep their door, or switch?
posted by 23skidoo at 2:49 PM on October 13, 2008 [1 favorite]

@ Green With You

Wow - what a great way to look at it. Thanks for that perspective.

My question's been answered, but here's just a little something to keep your mind active. Say you're out in the city and some hustler invites you to play a street version of this puzzle. Three cards, one is a king (winner), two are jokers (losers). Knowing nothing else, if you partake in this hustler's game and he gives you a chance to switch after revealing a joker, do you? Remember, you know nothing about his "rules". also, he's an honest guy and won't just run off with your money either way ;)
posted by siclik at 3:01 PM on October 13, 2008

@ 23 skidoo

What a great exercise - it's a lot easier to understand when you put emotion aside and look at this problem from such a large scale. That really opened my eyes.

Thanks!
posted by siclik at 3:11 PM on October 13, 2008