21 and Variable Change? No, No, No

21 was a movie based on a book about the MIT Blackjack Club, a group of students who were trained by a professor to count cards at Blackjack tables.

It’s a true story.  They made lots of money until they were caught.

The movie has a scene in which Professor Kevin (shown to the left) figures out that a student of his is a mathematical genius and therefore worthy of a spot on his team.

Something about that scene has been bothering me since I saw it.

I started to look up the answer, but I thought it might be interesting if I wrote what I was thinking before I looked it up.

I believe the scene where the professor demonstrates “variable change” in probabilities is complete crap.

Here’s what happens:

The professor says to the class to pretend they’re on a game show and the host presents them with three doors.  The good prize is behind one of them and there are goats behind the others.

He says that the host asks you to make a choice.

So you choose door 1.

Then the host says to make it interesting, he’s going to open door #3 — it’s a goat.

Then the host asks if you want to change your mind.

The student and hero of the story says, “YES.”

The professor asks why.

The student says, “To account for variable change.”

The idea is that by opening the 3rd door, that the host changed everything and now it needs to be figured out again.

I believe this is complete bullshit unless you are going to take into account why the host opened the 3rd door.

If it’s probability, then the chance on the first try is 33% each door.  On the second try, it’s still 50-50.

If you want to consider something like, “Why did he open that door?  Does he know I have it?  Does he want me to change?  Maybe I should stick?  Wait, he wants me to think I should stick, so I should change.”

Psychology aside… The location of the good prize DID NOT CHANGE.  There was no variable change, so the position is static, and unknown, and it’s behind one of two doors… FLIP A COIN!!!

** OK, NOW I AM SEARCHING THE INTERNET FOR THE ANSWER **

Ok, here is the answer from Yahoo Answers.

The key is we need to understand that the host knows where the car is. Ben picks door number 1, so at that point he has a 33.3% of being correct (so odds are against him because he has a 66.6% chance of being wrong). The host knows whether he is right or wrong, but he reveals one of the doors that was definately wrong (door 3).

Now it does seem like regardless to what the movie said there is a 50% chance of him getting it correct now.

But remember, before door 3 was revealed there was a 66.6% (in favor) that the car was in door 2 or 3. The host HAD to pick a wrong door out of 2 and 3, and since it is MOST LIKELY in either door 2 or 3 (before 3 was revealed), by eliminating door number 3 the chances are greater that is in door number 2 because ORIGINALLY the chances were better (66.6%) that it was in door 2 or 3.

If you still don’t understand, look at it this way:

Bens pick – Door 1 (33.3%)
Other pick – Door 2 and 3 (66.6%)
Now say your a bystander, are you going to bet that Ben is right? or that its in either door 2 and 3?
Your going to go with the door 2 and 3 because chances are better. The host CANT REVEAL Ben’s pick obviously, and the host CANT REVEAL the car, so he shows you that door number 3 doesn’t have the car. So theres still a 66.6% chance that Ben’s choice is wrong.

If the host didn’t know what was behind the doors and randomly picked number 3 and it happened to be a goat, then yes he would have a 50% of being right. Totally different story.

This is still complete CRAP.  This is exactly what was explained in the movie, but they’re just wrong.

It makes no difference that the host knows whether the choice was right or wrong.

True: There was a 66% chance that his choice was WRONG.  This is because there are two other choices.
True: The host knows where the good prize is, so he exposes the goat.
I know MATHEMATICALLY the student should change because if there is a 66% chance that the prize is behind one of the two other doors, and the host opens one, then there is a 66% chance the prize is behind the remaining other door.  
What they are saying does make sense, but it doesn’t MOVE the good prize to that other spot.  The prize was behind one of three doors and now there are two doors.  
It’s so easy to see.  
They include this crap to confound people.
This is a perfect example of why scientist types are incapable of understanding the world the way it really works.
Here’s another example:
A scientist visits a farmer and says that he wants to marry his daughter.  The farmer says, sure, but do this first:  Stand 50 feet from my daughter.  Move half the distance to her and stop.  Then move half the remaining distance to her and stop.  Then repeat.
The scientist says, “But I’ll never get to her!!!”
An engineer visits the farmer and takes the same test: “I’ll get close enough for practical purposes!!”
Bazinga, you scientific manchild!!
🙂
I love you, scientist!!

** UPDATE HERE IS A LONG DISSERTATION ON WHY THEY SAY I’M WRONG ON WIKIPEDIA: http://en.wikipedia.org/wiki/Monty_hall_problem

I still don’t believe it.  YES.  It makes sense mathematically.  I must be missing something here.  I understand the host is going to know where the car is.  I know the host is going to reveal a goat.

Does the host ALWAYS give the choice to switch?  If so, then it doesn’t change ANYTHING.

Oh wait… It’s starting to sink in.

Wait for it…

Wait for it…

I’ve got to think about it backwards.

Ok, fine.

I was getting ready to write a computer program to prove they’re wrong, but it’s going to prove they’re right over a number of tries.

When picking randomly one out of three choices, there is a 2 out of 3 chance you are wrong.  To the prize is most likely behind one of the other two.  Then they eliminate one of them, so it’s most likely behind that remaining one… you should switch.  

33% of random choices would hit the car on the first guess.  Switching would cause them to lose the car.

66% of the time the car is behind the other door.  Switching would cause then to win the car.

So if you run trials in your head, you see that 66% of the time you win by switching.

You should switch.

I’m sorry,  Mr. Scientist.

I’m going to steal your girlfriend!!


Comments

10 responses to “21 and Variable Change? No, No, No”

  1. >What if in the computer program suggested a door >always has to be picked after the first choice (no >matter what lies behind it)?
    I should have said "what if the computer program didn't always have to pick after first choice?", but then that's not the question, so yeah, I get it now too.

  2. Here’s the real question.

    The first contestant, Bob, switched his choice (cause he saw the movie the night before). Monty is thinking “Smart man!”. The second contestant, Jane, was in a sound proof room, has no idea what happened beforehand and is brought out. Monty asks her to pick between the two remaining and she flips a coin to help her decide. Its heads, she picks door 1.

    Monty snickers, and declares “Bob has a 66% chance of being right, Jane has a 33% chance of being right!”.

    What are the chances that Monty gets punched by the woman?

    Seriously, the accepted answer to this depends on Bob knowing what Monty knows. That is, the knowledge that Monty knows where the car is and will always show a door with a goat if he picked the other goat, and will not pick a door if he happened to pick a door with a car.

    What if in the computer program suggested a door always has to be picked after the first choice (no matter what lies behind it)?

    The question, as originally stated, is wrong.

    The presumptions are only given in the answers. But of course if you believe Monty was only interested in protecting the house, as in keeping with the theme of the movie, then of course these assumptions are smart ones.

    However, I always thought Monty was rooting for the contestants, though I do remember him getting punched by a woman dressed as a clown once.

  3. The odds are still 66% that it's the door the first contestant did not pick, but he doesn't know that… so it's 50%.

  4. The 2nd contestant was in a sound proof room. Came out after Monty opened the first door. Has no idea what happened beforehand. In fact, doesn't even know there is a door 3 (just to drive the point). The person uses a coin toss (just to drive home the point farther). What's the odds for that 2nd contestant?

  5. If you're asking whether the other person can choose your door or the other remaining door, he should choose the other remaining door… just like you should.

  6. Uh no, you are brainwashed…it happens when you think about this a long time. Here's the question, after the host shows the door with the goat, he also asks another contestant to pick one out of the remaining two. Whats the odds for the new player?

  7. "I was getting ready to write a computer program to prove they're wrong, but it's going to prove they're right over a number of tries."

    This made me laugh because this is exactly how I figured it out as well. I began typing up a small c++ program (because I'm masochistic like that) to prove that it must surely be a 50/50 proposition. But the very act of attempting to represent the problem algorithmically caused me to realize my error.

  8. You don't, but I think to take into account what you think the host wants you to think would be to fly in the face of the odds. At the time of the first pick, your choice is most likely wrong; and then when the host eliminates one of the choices, your first choice is still the same amount of likeliness of being wrong… and now there's only one other choice, so it's most likely right. But I see what you're saying — perhaps the host doesn't always eliminate a wrong answer… maybe they're just doing it because you got lucky.

  9. Anonymous Avatar
    Anonymous

    Hmmm interestng. I understand your views. I have a question tho how can you be so sure that the host isn't doing reverse psychology when he mentions the switch?

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