San Diego Housing Bubble News and Analysis

I did not see the movie, but there are two possibilities and the answer depends on which one is the case.
1) The host opens one of two remaining doors at random. In this case your logic is correct - odds remain 50/50.
2) The host knows which door has a car behind it, and he intentionally opens the one which does NOT have a car. In this case it's better to switch.

The host knows which door has a car behind it

Then switching doors will change odds to 1 i.e. certainty and not 2/3.

No, because it's possible that the car is behind the door that the player picked initially.

We have a contestant we'll call him 'TheBreeze' aka
'TB'.
TB choses door #1, the probability that TB chose the door with the car is 1/3 and, the probability that it is behind one of the other doors is 2/3. By showing TB one of doors #2 or #3 which does not hide the car but a goat (#2 for instance), the host is giving TB an good idea of what is behind those two doors. The probability is still 2/3 that one of them hides the car, but now you know which of the two it would be: Door 3.

The probability is still only 1/3 that the car is behind Door 1, but 2/3 that it is behind Door 3.

Anyone remember "Lets Make a Deal"?

duplicate deleted

Also known as the Monty Hall problem, an example of conditional probability.
http://en.wikipedia.org/wiki/Monty_Hall_...
http://en.wikipedia.org/wiki/Conditional...

You can literally count the outcomes if you need to and figure out your odds are better if you switch to avoid the mathematical proof.

There are many ways to explain why your probability of winning the car is in fact 66.6%. I find this easiest to understand:

Here is the key: Given the rules, if you always switch your pick, your first choice will always determine if you win or not.

(1) If your 1st choice is a goat, you will always win the car. This is because the host will reveal the other goat so when you change, it will ALWAYS be to the car.

(2) If your 1st choice is the car, you will always lose. This is because the host will reveal a goat so when you change, it will ALWAYS be to the other goat.

Therefore: Since your odds of initially picking a goat are 66.6%, your odds of winning the car is the same - 66.6%.

Not the simplest concept to understand but, it is obviously true.

Yep, 2/3 is correct. We studied this problem in probability.

I like Brian P's description of the problem. Makes it clear why the answer is 2/3's.

Oy vey..

It is assumed that the contestant is making a correct statement by saying "The contestant does decide to change from door number 1 to door #2. He states by changing from original chosen door #1 to door #2 his odds of winning the car increased to 66.7%." He is not. He is showing a complete lack of understanding in probability. The gameshows do this to up the 'excitement' because otherwise the game is played out. It allows them to clock another 5+ minutes on the same stupid choice with everyone else yelling and screaming. What it does tell you is that the host knows where the car is and a good poker player may be able to bluff it out of the game show host by watching the hosts responses to potential choices.

Basically, on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

After the choice, one of the potentials is removed by uncovering the goat. This makes 1 out of 2 possible or 50% any particular choice is right.

Whether the choice is;
1) He has already chosen one and gets to modify his choice
2) There is a new choice between two brand new doors...

It is the same thing. The underlying thing is that revealing what is under one of the doors does not alter what is behind any of the other doors. Also remember that the sum of probabilities on any single choice have to sum to 1 and you do not get to carry over probabilities from a previous run on to a new choice, in particular when a new piece of information is added (revealing of the goat).

I dare the doubters of this to write a program that allows you to pick one of three.. reveals one that it knows is not it and asks you to choose to keep your choice or switch. Track the frequency that the original choice is right versus switching was the right choice. The revealed choice in the first selection is not counted because it was not made by the person doing the selection.

And for a final nail:
Why would a gameshow give you a choice that could suddenly improve your odds dramatically and so easily, particularly since they have to foot the bill?

omg...

on the initial selection: 1 correct out of 3, making the odds 33% that any particular choice is right.

Exactly. So why do you think that the chance to have a car behind that door suddenly goes up from 33% to 50% when game show host reveals the goat?

Let's call the initial choice door #1.

If you do this game 100 times, in 33 cases the car will be behind door #1, in 67 cases it will be behind door #2 or #3.

If game show host always shows you the goat and never misses, he knows which door has the goat behind it. You will have 33 cases where the car is behind #1 and 67 cases when the car is behind the only other door that hasn't been opened yet. Chance to win when switching: 67/100.

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

The whole problem is that you added info to the system which altered the game. It also causes the probability calcs to be completely reshuffled. You can't carry over probability calcs when info gets added to the system. Example:

I have 10 doors. The prize is behind one: The odds of winning are 10%

I then open the door with the prize and then close it.
Are the odds of winning 10% per door, or 100% on the door I opened and then closed with 0% everywhere else. The information resets the calcs.

Ucodegen,

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

John

I agree with you however the games shows want people to win so that people will watch. The merchandise is donated by the manufacturers.

Yes but not easily. It is more important to have people on the edge of their seat than to make winning too easy which would be boring.

I can see where the 66% is coming from, the problem is the information being added to the system in the middle and between sets for selections. After the added information, are they independent runs?

If game show host opens the door at random: in 33 cases he will reveal the car and all will be over; in 33 cases the car will be behind #1; and in 33 cases it will be behind the other unopened door.

The example in wikipedia has the gameshow host always picking the goat and they come up with the 66.67%. What I have to convince myself is that the added information did not make them independent runs.

Now that I am finally waking up(dark chocolate kicking in).. they may not be independent runs because the entrance criteria to the second choice gives a 33.33% chance of being on the car and a 66.66% chance of being on the goat.

That does add another question which would take some gameshow observing. I notice that sometimes on these reveals, they throw in a completely different choice. If the probability on swap results of a 66% chance, would they give you the option to swap if you have picked the door with the car and a completely different set of choices if you haven't?

I consider them independent runs so I say 50/50. Your're right too they want to make it look hard but showing one of the doors that does not have the prize effectively makes it 50/50 so they know that over time there will be enough winna's to keep people tuned in etc.

John

I consider them independent runs so I say 50/50. Your're right too they want to make it look hard but showing one of the doors effectively makes it 50/50 so they know that over time there will be enough winna's to keep people tuned it etc.

I'm not convinced on their independence. I am coming up with 33% on a car and 66% chance on a goat going into the reveal. While the next choice is binary when independent, the entrance condition alters the probability.

I have to agree with Brian P. If there are 10 doors your chance of picking the car is only 10%. The host opens 8 doors that have goats. Your best bet is to switch doors, since the probability of having picked the car on your first choice is still only 10%.

I'm assuming the host knows which door has the car behind it.

I don't know why some of you people are arguing against the 2/3 answer. It IS the correct answer, whether you understand the logic or not.

People who study probabilities for a living have said this is the answer. All you amateur naysayers should just capitulate and move on...:-)

I don't know why some of you people are arguing against the 2/3 answer. It IS the correct answer, whether you understand the logic or not.

People who study probabilities for a living have said this is the answer. All you amateur naysayers should just capitulate and move on...:-)

Because the experts all said that real estate always goes up.. and guess what....

So it gets argued until it gets understood. We don't accept expert opinion as gospel here. We try to dissect it.

Comparing Math/Probability with Real Estate...

Novel approach. However, Mathematicians hold themselves to a much higher standard of PROOF. It's like Fermat's Last Theorem. You may not understand Andrew Wiles math but the experts who have verified his proof do and they've said it is correct. Debating it is pointless.

Next thing you'll be disputing 1+1=2

Interesting experiment to think of:

If we have 3 doors, 2 goats and 1 car.
You don't make a choice on the first sequence.. the announcer just opens one of the goat doors.

What is the probability of choice between the two remaining doors that either is a goat or a car, and how is this example different than the one given earlier.

The real importance it the independence of trails when doing the choosing. We are asking for the probability of the next decision, not the probability of the pattern of events leading up to the choice.

with a nod to flu
http://mathforum.org/dr.math/faq/faq.mon...

21 was the lamest movie ever. Get a big name star, throw in some people, make a crappy film and market it like it's tha shiznit... That should be the description on the cover of this movie. Or "amateurs' first try to make a movie, somehow get Kevin Spacey to star".

LOL...I chuckle a bit here, because I had the same feeling.But, thankfully, a lot of you aren't responsible for engineering things we depend on. Just kidding.

Anyway, don't mean anything by it. Take a probability class, and then take a stochastic processes class. Much of this is required especially for information theory and communication systems. I guess it's also applicable to finance, but I don't know much about that subject.

Here's a very simple explanation. You pick door 1 and since there is a 1/3 chance of picking car, then there must be 2/3 chance that car is behind door 2 or 3. When Monty shows you door 2, he will never show you the car. Thus, he just spilled the beans, that is the odds of the car behind door 2 or 3 is still 2/3, but since he showed it is not door 2, the odds that it is just behine door #3 is 2/3.

Here's the definitive explanation and simulation from nerds at UCSD math dept. Monty knows where the car is and that is why the odds from switching is 2/3. (He never open up the door with the car)

Simulation:
http://math.ucsd.edu/~crypto/Monty/monty...
explanation:
http://math.ucsd.edu/~crypto/Monty/monty...

To those who don't realize that switching is the right decision, it is extremely difficult to convince them.

The only way is to simulate it, really. Try the game about 1000 times. Switch half the time and don't switch the other half. You'll see you win more often if you switch.

Another way to look at it ...

Start with 100 doors. Choose 1 door.
Now - I am going to take away 98 of the 99 remaining doors, with the promise that none of the 98 I eliminate contain the prize.

If you play this game a few times, you'll soon realize that in 1 out of 100 cases, you will chose the right one in the first place, which means that if you switch you will be switching to the prize in 99 out of 100 cases.

I understand people who think it is 50/50 but they are wrong.

By the way, this logic provides some insight into poker as well.

If you are playing 10-handed no-limit hold 'em, for example, and one guy bets, then 8 guys fold pre-flop, and you have a mid-quality hand, you should fold. Even though it comes down to you against him, the fact that he is likely the best of 9 hands makes it very unlikely you have a better starting hand.

However, if there are only 3 players, one guy bets and the other guy folds, you should probably stay in as your maginal hand has a much better chance against the best of 2 random starting hands as opposed to the best of 9 random starting hands.

The act of folding in this case is like the host taking away the door with the goat.

I wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).

By doing the switch, you are saying: "the first door I chose is wrong", which is 2/3 of chance.

And that is why you must switch !

You have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it's not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50

it's much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn't pick is 999:1000.