Maths Please

[Sabre]

I was wondering if anyone here who is well versed in probability could help me out.

This come from a science podcast I listen to claimed that most people are ignorant of how chance and probablity work. I know that reverse probablity doesn't really work. eg, being delt a hand of cards might have a 1/5000 (or whatever) chance, but those were the card you got. The one I have problems with was demostrated with the following.

You are in a game show. There are 3 doors. One of them has a prize, the other 2 don't. So you pick a door, lets say door 1, the presenter then opens a door without a prize. Lets say its door 2. He then asks do you want to stay with door 1, or choose door 3.

Now, I would expect it wouldn't matter which you choose, as it's a 50-50 chance of a prize. They claim that this is wrong, and that you should always choose door 3 as because that will now give you a 2/3 chance of a prize. Me, and everyone I told in RL can't work out how that's the case.

...also I don't want to hear silly answers like 'look through the key hole' or 'choose the door that has a car sticking out of it'.

On a unrelated note. Anyone know of any interviews with the voice actors of left 4 dead please. There's some things I want to know about it.

Thanks.

[Mr. Krystal]

It's a 50-50 chance. If you have more than a 50-50 chance, there's something you're not telling us about the problem.

And, Alésia Glidewell was a voice actor on Left 4 Dead. You could try contacting her.

[DZComposer]

This is called the "Two Goat Problem," "Car and Goats Problem," or the "Monty Hall Problem." It actually came from an old TV game show called "Let's Make a Deal." Monty Hall was the host. There were 3 doors, two had goats behind them and one had a new car. The contestant would play in the manner you described.

I borrowed some of the specifics from a couple of sources, but I have heard of this before.

This is the show that spawned the saying "Let's see what's behind door number 1."

There actually was a lot of debate over this one. The 50/50 idea was solid for a long time. Until the 90s, actually. But, it turns out that you actually do have a 2/3 chance by switching.

1( ) 2( ) 3( )

X represents a goat, * represents a car.

I have a 1 in 3 chance of picking the right blank, so I pick #1. #3 is revealed to be a X:

1( ) 2( ) 3(X)

The principal of the argument is this: I had a 1-in-3 chance when I picked, and STILL have a 1-in-3 chance that I picked correctly. Nothing changed because it is a fact that either or both 2 and 3 have an X.

Therefore, I should change my pick.

Here is a visual representation:

Player chooses blank 1:

winning blank	1	1	2	2	3	3
Opened	2	3	3	3	2	2
Revealed	X	X	X	X	X	X
Switch Result	X	X	*	*	*	*

It works out the same if you pick 2 or 3 first.

As you can see, I have a 4/6 (AKA 2/3) chance of winning.

A key component if this problem is it's wording: The host ALWAYS reveals a goat! That is why it is not a 50/50. If it was a random door, then it would be a 50/50:

Player chooses blank 1:

winning blank	1	1	2	2	3	3
Opened	2	3	2	3	2	3
Revealed	X	X	*	X	X	*
Switch Result	X	X	?	*	*	?

? represents an undefined condition. As you can see, you have a 50/50 chance in this scenario.

Notice how this gets sticky, though? That is why it is ALWAYS a goat that is revealed. That stacks the odds.

If you still don't get it, Google "car and goats problem."

[Sabre]

I googled it, and I found an interesting way of looking at it. Basically, by switching, the only way to lose would be if you chose the prize, thus making the chance of wining 2/3. Clever.

However, I don't think that any or the sites I read about the problem were compelling enough for me to change my opinion on this. The chance of drawing an ace of clubs from a pack of cards is 1/52, if you said you try again if you get a 10, then you might as well emove the 10s and just have a 1/48 chance.

I think, whether 3 doors or a million, once your real option is 2 doors, then it's a 50/50 chance. I remain unconvinced.

On a side note. I none of the sites I read had tried it out practically. Maybe it's just not worth the time.

[DZComposer]

When you are drawing cards out of a deck, you are, usually, doing a random act.

That is the key to why this is not a 50/50.

The host ONLY reveals a goat. That is like saying that every card until the last two picked CAN NOT be the ace of clubs.

If the host knows where the car is, it is not 50/50.

Look at the tables in my previous post again. You can clearly see that you have a 4 of 6 chance of winning, which reduces to 2 of 3.

[Mr. Krystal]

I googled it, and I found an interesting way of looking at it. Basically, by switching, the only way to lose would be if you chose the prize, thus making the chance of wining 2/3. Clever.

However, I don't think that any or the sites I read about the problem were compelling enough for me to change my opinion on this. The chance of drawing an ace of clubs from a pack of cards is 1/52, if you said you try again if you get a 10, then you might as well emove the 10s and just have a 1/48 chance.

I think, whether 3 doors or a million, once your real option is 2 doors, then it's a 50/50 chance. I remain unconvinced.

On a side note. I none of the sites I read had tried it out practically. Maybe it's just not worth the time.

DZ is right, I'm afraid. There are a couple other ways of presenting the reason why it's not 50/50 that they didn't mention on Wikipedia (another way of looking at it, if you will). Notice that there are more empty doors than cars at the beginning, so the likelihood that you picked the wrong door is 2/3. Then, you remove one door from the equation, one that is not the car. Your likelihood of picking the wrong door is now reduced to 1/3. That leaves 2/3s probability that the other door contains the car.

[Sabre]

But since he opens a wrong door anyway, it might as well be just choosing between 2 doors, thus 50/50.

Another thing. How come none of the sites I've read have done this for real? Do the problem and see if it makes a statistically significant difference.

[PuffNStuff]

^Because it isn't worth the effort. IMO this is interesting and all, but the subject is so circumstantial that it is a waste of brainpower to sit and think about this in depth. Let alone to actually attempt to find the exact statistics.

[Mr. Krystal]

But since he opens a wrong door anyway, it might as well be just choosing between 2 doors, thus 50/50.

Another thing. How come none of the sites I've read have done this for real? Do the problem and see if it makes a statistically significant difference.

They have in simulation. They've taken every possible scenario, then added up the number of wins versus losses, compared the number of wins relative to whether or not switching would be effective, and found that that chance adds up to 2/3. Placing actual cars and goats behind doors wouldn't actually make the statistics different. The only other thing that might play into this problem is that people do not make random choices. They will tend towards one door over another, so the chance is not exactly 2/3, but close to it.

[DZComposer]

But since he opens a wrong door anyway, it might as well be just choosing between 2 doors, thus 50/50.

Another thing. How come none of the sites I've read have done this for real? Do the problem and see if it makes a statistically significant difference.

Your major error is that you are tossing out the host's door as extraneous. That is not the case. It is a fundamental part of the problem. You STILL have a 1/3 chance that you picked the right door at the time of reveal. If you switch, you have a 1/3 chance of picking a goat, and a 2/3 chance of the car. The only way you lose if you switch is if you had originally guessed correctly, which was a 1/3 chance.

Think of it this way: Even though one of the doors was opened, there are STILL three doors!

Wikipedia actually has a detailed Bayesian proof of it.

http://en.wikipedia.org/wiki/Monty_Hall_problem

If you want to actually try it, here'e the New York Times' interactive article on the problem:

http://www.nytimes.com/2008/04/08/science/08monty.html?_r=1&oref=slogin

This site has a computer do the problem hundreds of times and shows a graph:

http://demonstrations.wolfram.com/MontyHallParadox/

[Sabre]

Your major error is that you are tossing out the host's door as extraneous. That is not the case.

I think it is. Eg. What if the door was already open and you only had to pick between 2.

IMO this is interesting and all, but the subject is so circumstantial that it is a waste of brainpower to sit and think about this in depth. Let alone to actually attempt to find the exact statistics.

Maybe. Then again, people spend more time on football and none news.

<-- remains unconvinced.

[DZComposer]

I think it is. Eg. What if the door was already open and you only had to pick between 2.

Then it would be 50/50 because you have eliminated the first choice.

The first choice IS ESSENTIAL to the odds. It IS NOT insignificant.

Once you make your first choice, it determines whether or not switching will win before the host even reveals a door.

Because of that, the first choice IS NOT insignificant, as it is what determines whether or not you will win if you switch.

If the door is opened before you make a pick, or doesn't exist, then it changes everything.

Look at the charts in my third link. If you run it a thousand times, switching will make you win pretty close to 330 times. Or play the New York Times one for about an hour and it will be obvious that you have about a 66% chance of winning if you switch.

[PuffNStuff]

Chances are decided on the amount of possible outcomes. Not by how it is chosen.