do you switch your door?

[dullard]

Originally posted by: GeekDrew
How is that a choice? How does A+B=C?

That is the way the game is played. It is the rules. The text is written to confuse you to that fact, but that is the way the game is played. So, do you want C or A+B?

 

[KillyKillall]

Originally posted by: ElFenix

Originally posted by: KillyKillall

Watching "Numbers" again or reading the sites?

just had this explanation given to my m&a class. it's only the first class day so he's explaining odds to law students.

This was the EXACT explanation shown and given on that show "Numbers" about 2 months ago.

 

[KillyKillall]

Originally posted by: dullard

Originally posted by: GeekDrew
How is that a choice? How does A+B=C?

That is the way the game is played. It is the rules. The text is written to confuse you to that fact, but that is the way the game is played. So, do you want C or A+B?

That's not how it is played.

He chose C.... you show him either A or B...Say you show him A is a goat...

Then does he want B or C is his choice. That is the question on when to change based on 50% of what you didn't choose is definitely a loser.

Read the rules.

 

[dullard]

Originally posted by: KillyKillall
That's not how it is played.

Yes, it is. Let me go on, and I'll show you why.

 

[mugs]

Someone in another thread had an example that made it more obvious.

Suppose there are 10 doors. You pick one. 8 doors that are known to be empty are opened, leaving your door and one other. Do you still think your odds are 50/50? You had a 1 in 10 chance that you picked the right door at the beginning, you still have a 1 in 10 chance that you have the right door.

There was a 9/10 chance that the prize was behind one of the other 9 doors. Since they eliminated 8 of them, the last remaining door has a 9/10 chance of being the right one.

 

[KillyKillall]

Originally posted by: dullard

Originally posted by: KillyKillall
That's not how it is played.

Yes, it is. Let me go on, and I'll show you why.

Ever watch "Let's Make a Deal?" That's the reference.

Now read this and tell me I'm wrong again:

Link to example

 

[dullard]

Originally posted by: KillyKillall
...

Why ruin GeekDrew's chance of understanding the game? I'm simply doing things in reverse to show him the reason that you should switch. The fact that I reverse the game doesn't mean my presentation is incorrect. Its like doing a puzzle backwards (often far easier) doesn't mean the puzzle pieces are suddenly false.

Haven't you ever tried backsolving a problem from the answer back to the original question?

 

[GeekDrew]

Originally posted by: mugs
Someone in another thread had an example that made it more obvious.

Suppose there are 10 doors. You pick one. 8 doors that are known to be empty are opened, leaving your door and one other. Do you still think your odds are 50/50? You had a 1 in 10 chance that you picked the right door at the beginning, you still have a 1 in 10 chance that you have the right door.

There was a 9/10 chance that the prize was behind one of the other 9 doors. Since they eliminated 8 of them, the last remaining door has a 9/10 chance of being the right one.

OK.. with this example -- how is it that you still have a 1/10 chance of being the right one? There are only two doors remaining, and one is a car, and one is nothingness (to add to the example). I don't understand why it would still be a 1/10 chance at all... that's what I'm trying to understand. When there are only two doors remaining, why do you continue to consider the other 8 (now removed) doors?

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: KillyKillall
...

Why ruin GeekDrew's chance of understanding the game? I'm simply doing things in reverse to show him the reason that you should switch. The fact that I reverse the game doesn't mean my presentation is incorrect. Its like doing a puzzle backwards (often far easier) doesn't mean the puzzle pieces are suddenly false.

Haven't you ever tried backsolving a problem from the answer back to the original question?

K. If I can chose either A+B or C, then I would obviously want A+B, since my chances would be 1/3 better.

Yeah, I've tried backsolving a problem from the answer... and have yet to succeed, once in my life.

 

[dullard]

Originally posted by: GeekDrew
If I can chose either A+B or C, then I would obviously want A+B, since my chances would be 1/3 better.

So, you'd rather choose A+B. You agree, that choosing A+B is twice as likely to have the car.

Suppose I offer to tell you the name of my cat, would you switch from A+B to C?

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
If I can chose either A+B or C, then I would obviously want A+B, since my chances would be 1/3 better.

So, you'd rather choose A+B. You agree, that choosing A+B is twice as likely to have the car.

Suppose I offer to tell you the name of my cat, would you switch from A+B to C?

No, because that information is irrelevant.

 

[dullard]

Originally posted by: GeekDrew
No, because that information is irrelevant.

If I agree to tell you the mileage on my car would you switch from A+B to C?

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
No, because that information is irrelevant.

If I agree to tell you the mileage on my car would you switch from A+B to C?

No. Same reason.

 

[dullard]

Originally posted by: GeekDrew
No. Same reason.

Suppose I put one penny behind one door, a nickel behind another door, and a dime behind a third door. But the car doesn't change.

Would you now switch from A+B to C?

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
No. Same reason.

Suppose I put one penny behind one door, a nickel behind another door, and a dime behind a third door. But the car doesn't change.

Would you now switch from A+B to C?

Of course not.

 

[dullard]

Originally posted by: GeekDrew
Of course not.

Ok then. We know know the value of the prizes behind all doors is different. Thus, one of the doors A or B is more valuable than the other.

Suppose I told you an irrelevant piece of information. Suppose that irrelevant information is to tell you which of A or B is the most valuable.

Now will you switch from A+B to C?

 

[DAGTA]

This is one of those things with numbers that doesn't agree with logic.

When Monty opens a door (any door) he is simply narrowing the choices from three doors down to two. He'll never open the door with the car. So you're odds will be 50/50, just like they were at the beginning because you knew going into the game that Monty was going to reveal one of the goats.

Saying your odds increase is a prime way to learn how to talk like a lawyer. Basically twist statistics so you can BS.

 

[mugs]

Originally posted by: GeekDrew

Originally posted by: mugs
Someone in another thread had an example that made it more obvious.

Suppose there are 10 doors. You pick one. 8 doors that are known to be empty are opened, leaving your door and one other. Do you still think your odds are 50/50? You had a 1 in 10 chance that you picked the right door at the beginning, you still have a 1 in 10 chance that you have the right door.

There was a 9/10 chance that the prize was behind one of the other 9 doors. Since they eliminated 8 of them, the last remaining door has a 9/10 chance of being the right one.

OK.. with this example -- how is it that you still have a 1/10 chance of being the right one? There are only two doors remaining, and one is a car, and one is nothingness (to add to the example). I don't understand why it would still be a 1/10 chance at all... that's what I'm trying to understand. When there are only two doors remaining, why do you continue to consider the other 8 (now removed) doors?

Regardless of what you pick, there will always be at least 8 doors that are empty. They are meaningless, opening them doesn't increase the odds that you picked the right one to begin with.

Look at it in reverse: 9 times out of 10, you will pick the wrong door on the first try. Right? Even after opening the other 8 doors, 9 times out of 10 you still have the wrong door. They just made it easy for you by showing you which of the other 9 doors has the prize in it.

As someone else suggested, have someone else play the game with you. Or write a computer program to simulate it.

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
Of course not.

Ok then. We know know the value of the prizes behind all doors is different. Thus, one of the doors A or B is more valuable than the other.

Suppose I told you an irrelevant piece of information. Suppose that irrelevant information is to tell you which of A or B is the most valuable.

Now will you switch from A+B to C?

nope

 

[dullard]

Originally posted by: GeekDrew
nope

Ok. Suppose I even go so far as to show you the least valuable of A or B (it'll be just a lonely coin). Now will you switch from A+B to C?

 

[mugs]

Here, I wrote a quick PHP program to simulate it:

http://www.savemejebus.net/index.php

If you'd like the code I can send it to anyone who wants or I could post it in the Software Forum

Edit: I'm improving it at the moment...

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
nope

Ok. Suppose I even go so far as to show you the least valuable of A or B (it'll be just a lonely coin). Now will you switch from A+B to C?

no.

 

[dullard]

Originally posted by: GeekDrew
no.

Suppose a goat comes in and eats that coin (you aren't reimbursed for this minimal loss). Now will you switch to C?

 

[GeekDrew]

Originally posted by: dullard

Originally posted by: GeekDrew
no.

Suppose a goat comes in and eats that coin (you aren't reimbursed for this minimal loss). Now will you switch to C?

Christ. Enough of the irrelevance. NO. At that point, I have an equal chance of winning with either A or C. I just have NO reason to chose C, because A is just as likely to have the answer. There are only two options left.

 

[dullard]

Originally posted by: GeekDrew
Christ. Enough of the irrelevance. NO. At that point, I have an equal chance of winning with either A or C. I just have NO reason to chose C, because A is just as likely to have the answer. There are only two options left.

But A+B is twice as likely as C, you said so yourself. At which point did the irrelevant information make A+B less likely?