You guessed it, ANOTHER MATHS THREAD :D

[Solaris]

Right, here's a maths problem I saw in the newspapers a few years ago, and then saw in a book last year, it's pretty intresting.

You're on a gameshow, in front of you are 3 garages, one of these garages contains a car, the other two contain nothing. The doors are closed and you don't know which one has a car in and which two don't. You are asked to pick a garage which you think has the car inside. You pick a garage. That garage door remains closed, as the gameshow host opens a different garage door to the one you picked, to reveal it is empty.

You are now left with two garages, one of which was your original choice, you now know one of these garages contains a car. The gameshow host gives you the option to switch your choice to the other garage. If you swap to the other garage, are you more likely to win the car?

 

[DreadLord]

No. Showing that the other garage is empty just makes it a 50/50 chance now.

If the question is "are you more likely to win the car than you were before he showed an empty one?", then yes. Because it's now a 1 in 2 chance, rather than a 1 in 3 chance.

 

[Solaris]

Yes

 

[Stieffers]

If Gordon boards a train headed for Nova Prospekt at 0800 hours, the train's speed is a constant 100mph and Nova Prospekt is 365 miles from City 17, and exactly 10 minutes after he leaves, Alyx boards a hijacked gunship headed for City 17 that gains 10mph every minute (with an initial velocity of 0, no max speed), at what time do their paths meet, and how far are they from Nova Prospekt?

 

[Canadian Gunner]

Yes. (original question)

 

[MiccyNarc]

When you picked the first time you had a 33% chance of choosing the right door. Now you have a 50% chance of picking the right one.
Somehow that increases your odds.

 

[Dalamari]

That sure was short lived, gg n00b

 

[Solaris]

When you picked the first time you had a 33% chance of choosing the right door. Now you have a 50% chance of picking the right one.
Somehow that increases your odds.

Nope.
You're right, but that isn't why.

 

[Canadian Gunner]

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

 

[Llama]

I dont get that...(Im stupid ) why does the chance increase to 2/3 if there are only 2/3 doors left and 1/3 is empty?

 

[DreadLord]

At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

• The player originally picked the door hiding goat number 1. The game host has shown the other goat.
• The player originally picked the door hiding goat number 2. The game host has shown the other goat.
• The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three.

That's an incredibly stupid way of looking at it. I suppose it makes sense in that regard, but it still doesn't change the fact that there's a 50% chance door A has the car, and a 50% chance door B has the car.

I still say no.

 

[Llama]

*nevermind*
That actually makes sense to me now

 

[Solaris]

That's an incredibly stupid way of looking at it. I suppose it makes sense in that regard, but it still doesn't change the fact that there's a 50% chance door A has the car, and a 50% chance door B has the car.

I still say no.

How's it an incredably stupid way if its right and you're wrong, unless of course you don't accept you're wrong. Have a look at the Probablity tree on the wiki.

 

[dfc05]

Another way to look at it would be to consider if you had say 99 empty doors and 1 with a car, and you picked a door and they took away all the empty doors except one... of course you'd switch.

The same holds true for 3 doors but on a smaller scale. The first time you picked, it was twice as likely that you got one without anything in it than getting one with the car, so you want to switch.

 

[63]

Original question: Yes, you change.

Solaris is right.

 

[dfc05]

For the train question:

time = 65.422 minutes
distance from City17 = 255.964

Physics ftw?

That's if Alyx is constantly accelerating. If she's at a constant speed for one minute then suddenly jumps to a faster speed, then it's different. And harder to solve.

 

[Dan]

You should swap it. The first pick you made is 1/3 correct. If you swap it, you are upping your odds to 1/2 correct.

 

[63]

You should swap it. The first pick you made is 1/3 correct. If you swap it, you are upping your odds to 1/2 correct.

Yes.

 

[dfc05]

Your odds are always 2/3 correct from the start. They never change.

Like if you had started with 99 bad doors, took all but one away, and then switched, your odds would still be 99/100 that you get the right one. You wouldn't say your first pick is 1/100 correct and changed to 1/2 as it is still much much more likely that you get the right door by switching.

Also by the logic that your odds change to 1/2 correct, then what would be the point in switching? 1/2 correct means that you are equally likely to win by either switching or staying.

 

[DreamThrall]

The answer, of course, is 42.

Oh, and 1 != .999...

 

[JNightshade]

1 + 2 = ?

solve.

 

[DreadLord]

You should swap it. The first pick you made is 1/3 correct. If you swap it, you are upping your odds to 1/2 correct.

Not true according to Solaris and Wiki's answer. By their math you are upping to 2/3 correct. Which makes no sense really.

Yes.

No.. You said you stood with Solaris before. It's one or the other.

Your odds are always 2/3 correct from the start. They never change.

No, you have a 1/3 chance before the first door is opened.

1 + 2 = ?

solve.

7.

 

[Angry Lawyer]

I've never quite understood the semantics, but I know full well the answer is that it improves.

-Angry Lawyer

 

[dfc05]

No, you have a 1/3 chance before the first door is opened.

If you consider the entire problem as a whole rather than 2 distinct steps, and take the choice of switching as a given, then you start out with a 2/3 chance. Probabilities are determined from the initial set-up. All you are choosing between is a 2/3 chance or the 1/3 chance. The 1/2 chance doesn't exist.

 

[DreadLord]

If you consider the entire problem as a whole rather than 2 distinct steps, and take the choice of switching as a given, then you start out with a 2/3 chance. Probabilities are determined from the initial set-up. All you are choosing between is a 2/3 chance or the 1/3 chance. The 1/2 chance doesn't exist.

Then by the same logic we can deduce that there is in FACT a 1/2 chance of there being a CAR behind the door you choose. Ignoring the other parts you are undoubtedly left with TWO DOORS, one has an ASS, one has a CAR.

See the discrepancy?

 

[dfc05]

Then by the same logic we can deduce that there is in FACT a 1/2 chance of there being a CAR behind the door you choose. Ignoring the other parts you are undoubtedly left with TWO DOORS, one has an ASS, one has a CAR.

See the discrepancy?

Sure if you consider the second step of picking between two doors as an isolated case. But the first step influences the second step because the door you chose came from the first step. The steps are inseparable! Going back to the 99 door extension, so if you picked one door out of 100 and only one had a car, and they took away 98 other doors without cars, and you're left with your first pick or the other door, would you still say you had a 50% chance of having the car in your original pick? Sure there are now only two doors left, one of which has a car behind it, but the unlikeliness of you picking the correct door the first time around bleeds over into the two-door part.

 

[DreadLord]

There are still two doors. There is still one car. My guess has just as much potential as the other door.

 

[63]

Original question: Yes, you change.

Solaris is right.

Yes.

My answer remains the same Dread... You can try it with cards, see the result for yourself.

 

[DreadLord]

My answer remains the same Dread...

Solaris says your chances increase to 2/3.

Dan said they increase to 1/2.

You can't agree with both.

 

[63]

I'm too lazy to reread the thread, you win dread.

 

[Dan]

I think a better way to look at it is without any definite car. You start with 1/3 of a car behind each of the doors. When you open an empty door, what you are doing is essentially taking 1/3 of a car and moving it into the last door. So you have the door you chose first with 1/3 of a car in it and the other door with 2/3 of a car in it.

 

[DreadLord]

I think a better way to look at it is without any definite car. You start with 1/3 of a car behind each of the doors. When you open an empty door, what you are doing is essentially taking 1/3 of a car and moving it into the last door. So you have the door you chose first with 1/3 of a car in it and the other door with 2/3 of a car in it.

:|:|

You win Dan.

Thanks for breaking my car.

 

[Triggerhappy41]

dread, you are looking at it the wrong way.

you would be right about the 1/2 chance if the host revealed a goat before you made a choice, since one door would have a car and another wouldn't. Simple enough.

But you have to take the past into consideration. Just like you posted:

-The player originally picked the door hiding goat number 1. The game host has shown the other goat.
-The player originally picked the door hiding goat number 2. The game host has shown the other goat.
-The player originally picked the door hiding the car. The game host has shown either of the two goats.

these are the three possibilities of the choices you can make in this game.

Now, assume that you ALWAYS switch. You would always win in the first two cases, and in the last one you would lose. 2/3 probability.

Assume that you never switch. You lose the first 2 and win the third. 1/3 probability.

The point is, the 1/3 or 2/3 probability claim is taking the WHOLE problem into account, based on what we know about the host's revealing strategy, and it works.

Your 1/2 claim is only taking into account the fact that there is a car and a goat behind two doors; it leaves out the information that the host gives us by revealing a goat door.

 

[theotherguy]

discover magazine had an entire section on this.

Think of it this way:

Situation 1:If you had picked the car, that means there are two with no cars;
(1/3 chance)

Situation 2:If you had not picked a car, that means there is a car in one of the two;
(2/3 chance)

One of the garage doors open, it HAS to be the one without the car.

Therefore, if you switch, you have a 2/3 chance of it being situation 2, and you will get the car.

 

[dfc05]

dread, you are looking at it the wrong way.

you would be right about the 1/2 chance if the host revealed a goat before you made a choice, since one door would have a car and another wouldn't. Simple enough.

But you have to take the past into consideration. Just like you posted:

-The player originally picked the door hiding goat number 1. The game host has shown the other goat.
-The player originally picked the door hiding goat number 2. The game host has shown the other goat.
-The player originally picked the door hiding the car. The game host has shown either of the two goats.

these are the three possibilities of the choices you can make in this game.

Now, assume that you ALWAYS switch. You would always win in the first two cases, and in the last one you would lose. 2/3 probability.

Assume that you never switch. You lose the first 2 and win the third. 1/3 probability.

The point is, the 1/3 or 2/3 probability claim is taking the WHOLE problem into account, based on what we know about the host's revealing strategy, and it works.

Your 1/2 claim is only taking into account the fact that there is a car and a goat behind two doors; it leaves out the information that the host gives us by revealing a goat door.

Correct. Yay. Good explanation . Finally someone who looks at big-picture situations.