Fun math problem...

[huntedannoyed]

OK, we know the one about the dog, and we know the crazy percentage one (thank you whoops1995)... but how about the one from Die Hard with a Vengence? They never really explain how to do it in the movie, but it can be done.
You are at a fountain (unlimited water supply) and you have two empty water jugs. One is a five (5) gallon jug and the other is a three (3) gallon jug... how can you measure EXACTLY four (4) gallons?

*It isn't impossible, so don't look down... Try to figure it out!

 

[Amnestic]

You are at a fountain (unlimited water supply) and you have two empty water jugs. One is a five (5) gallon jug and the other is a three (3) gallon jug... how can you measure EXACTLY four (4) gallons?

This was on KotOR 1. I always manage to do it and always forget how.

I'll refer to 5 Gallon as jug A and 3 gallon as jug B.

Spoiler: Solution

Fill Jug A.
Pour it into Jug B. (2 Gallons in Jug A, 3 in Jug B).
Empty Jug B.
Transfer A to B (2 Gallons in Jug B, 0 in Jug A)
Fill Jug A (2 Gallons in Jug B, 5 in Jug A)
Transfer A to B (3 Gallons in Jug B, 4 in Jug A)

There may be a faster way, but that's how I do it.

 

[Novajam]

Spoiler

Fill the 5 litre jug from the fountain, and then fill the 3 litre jug with the 5 litre jug. The 5 litre jug will now be 2/5 full. Empty the 3 litre jug and refill it with the 5 litre jug. The 5 litre jug is now empty and the 3 litre jug is 2/3 full. Fill the 5 litre jug from the fountain again, and fill the 3 litre jug with the 5 litre jug. The 5 litre jug now has 4 litres of water in it. Then you realise the markings on the side. Damn.

 

[huntedannoyed]

Haha, stuiped markings! Ok, now for the hard part... how many Oz. are in four gallons? No google!

 

[Steve Dark]

Easy. Fill the 3, put it in the 5. Fill the 3 again and top up the 5 from it. You now have 1 in the 3. Empty the 5 and put the 1 in the 5. Then fill up the 3 again. Add the 3 to the 1 in the 5 and you get 4 in the 5.

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

 

[huntedannoyed]

Easy. Fill the 3, put it in the 5. Fill the 3 again and top up the 5 from it. You now have 1 in the 3. Empty the 5 and put the 1 in the 5. Then fill up the 3 again. Add the 3 to the 1 in the 5 and you get 4 in the 5.

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

This was popularized in the movie "21." You should change your answer if given the choice because now your odds are 50% instead of 33.33% Funny enough, I tried this with three playing cards and it is totally accurate. Which leads to this thought... If you ever get down to two cases on "Deal or no Deal," should you switch your case if you get down to two... Yes, you should.

 

[Ronwue]

Stick with your choice. No reasoning

 

[huntedannoyed]

Stick with your choice. No reasoning

Thats the G. W. Bush mathamatical standard!

 

[Novajam]

Haha, stuiped markings! Ok, now for the hard part... how many Oz. are in four gallons? No google!

Four gallons worth? (Hur hur)

I'm going to guess 64.

 

[mintsauce]

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

Surely the answer is that it doesn't matter whether you swap or not, since your odds of winning have increased from 33% to 50%, but are not affected by which door you choose?

 

[Lukeje]

Easy. Fill the 3, put it in the 5. Fill the 3 again and top up the 5 from it. You now have 1 in the 3. Empty the 5 and put the 1 in the 5. Then fill up the 3 again. Add the 3 to the 1 in the 5 and you get 4 in the 5.

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

This was popularized in the movie "21." You should change your answer if given the choice because now your odds are 50% instead of 33.33% Funny enough, I tried this with three playing cards and it is totally accurate. Which leads to this thought... If you ever get down to two cases on "Deal or no Deal," should you switch your case if you get down to two... Yes, you should.

No... it goes from 1/3 to 2/3 if you swap (you are doubling your chance of winning, as you are now betting on TWO doors, not just one).

 

[Sgt. Dante]

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

Someone was watching 21...

 

[mintsauce]

No... it goes from 1/3 to 2/3 if you swap (you are doubling your chance of winning, as you are now betting on TWO doors, not just one).

Not possible. Your chance of winning cannot be 2/3 if you are choosing between two doors.

 

[Lukeje]

No... it goes from 1/3 to 2/3 if you swap (you are doubling your chance of winning, as you are now betting on TWO doors, not just one).

Not possible. Your chance of winning cannot be 2/3 if you are choosing between two doors.

http://en.wikipedia.org/wiki/Monty_Hall_problem
Thats kind of the point; in the second round you are choosing between one door (the original door you picked) or two doors (the two that you didn't pick, for which you know that at least one doesn't have the prize).

 

[Unknower]

This has something to do with mathematics. Very, very simpole mathematics, but at least it gives me an excuse to post this:

Count the fs in this sentence. Count them only once.

FINISHED FILES ARE THE RESULT OF YEARS OF SCIENTIFIC STUDY COMBINED WITH THE EXPERIENCE OF YEARS.

Spoiler: Answer

There are 6 fs. Some get this right the first time, but lots of people count only 3.

 

[000Ronald]

...Math...

I was never any good at it.

Oh well. Have fun debating.

Apologies for apathy

 

[hemahemahema]

Monty Hall problem, you should always switch to increase the odds from 33.33% to 66.66%, i.e. 1/3 to 2/3

think of it this way

choose goat-switch-get car
choose car-switch-get goat

since you are more likely to start with a goat (2/3), you should always switch

Not possible. Your chance of winning cannot be 2/3 if you are choosing between two doors.

Yes it can. Think of it as a biased coin - the chance of getting a head and a tail is not 50/50, and you know it is biased from the 1st part of the game. Just because there are 2 outcomes doesn't mean they have to divide the possibilities between them evenly.

Another way to think about it is, if you started with the car, the game host HAS a choice of which goat he reveals, but if you started with a goat, the game host has no choice but to reveal the other one. This is a direct consequence of your initial choice among 3 doors. If he has a choice, you should not switch, otherwise, switch. The latter is obviously more likely than the former.

And the water Jug one: fill up 5 gallon jug, use it to fill up the 3 gallon, so you have 2 gallons in the 5 gallon one. Empty the 3 gallon one and transfer the 2 gallon into it. Now fill up the 5 gallon jug and use it to fill up the 3 gallon one. You need 1 gallon to do it so you are left with 4 gallon in the 5 gallon jug.

 

[mintsauce]

Yeah, I've got it now. I had a read through the Wikipedia article and tried the NY Times demonstration as well. It's the strangest thing though, possibly the most counter-intuitive thing I have ever encountered!

 

[huntedannoyed]

Easy. Fill the 3, put it in the 5. Fill the 3 again and top up the 5 from it. You now have 1 in the 3. Empty the 5 and put the 1 in the 5. Then fill up the 3 again. Add the 3 to the 1 in the 5 and you get 4 in the 5.

Here's a better Maths Question:
You're on a game show and there are three doors to choose from, only one of which has a prize behind it. You choose one of the doors, but before it's opened the Gameshow host opens one of the other doors with nothing behind it and then asks you if you want to change your mind. Should you stick with the one you chose initially or swap to the other one?

This was popularized in the movie "21." You should change your answer if given the choice because now your odds are 50% instead of 33.33% Funny enough, I tried this with three playing cards and it is totally accurate. Which leads to this thought... If you ever get down to two cases on "Deal or no Deal," should you switch your case if you get down to two... Yes, you should.

No... it goes from 1/3 to 2/3 if you swap (you are doubling your chance of winning, as you are now betting on TWO doors, not just one).

If the guy opens one door, there are now only two doors to choose from; 50-50, up from 1/3.

 

[SkinnySlim]

I've got one for you, guess the next sequence of numbers in this group...


1
11
21