Can anyone get this tricky math problem?

[swiftaw]

Originally Posted by queshy

is the answer to the OPs problem 10990?

it is

[queshy]

Ok, fun question:

Suppose you have two unmarked buckets: one has a capacity of 3 L and the other has a capacity of 5 L. Let's say you're in a room with just a hose - no measuring instruments, no scale, etc, nothing, just a hose with running water, and these two empty, unmarked buckets. How could you obtain EXACTLY 4 L using the two buckets?

Don't look this one up on the net because I'm sure it's a popular one - try and figure it out!

[swiftaw]

Originally Posted by queshy

Ok, fun question:

Suppose you have two unmarked buckets: one has a capacity of 3 L and the other has a capacity of 5 L. Let's say you're in a room with just a hose - no measuring instruments, no scale, etc, nothing, just a hose with running water, and these two empty, unmarked buckets. How could you obtain EXACTLY 4 L using the two buckets?

Don't look this one up on the net because I'm sure it's a popular one - try and figure it out!

Someone's been watching Die Hard with a Vengence,

Ok, fill the 5 L, then pour 3 L from the 5L to the 3L (so there is now 3L in the 3L and 2L in the 5L.

Now, empty the 3L, and pour the 2L remaining in the 5L into the 3L (so there is now 2L in the 3L and 0 in the 5L)

Now, fill the 5L, and pour 1L into the 3L, leaving 4L in the 5L

[queshy]

Originally Posted by swiftaw

Someone's been watching Die Hard with a Vengence,

Ok, fill the 5 L, then pour 3 L from the 5L to the 3L (so there is now 3L in the 3L and 2L in the 5L.

Now, empty the 3L, and pour the 2L remaining in the 5L into the 3L (so there is now 2L in the 3L and 0 in the 5L)

Now, fill the 5L, and pour 1L into the 3L, leaving 4L in the 5L

Yes and yes!

[CalBoy]

Originally Posted by queshy

Ok, fun question:

Suppose you have two unmarked buckets: one has a capacity of 3 L and the other has a capacity of 5 L. Let's say you're in a room with just a hose - no measuring instruments, no scale, etc, nothing, just a hose with running water, and these two empty, unmarked buckets. How could you obtain EXACTLY 4 L using the two buckets?

Don't look this one up on the net because I'm sure it's a popular one - try and figure it out!

That one's kind of old.


I still like the 12 unmarked coins with 1 counterfeit and only a basic "heavy" "light" scale problem. That one required good old fashioned thinking power.

[swiftaw]

Originally Posted by CalBoy

That one's kind of old.


I still like the 12 unmarked coins with 1 counterfeit and only a basic "heavy" "light" scale problem. That one required good old fashioned thinking power.

Yeah, I like that one too, of course, the hard part is that you are restricted to a small number of weighings

[CalBoy]

Originally Posted by swiftaw

Yeah, I like that one too, of course, the hard part is that you are restricted to a small number of weighings

Yeah, but you knew which one I was talking about.

It took me a while to get that one the very first time I had heard it.

[Ish]

Originally Posted by Doctor Q

A bus has stopped and the driver and all passengers, pets, and other living creatures have gotten off, except that one centipede remains on the floor under a seat. On top of the same seat is a take-out food carton, in which a passenger accidentally left a turkey leg and a lucky rabbit's foot.

How many legs are on the bus?

All the buses I've seen run on wheels.

[Unspeaked]

Originally Posted by Ish

All the buses I've seen run on wheels.

You've never watched the Flintstones?

[gnasher729]

Originally Posted by CalBoy

There are 3 doors. Behind 1 door lies immense wealth and riches. Behind the other two lie certain death.

A guard asks you to choose a door and you pick #2. The guard then reveals to you what's behind Door #1 and you see that behind it was one of two chances of certain death.

The guard now asks you if you would like to change your selection from Door #2 to Door #3.

Mathematically is there a probability increase that benefits you if you change doors?

This problem comes up regularly on sci.math. Unfortunately, we don't have enough information about the rules to be able to answer. I'll assume that your choices as the player are to either take what is behind door #2, or to take what is behind door #3.

The first question is: What possible choices did the guard have when he opened the door? Did he have the choice to open a door or not open a door? Did he have the choice to open a door that showed wealth (and you are stuck with death ), did he have the choice to show you door #3 or door #1?

Someone posted that you should always switch. Now assume the guard really doesn't like you. Here is what he does: If your door holds wealth, he opens a door showing death. If your door holds death, he opens a door showing wealth. If he is allowed to do this, and you follow the advice you were given if possible, you will certainly die.

Another possibility: If your door holds wealth, he opens a door showing death. If your door holds death, he doesn't open a second door. So if he opens a door and you follow the advice, you die. But if you know that this is the strategy he follows, then you can beat him: If he opens a door sowing death, you stay with your choice. Otherwise you switch. Now your chances are one in three to win.

[Ish]

Originally Posted by Unspeaked

You've never watched the Flintstones?

Not voluntarily!

[DALLASFAN]

10,990

[Jade Cambell]

That's a tricky math problem? Wow... I feel like i'm losing IQ points as i'm writing this out but: 14 girls legs, 1372 big cat legs, and 9604 little cat legs. Total = 10990 legs.


edit: Nevermind. I looked at the original post again and it says it's a 5th grade math problem. Now i'm not so confused anymore.

[KINABUCH]

Originally Posted by EricNau

I love this one:

Give the next two numbers in this sequence:

9, 10, 11, 12, 13, 14, 21, ...

i guess 28?

[CityGurl]

Originally Posted by Frisco

Just for fun, not for class. I can't get it.

Good luck with this and have fun! This is a 5th grade math problem.
If you can't stand word math problems, just delete now. If you can open the
spreadsheet, you'll see it's a very small list of people who have gotten
the correct number. This is not a trick question. This is a real math
problem so don't say that a bus has no legs.


There are 7 girls in a bus

Each girl has 7 backpacks

In each backpack, there are 7 big cats

For every big cat there are 7 little cats

Question: How many legs are there in the bus?


The number of legs is the password to unlock the Excel sheet. (Do not have to spell out #)
If you open it.

Spreadsheet

8

[CitiXen]

Originally Posted by Doctor Q

That puzzle sounds like yet another variation on this classic, thought to have originated in 1650 BC!

How 'bout this puzzle? -

A bookshelf has three encyclopedia volumes: A-I, J-R, and S-Z. They are in the usual order, left to right on the shelf. The covers are 1/16" thick. Each book has pages numbered 1 to 1000. Each sheet of paper is 1/250 of an inch thick.

If a bookworm chews its way from page 1 of the A-I volume through page 1000 of the S-Z volume, how far did it travel?

(You may decide whether or not the bookworm ate the starting page and the ending page; use whichever assumption makes your computation easier.)

If it's eating through the 1,000th page it's not eating through the last book cover (atleast it's not stated in the question). So it would be 1/16" less distance.

[fireshot91]

How would 1+1 equal a window?





Only one I know

[AngryApple]

Originally Posted by fireshot91

How would 1+1 equal a window?





Only one I know

Impossible. You need an equal sign at the end to do that! That's an old one. The ones are the left and right sides of the window and the equal sign is the top and bottom. The plus is that middle thing that some people have in their windows. I don't have em so I dunno the name.

[Doctor Q]

Originally Posted by AngryApple

The plus is that middle thing that some people have in their windows. I don't have em so I dunno the name.

It's called a muntin.

[Frisco]

Wiki Answers has the solution

[Doctor Q]

Originally Posted by Frisco

Wiki Answers has the solution

I'm not sure about that solution. For some reason it seems to assume that all children are girls.

[Shaun.P]

Originally Posted by CalBoy

Ok, since you asked.

There are 3 doors. Behind 1 door lies immense wealth and riches. Behind the other two lie certain death.

A guard asks you to choose a door and you pick #2. The guard then reveals to you what's behind Door #1 and you see that behind it was one of two chances of certain death.

The guard now asks you if you would like to change your selection from Door #2 to Door #3.

Mathematically is there a probability increase that benefits you if you change doors?

BTW, I don't know the answer to this, but I've seen it thrown around a few times

Yes - there is a probability increase if you switch. The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3.

[Doctor Q]

Originally Posted by ShaunPriest

Yes - there is a probability increase if you switch. The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3.

I disagree.

The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3. But... keeping door #2 will also give you odds of 1 in 2 rather than 1 in 3. So your chances improved when you were told what's behind door #1, but your chances won't improve further by switching from door #2 to door #3.

You can also see this by enumerating the cases, based on assumption that the three arrangements are equally likely:

Case 1: {wealth and riches, certain death, certain death}
Case 2: {certain death, wealth and riches, certain death}
Case 3: {certain death, certain death, wealth and riches}

Case 1 is ruled out in the problem statement. So it's Case 2 or Case 3 with equal probability. You pick door #2. Door #1 is shown to be certain death. If you change to door #3, you win half the time (Case 3) and lose half the time (Case 2). If you stick with door #2, you win half the time (Case 2) and lose half the time (Case 3).

It could be that there are two types of certain death, making 6 equally likely cases:

Case 1: {wealth and riches, certain death by tickling, certain death by accordion music}
Case 2: {wealth and riches, certain death by accordion music, certain death by tickling}
Case 3: {certain death by tickling, wealth and riches, certain death by accordion music}
Case 4: {certain death by accordion music, wealth and riches, certain death by tickling}
Case 5: {certain death by tickling, certain death by accordion music, wealth and riches}
Case 6: {certain death by accordion music, certain death by tickling, wealth and riches}

But you get the same answer. Cases 1 and 2 are ruled out in the problem statement. So it's Case 3, 4, 5, or 6 with equal probability. You pick door #2. Door #1 is shown to be certain death by either tickling or accordion music. If you change to door #3, you win half the time (Cases 5 and 6) and lose half the time (Cases 3 and 4). If you stick with door #2, you win half the time (Cases 3 and 4) and lose half the time (Cases 5 and 6).

[Shaun.P]

Originally Posted by Doctor Q

I disagree.

The odds of picking the right door initially is 1 in 3. You are more likely to pick certain death than you are the correct answer. If another door is opened and it is certain death, switching your bet would give you odds in 1 in 2 rather than 1 in 3. But... keeping door #2 will also give you odds of 1 in 2 rather than 1 in 3. So your chances improved when you were told what's behind door #1, but your chances won't improve further by switching from door #2 to door #3.

You can also see this by enumerating the cases, based on assumption that the three arrangements are equally likely:

Case 1: {wealth and riches, certain death, certain death}
Case 2: {certain death, wealth and riches, certain death}
Case 3: {certain death, certain death, wealth and riches}

Case 1 is ruled out in the problem statement. So it's Case 2 or Case 3 with equal probability. You pick door #2. Door #1 is shown to be certain death. If you change to door #3, you win half the time (Case 3) and lose half the time (Case 2). If you stick with door #2, you win half the time (Case 2) and lose half the time (Case 3).

It could be that there are two types of certain death, making 6 equally likely cases:

Case 1: {wealth and riches, certain death by tickling, certain death by accordion music}
Case 2: {wealth and riches, certain death by accordion music, certain death by tickling}
Case 3: {certain death by tickling, wealth and riches, certain death by accordion music}
Case 4: {certain death by accordion music, wealth and riches, certain death by tickling}
Case 5: {certain death by tickling, certain death by accordion music, wealth and riches}
Case 6: {certain death by accordion music, certain death by tickling, wealth and riches}

But you get the same answer. Cases 1 and 2 are ruled out in the problem statement. So it's Case 3, 4, 5, or 6 with equal probability. You pick door #2. Door #1 is shown to be certain death by either tickling or accordion music. If you change to door #3, you win half the time (Cases 5 and 6) and lose half the time (Cases 3 and 4). If you stick with door #2, you win half the time (Cases 3 and 4) and lose half the time (Cases 5 and 6).

Sorry, I mixed up my odds slightly, but the fact remains your chances double when you switch.
http://www.youtube.com/watch?v=P9WFKmLK0dc

[Doctor Q]

Originally Posted by ShaunPriest

Sorry, I mixed up my odds slightly, but the fact remains your chances double when you switch.
http://www.youtube.com/watch?v=P9WFKmLK0dc

Sorry, still not true. The odds are the same whether you switch doors or not, in the problem stated above. The odds might change if the other party gave you new information, e.g., by picking which door to show you based on already knowing what's behind them. That's not the case here, where what's revealed to you is predetermined to be door #1.

As additional confirmation, I wrote a test program based on random numbers and ran one million trials eight times, four of them keeping door #2 and four of them switching to door #3.

Here are my results:

If I kept the same door, I won this percentage of the time in the four trial runs: 49.9995%, 50.0665%, 50.0933%, 50.0150%

If I changed doors, I won this percentage of the time in the four trial runs: 50.0103%, 50.0423%, 49.8875%, 50.0214%

In other words, 50% either way.