MacRumors Forums Can anyone get this tricky math problem?
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Feb 9, 2008, 11:27 PM   #101
swiftaw
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Quote:
 Originally Posted by queshy is the answer to the OPs problem 10990?
it is
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 Feb 9, 2008, 11:30 PM #102 queshy macrumors 68040     Join Date: Apr 2005 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! 0
Feb 9, 2008, 11:33 PM   #103
swiftaw
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Quote:
 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
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Feb 10, 2008, 12:54 AM   #104
queshy
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Quote:
 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!
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Feb 10, 2008, 02:54 AM   #105
CalBoy
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Quote:
 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.
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Feb 10, 2008, 02:55 AM   #106
swiftaw
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Quote:
 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
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Feb 10, 2008, 02:58 AM   #107
CalBoy
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Quote:
 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.
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Feb 11, 2008, 10:20 AM   #108
Ish
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Quote:
 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.
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Feb 11, 2008, 10:48 AM   #109
Unspeaked
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Quote:
 Originally Posted by Ish All the buses I've seen run on wheels.
You've never watched the Flintstones?
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Feb 11, 2008, 10:53 AM   #110
gnasher729
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Quote:
 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.
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Feb 12, 2008, 02:43 AM   #111
Ish
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Quote:
 Originally Posted by Unspeaked You've never watched the Flintstones?
Not voluntarily!
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 Feb 19, 2008, 11:56 AM #112 DALLASFAN macrumors newbie   Join Date: Feb 2008 The Answer Is.... 10,990 0
 Feb 19, 2008, 02:20 PM #113 Jade Cambell Banned   Join Date: Sep 2007 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. 0
Dec 4, 2008, 12:01 PM   #114
KINABUCH
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Quote:
 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?
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Mar 12, 2009, 09:39 AM   #115
CityGurl
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Quote:
 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
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Mar 12, 2009, 02:44 PM   #116
CitiXen
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Quote:
 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.
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 Mar 12, 2009, 03:57 PM #117 fireshot91 macrumors 601     Join Date: Aug 2008 Location: Northern VA How would 1+1 equal a window? Only one I know 0
Mar 12, 2009, 10:50 PM   #118
AngryApple
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Quote:
 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.
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Mar 13, 2009, 02:02 PM   #119
Doctor Q

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Quote:
 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.
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 Mar 13, 2009, 02:46 PM #120 Frisco Thread Starter macrumors 68020   Join Date: Sep 2002 Location: Jersey Wiki Answers has the solution __________________ Frisco 0
Mar 13, 2009, 03:15 PM   #121
Doctor Q

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Quote:
 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.
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Mar 13, 2009, 03:32 PM   #122
Shaun.P
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Quote:
 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.
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Last edited by Shaun.P; Mar 13, 2009 at 04:48 PM.
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Mar 13, 2009, 04:15 PM   #123
Doctor Q

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Quote:
 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).
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Mar 13, 2009, 04:45 PM   #124
Shaun.P
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Quote:
 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.
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Mar 13, 2009, 05:47 PM   #125
Doctor Q

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Quote:
 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.
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