Thread: Logic Puzzles

I am a fan of logic puzzles; problems that, once you know the answer, are sooooooo simple, yet at times very difficult to solve. I don't mean " if two trains.." or "if my aunt is my mother's brother's barber...", but something like the following:


(This is one I used to hear when applying for programming jobs)

There is a cake, with one piece already cut from it. How would you cut the remaining cake in half ?

Everything you need to know is in those two sentences. There is no 'trick', just a simple answer. If you like, I can post more later. Oh, if after a week or so, no one gets the answer, for a small fee I will post it!

Enjoy, and please post your favs.

With a knife, probably horizontally, starting at the midpoint of one side.

With a knife?

(You usually cut cakes with a knife, don’t you? )

Originally Posted by JaneBennet

With a knife?

How would you cut the remaining cake in half

The question is not specific as to the method or instrument to be used.

Draw a line through the cake that halves the angle created by the missing piece. Cut...along...it :? ? Can it be as simple as that?

Originally Posted by KALSTER

Draw a line through the cake that halves the angle created by the missing piece. Cut...along...it :? ? Can it be as simple as that?

The questions do not mention the shape of the piece that was cut from the cake. They don’t even state what the shape of the cake is! You’re only assuming the cake is round and the piece cut from it is in the form of wedge.

Originally Posted by zinjanthropos

With a knife, probably horizontally, starting at the midpoint of one side.

Yep. Most people do not think that a cake has 3 dimentions, including depth. By cutting the depth in half, we are left with two equal parts.

No other information is required, i.e. method of cutting etc. It is a simple test for 'lateral thinking'

Originally Posted by CShark

Originally Posted by zinjanthropos

With a knife, probably horizontally, starting at the midpoint of one side.

Yep. Most people do not think that a cake has 3 dimentions, including depth. By cutting the depth in half, we are left with two equal parts.

No other information is required, i.e. method of cutting etc. It is a simple test for 'lateral thinking'

What if the cake was shaped like a pyramid, or some irregular 3D shape? Cutting horizontally still won’t work.

Good point Jane :wink: .

Assuming "cut in half" means geometrically:
So try and determine an axis of symmetry and cut along it, halving the missing bit.

Assuming "cut in half" means volumetrically:
Build a mold, submerge in water to determine volume, cut a regular shape (cube) out that equates to half the volume.

Another one of those questions where it is a perfectly reasonable question to ask to a programmer, but gets (rightly) complicated by a scientifically inclined person's scrutiny, like the plane-on-conveyorbelt question.

I found a more well-posed version of this question. Suppose both the original cake and the piece removed are rectangular--how can you cut the cake in half with one swipe of a knife? And let's look for a different (or more general) solution than cutting horizontally.

All right, let’s analyse the problem a little. Two points:

(i) Cakes usually have uniform horizontal cross sections. Are we allowed to assume this for this problem?

(ii) When you cut a piece from a cake, you usually do so with slices parallel to the vertical sides of the cake (you don’t e.g. slice it obliquely); the slices also run all the way from the top to base. Are we allowed to assume this as well?

If we can assume both (i) and (ii), then Zinjanthropos’s solution would be perfect. But …

Originally Posted by CShark

Everything you need to know is in those two sentences.

And CShark didn’t mention either of the two points above. :?

As for me, I think the “in half� part is a bit of a red herring. I merely read the second statement as “how would you cut the remaining cake� – full stop (or rather question mark). A bit like Alexander and the Gordian knot, really. :P

Originally Posted by serpicojr

I found a more well-posed version of this question. Suppose both the original cake and the piece removed are rectangular--how can you cut the cake in half with one swipe of a knife? And let's look for a different (or more general) solution than cutting horizontally.

I didn’t see your post while posting the above. Well, this makes the problem much more tractable. :-D

I presume “rectangular� means that the shape is a hexahedron* with three pairs of parallel faces, the different pairs being mutually perpendicular and all the faces being rectangular, right? Just want to be absolutely certain.

*hexahedron = solid figure with exactly 6 plane faces.

Originally Posted by serpicojr

I found a more well-posed version of this question. Suppose both the original cake and the piece removed are rectangular--how can you cut the cake in half with one swipe of a knife? And let's look for a different (or more general) solution than cutting horizontally.

Ok, someone else can do the actual math (I don't know it ). Formulate an equation to work out the area of each half, i.e. area of triangle + area of quadrangle and then equate them to each other:

Area of A = Area of B

and then limits again? :?

Originally Posted by JaneBennet

All right, let’s analyse the problem a little. Two points:

(i) Cakes usually have uniform horizontal cross sections. Are we allowed to assume this for this problem?

(ii) When you cut a piece from a cake, you usually do so with slices parallel to the vertical sides of the cake (you don’t e.g. slice it obliquely); the slices also run all the way from the top to base. Are we allowed to assume this as well?

If we can assume both (i) and (ii), then Zinjanthropos’s solution would be perfect. But …

Originally Posted by CShark

Everything you need to know is in those two sentences.

And CShark didn’t mention either of the two points above. :?

I did not realize this would be picked apart as much as it has been . You are correct: I did not state the (what I believe to be obvious) fact that the cake was not an irregular shape, nor that the missing piece was indeed cut at precisely 90degrees to the horizontal....it was a simple brain tester, nothing more.

An example of my rant would be, say, the math question pertaining to two trains. Nothing states that each train is identical in every way, nor that the tracks are exactly the same in terms of condition, friction, etc. There is nothing to be gained by adding speculation and unneeded detail to a trivial time-waster.

I will think twice before posting more of these...

Originally Posted by CShark

There is nothing to be gained by adding speculation and unneeded detail to a trivial time-waster.

Problems like these are not time-wasters. A lot can be learnt from them – at any rate, from the way people generally go about tackling them. They test the mettle of mathematicians, and also provide psychologists with insights into human problem-solving processes.

Please post more of these problems. :-D When I get used to the way you present your problems, I won’t need to be so anally pedantic again.

Also, you would HAVE to assume that you would be able to take measurements, as you can't cut anything in half without it. With serpicojr's example it is the easyest.

The cut-out piece of CShark though can have any shape and can be quite complicated (barring the horizontal option, which was the answer).

[quote="JaneBennet"]

Originally Posted by CShark

Please post more of these problems. :-D When I get used to the way you present your problems, I won’t need to be so anally pedantic again.

Sorry to be so sensitive (sniff, sniff...) must be my time of the month.

Ok, just remember, take these at their face value. There is no need to get out the sliderule or work out complex calculus; just have fun!

Originally Posted by KALSTER

Also, you would HAVE to assume that you would be able to take measurements, as you can't cut anything in half without it. With serpicojr's example it is the easyest.

The cut-out piece of CShark though can have any shape and can be quite complicated (barring the horizontal option, which was the answer).

NONONONONO! you don't need to take measurments. See 'a cake'. See a missing piece (how people normally cut a piece of cake, not something strange). Now, by cutting the remaining cake laterrally, you have two equal parts. No math required.

Ok, as requested, I will try this again. Please do not read any more into this than there is.

You are standing in front of a closed door, which leads into a room. There is no way to see into the room until you open the door (no cheating, no keyhole, no way). On the wall by the door (still outside the room) you see a light switch, with three (3) switches.

In the room there are three light bulbs. One switch controls one bulb. Don't forget, you cannot see into the room, nor the light bulbs, until you open the door.

You can turn on or off any or all the switches you like, until you open the door. Once you open the door, you cannot go back out and play with the switches.

Your task is to figure out which switch activates which light bulb.

Originally Posted by CShark

Originally Posted by KALSTER

Also, you would HAVE to assume that you would be able to take measurements, as you can't cut anything in half without it. With serpicojr's example it is the easyest.

The cut-out piece of CShark though can have any shape and can be quite complicated (barring the horizontal option, which was the answer).

NONONONONO! you don't need to take measurments. See 'a cake'. See a missing piece (how people normally cut a piece of cake, not something strange). Now, by cutting the remaining cake laterrally, you have two equal parts. No math required.

Yes, I got that and agreed to as much. :wink:

It is only when you want to look deeper into the problem (a real-world scientific approach) would all the complication be necessary.

Originally Posted by KALSTER

Yes, I got that and agreed to as much. :wink:

It is only when you want to look deeper into the problem (a real-world scientific approach) would all the complication be necessary.

I should have thought of that before I posted in a scientific forum ! My fault (live and learn).

Flip the first and the second switch. Wait a while and then turn off the second switch. Enter the room. One bulb will burn (first switch), one will be hot, but off (second switch) and one will be off and cold (third switch).

Originally Posted by KALSTER

Flip the first and the second switch. Wait a while and then turn off the second switch. Enter the room. One bulb will burn (first switch), one will be hot, but off (second switch) and one will be off and cold (third switch).

Well done.

Do you have any of these ?

No, are they on a website?

Originally Posted by KALSTER

No, are they on a website?

Not that I am aware of. I just picked these up over the years. I have a few more, but will wait a bit to see if anyone else picks this up.

In the meantime, I have a simple one.

How can this equation be true? 111 − 1 = 11.

Originally Posted by JaneBennet

In the meantime, I have a simple one.

How can this equation be true? 111 − 1 = 11.

Without googling I'd say you're talking Roman Numerals.

Well, I don’t think Googling would be much help because I made the puzzle up myself.

But you’re right, they are Roman numerals. :-D

Ok ,once more here

If you are in a bus and have 7 cm golden bar , you must give the driver 1 cm at every station , you have only 2 saws which are used only for one time ????

I suggest the competition to be a chain : Who solve the puzzle give the question , what is your opinion?

Saw the bar into three pieces of 1*cm, 2*cm, and 4*cm respectively. Now you can make n*cm from any combination of these three pieces, where n ∈ {0,1,2,3,4,5,6,7}.

In general, if you have k numbers 1, 2, 4, …, 2^k−1, you can make any number n, where n is an integer from 0 to 2^k−1, from the sum of any combination of them.

Originally Posted by raed

Ok ,once more here

If you are in a bus and have 7 cm golden bar , you must give the driver 1 cm at every station , you have only 2 saws which are used only for one time ????

I suggest the competition to be a chain : Who solve the puzzle give the question , what is your opinion?

Get off after the second station and keep the remaining 5 cm.

Save your gold for a better saw.

Ok

Jene is right

Originally Posted by JaneBennet

Saw the bar into three pieces of 1Â*cm, 2Â*cm, and 4Â*cm respectively. Now you can make nÂ*cm from any combination of these three pieces, where n ∈ {0,1,2,3,4,5,6,7}.

In general, if you have k numbers 1, 2, 4, …, 2^k−1, you can make any number n, where n is an integer from 0 to 2^k−1, from the sum of any combination of them.

Well done! Tell me Jane, are you a math major ? (serious question)

Yes, I am.

One final comment on the last puzzle: the fact that any positive integer can be expressed as a sum of powers of 2 (or is a power of 2 itself) is the reason why the binary (base-2) system works. 8)

Originally Posted by JaneBennet

Saw the bar into three pieces of 1 cm, 2 cm, and 4 cm respectively. Now you can make n cm from any combination of these three pieces, where n ∈ {0,1,2,3,4,5,6,7}.

In general, if you have k numbers 1, 2, 4, …, 2^k−1, you can make any number n, where n is an integer from 0 to 2^k−1, from the sum of any combination of them.

Very clever, but how does that allow you to give the driver 1 cm at each station?

You give the driver a net total of 1*cm gold every station.

Originally Posted by Harold14370

Originally Posted by JaneBennet

Saw the bar into three pieces of 1 cm, 2 cm, and 4 cm respectively. Now you can make n cm from any combination of these three pieces, where n ∈ {0,1,2,3,4,5,6,7}.

In general, if you have k numbers 1, 2, 4, …, 2^k−1, you can make any number n, where n is an integer from 0 to 2^k−1, from the sum of any combination of them.

Very clever, but how does that allow you to give the driver 1 cm at each station?

Have to agree with Harold on that one.

Originally Posted by zinjanthropos

Originally Posted by Harold14370

Originally Posted by JaneBennet

Saw the bar into three pieces of 1 cm, 2 cm, and 4 cm respectively. Now you can make n cm from any combination of these three pieces, where n ∈ {0,1,2,3,4,5,6,7}.

In general, if you have k numbers 1, 2, 4, …, 2^k−1, you can make any number n, where n is an integer from 0 to 2^k−1, from the sum of any combination of them.

Very clever, but how does that allow you to give the driver 1 cm at each station?

Have to agree with Harold on that one.

Perhaps I can help: you may be assuming the passenger has to pay 1 cm at each stop. He only pays once, when he reaches his destination; hence, by having 1, 2, and 4 cm pieces, he can 'give exact change' for any trip from 0 to 7 stops.

You can pay at each stop. The point is you can always get change from the bus driver:

You will have 1 cm , 2 cm and 4 cm pieces

Station 1: give him 1 cm
Station 2 : give him 2 cm and take the 1 cm piece
Station 3 : give him the 1 cm piece in addition to 2 cm piece
Station 4 : give him 4 cm piece and take the 3 cm piece
Station 5 : give him 1 cm piece in addition to 4 cm piece
Station 6 : give him 2 cm and take the 1 cm piece
Station 7 : give him the 1 cm piece

Originally Posted by serpicojr

You can pay at each stop. The point is you can always get change from the bus driver:

(slaps himself) Change! I never thought about change!

Originally Posted by serpicojr

You can pay at each stop. The point is you can always get change from the bus driver:

Gotcha!

How do you know he didn't give your cm in change to somebody else?

Don't mind me, I'm just being a PITA.

Easy....from a personal experience. Hope I don`t screw up but what the hell.

A group sold eight thousand 50-50 tickets hoping to raise 4 G's. The tickets were numbered consecutively starting from #0001. The digits are as they appear here1234567890). The draw was made and much to the organizer's surprise they had to pay out 2 winning tickets. What was the winning number, which by the way does not contain a zero nor does it contain consecutive repeating numbers (i.e....11,222,3333, etc).

I think I`ve covered everything.

6989/6869

Originally Posted by serpicojr

6989/6869

I should have also said no 8's but that will do. I had 6969.

6969 is not a solution. When you rotate it, you get... 6969. So only one person has a ticket that can be interpreted as 6969. If you impose the condition that there is no 8, then there are no solutions. Let me show you how my solution is unique assuming 8's are cool.

1. Consider only digits which look like another digit when you rotate 180 degrees. These are 0, 6, 8, and 9. The rules say 0 isn't cool, so we're stuck with 6, 8, and 9.

2. Only 8000 tickets were sold, starting at 0001, and so our numbers must be less than 8000. Thus they must both start with 6, and so our numbers are of the form 6xx9.

3. The next digit after 6 cannot be a 6 by the rules, so try the other possibilities.

a. If the second digit is an 8, then the 3rd digit must be a 6 by the rules. So we get 6869. Flipping, we get 6989.

b. If the second digit is a 9, then the 3rd digit can be 6 or 8. If it's 6, we run into the problem above. So it must be an 8, and we again get 6989.

Originally Posted by serpicojr

assuming 8's are cool

What about 1’s? :?

I guess I was assuming the 1 has serifs. I'm going to guess that 1's are not allowed, as that would destroy uniqueness of the solution.

Well, since no one is adding a new puzzle, I will jump in with a very simple one.

A policeman is told to arrest a fireman for stealing from fire victims. He is told the fireman's name, Joe, and the fact that Joe is playing poker with three others, a lawyer, a bank manager, and a baseball player.

As soon as the cop enters the poker room, he immediately walks over and places Joe under arrest.

How does the cop know which poker player is Joe ?

Originally Posted by CShark

Well, since no one is adding a new puzzle, I will jump in with a very simple one.

A policeman is told to arrest a fireman for stealing from fire victims. He is told the fireman's name, Joe, and the fact that Joe is playing poker with three others, a lawyer, a bank manager, and a baseball player.

As soon as the cop enters the poker room, he immediately walks over and places Joe under arrest.

How does the cop know which poker player is Joe ?

Hmmm.....perhaps a hint ? Nah...I'll give it a few more days.

Originally Posted by CShark

Well, since no one is adding a new puzzle, I will jump in with a very simple one.

A policeman is told to arrest a fireman for stealing from fire victims. He is told the fireman's name, Joe, and the fact that Joe is playing poker with three others, a lawyer, a bank manager, and a baseball player.

As soon as the cop enters the poker room, he immediately walks over and places Joe under arrest.

How does the cop know which poker player is Joe ?

Are the other three women?

Originally Posted by Harold14370

Originally Posted by CShark

Well, since no one is adding a new puzzle, I will jump in with a very simple one.

A policeman is told to arrest a fireman for stealing from fire victims. He is told the fireman's name, Joe, and the fact that Joe is playing poker with three others, a lawyer, a bank manager, and a baseball player.

As soon as the cop enters the poker room, he immediately walks over and places Joe under arrest.

How does the cop know which poker player is Joe ?

Are the other three women?

Nice work Harold!

Women can be named Jo.

Originally Posted by CShark

There is a cake, with one piece already cut from it. How would you cut the remaining cake in half ?

Everything you need to know is in those two sentences. There is no 'trick', just a simple answer. If you like, I can post more later. Oh, if after a week or so, no one gets the answer, for a small fee I will post it!

Enjoy, and please post your favs.

'Piece of cake...'

Cut it with a blender and weigh out two equal portions. Then the consumers can drink their share.


Cheers,
william

Originally Posted by serpicojr

Women can be named Jo.

Yes but they would probably be called firefighters instead of firemen,

Next problem:
A creeper plant is climbing up and around a cylindrical tree trunk in a helical manner. The tree trunk has a height of 720 inches and a circumference of 48 inches.

If the creeper covers a vertical distance of 90 inches in one complete twist around the tree trunk, what is the total length of the creeper?

Originally Posted by Harold14370

Originally Posted by serpicojr

Women can be named Jo.

Yes but they would probably be called firefighters instead of firemen,

Next problem:
A creeper plant is climbing up and around a cylindrical tree trunk in a helical manner. The tree trunk has a height of 720 inches and a circumference of 48 inches.

If the creeper covers a vertical distance of 90 inches in one complete twist around the tree trunk, what is the total length of the creeper?

720/90 * 48 = 384

Circumference = pi.D = 150.8"
Helix angle = arctan (90/150.8 ) = 30.8 degrees
Length of vine per rotation = 150.8/cos 30.8 = 175.5
No. of turns = 720/90 = 8
So length of creeper = 8 x 175.5 = 1404.5"

(assuming it goes to the top, which was not stated)

Bunbury, the circumference is 48.

If you unwrap the cylinder of a 90in vertical section and use pythagoras, you get that the creeper length is 102in per 90in vertical section. (720/90)*102 gets 816in.

So the vine is described by the parametric equations:

x(t) = 24cos(t)/π
y(t) = 24sin(t)/π
z(t) = 45t/π

with 0 ≤ t ≤ 16π (8 twists), and so we integrate from 0 to 16π the quantity:

sqrt((x'(t))²+(y'(t))²+(z'(t))²)
= sqrt(24²/π²+45²/π²)
= 51/π

So we get 16π*51/π = 816.

Jeez, guess who is a math wizz! :-D

Shouldn’t z(t) be (45*⁄*π)*×*t?

Bunbury, the circumference is 48.

oops read too fast.

Originally Posted by JaneBennet

Shouldn’t z(t) be (45Â*⁄Â*π)Â*×Â*t?

It is, what're you talking about?

Sorry, I thought you wrote 45Â*⁄Â*π. I read too fast as well.

Ok, so just to be clear, the 816in I got first was correct?

Originally Posted by KALSTER

Ok, so just to be clear, the 816in I got first was correct?

Yes, but you don't get the extra credit. You did it the easy way.

Originally Posted by Harold14370

Originally Posted by KALSTER

Ok, so just to be clear, the 816in I got first was correct?

Yes, but you don't get the extra credit. You did it the easy way.

The only way I know how :|

Originally Posted by serpicojr

So the vine is described by the parametric equations:

x(t) = 24cos(t)/π
y(t) = 24sin(t)/π
z(t) = 45t/π

with 0 ≤ t ≤ 16π (8 twists), and so we integrate from 0 to 16π the quantity:

sqrt((x'(t))²+(y'(t))²+(z'(t))²)
= sqrt(24²/π²+45²/π²)
= 51/π

So we get 16π*51/π = 816.

Impressive!

Ready for next puzzle?

Same as Raed’s problem earlier, except that you now have 11 cm of gold bar. The driver still demands an additional cm of gold every stop, and you have the same two saws as before (each saw can only be used once). How do you make two cuts and keep the driver happy?

Originally Posted by JaneBennet

Ready for next puzzle?

Same as Raed’s problem earlier, except that you now have 11 cm of gold bar. The driver still demands an additional cm of gold every stop, and you have the same two saws as before (each saw can only be used once). How do you make two cuts and keep the driver happy?

Cut at the 4 cm mark. Then lay the two pieces side by side and cut the 4 cm piece into pieces of 1 and 3 and cut the 7 cm piece into 2 and 5 You now have pieces of 1, 2, 3, and 5 cm.
Stop 1 pay 1 cm
Stop 2 pay 2 cm get 1
Stop 3 pay 3 get 2
Stop 4 pay 1
Stop 5 pay 5 get 1+3
Stop 6 pay 1
Stop 7 pay 2 get 1
Stop 8 pay 3 get 2
Stop 9 pay 1
stop 10 pay 2 get 1
stop 11 pay 1

You got it! :-D

There are other solutions, e.g. cutting into pieces of 1*cm, 1*cm, 3*cm, and 6*cm also works. But the idea is the same: cut two pieces together with the second saw. 8)

Okay, here's the next one. You may have seen it before, the Monty Hall problem. You are contestant on Let's Make a Deal. Monty offers you a pick of 3 doors. Behind one door is a new car, nothing worthwhile behind the other doors. You pickDoor 1. Monty reveals that there is nothing behind Door 2 and offers to let you switch to Door 3 or stick with 1. Can you improve your odds by switching?

I’ve seen that one before.

Originally Posted by Harold14370

Okay, here's the next one. You may have seen it before, the Monty Hall problem. You are contestant on Let's Make a Deal. Monty offers you a pick of 3 doors. Behind one door is a new car, nothing worthwhile behind the other doors. You pickDoor 1. Monty reveals that there is nothing behind Door 2 and offers to let you switch to Door 3 or stick with 1. Can you improve your odds by switching?

I know the answer to this one. But my brain gets totally messed up when explaining it. Whenever I see the explanation, I say "Yep. Right." But still I get confused, when I start thinking about it again.

Originally Posted by Harold14370

Okay, here's the next one. You may have seen it before, the Monty Hall problem. You are contestant on Let's Make a Deal. Monty offers you a pick of 3 doors. Behind one door is a new car, nothing worthwhile behind the other doors. You pickDoor 1. Monty reveals that there is nothing behind Door 2 and offers to let you switch to Door 3 or stick with 1. Can you improve your odds by switching?

Like some others, I've seen it before, so it would be unfair for me to explain. I'll leave it to someone for whom it's new.

FWIW: (Because this thing is like an itch you have to scratch,eh?) 'Tis better to switch, but the explanation as to why is not the simplest to grasp, and the problem has caused oodles of controversy even amongst professional mathematicians, statisticians and probabilists!

It is a pretty interesting problem! At first glance everyone will think it doesn't matter whether you stay or not. I cheated and took a look at the Wiki Page. They make it a bit easier by explaining that it makes a difference whether the host knows which is which and also that it becomes clear when 1 000 000 doors are considered, with you choosing one and the host opening 999 998 others.

I don't think I gave away too much as even Nobel laureates get this wrong!
I'll remove the clue if requested.

Originally Posted by Dishmaster

Originally Posted by Harold14370

Okay, here's the next one. You may have seen it before, the Monty Hall problem. You are contestant on Let's Make a Deal. Monty offers you a pick of 3 doors. Behind one door is a new car, nothing worthwhile behind the other doors. You pickDoor 1. Monty reveals that there is nothing behind Door 2 and offers to let you switch to Door 3 or stick with 1. Can you improve your odds by switching?

I know the answer to this one. But my brain gets totally messed up when explaining it. Whenever I see the explanation, I say "Yep. Right." But still I get confused, when I start thinking about it again.

This one bugged me the first time I heard it. I imagined a viewer at home playing along. Suppose the player picked door 1, and the viewer at home picked door 3. Let's say door 2 was opened. Then if each switched, it would seem the odds can't change.

However, if one creates a probability table with all the options, one sees that the odds are indeed increased by switching.

However, this doesn't explain the viewer at home playing along. That's another puzzle to solve.


Cheers,
william

However, this doesn't explain the viewer at home playing along. That's another puzzle to solve.

I'd say his odds increase as well by switching. :wink:

Originally Posted by william

This one bugged me the first time I heard it. I imagined a viewer at home playing along. Suppose the player picked door 1, and the viewer at home picked door 3. Let's say door 2 was opened. Then if each switched, it would seem the odds can't change.

However, if one creates a probability table with all the options, one sees that the odds are indeed increased by switching.

However, this doesn't explain the viewer at home playing along. That's another puzzle to solve.


Cheers,
william

I think the difference there is that Monty is going to reveal a door without the prize, and his choice will (sometimes) depend on which door the contestant picks, i.e. he won't reveal that one. He cannot react to a player playing from home.

The way I think of it is, on your initial pick you have 1 chance in 3 of picking the right door. That means there is 2/3 chance of the prize being behind one of the other 2 doors. That fact does not change when Monty opens door 2. Monty has just told you which of the two remaining doors has that 2/3 probability.

I agree with Harold, the presenters decision of which door to open depends on which door the player picks, and the presenter can't know which one you picked from home. Also there's a chance he could open the door you picked, then your odds would be 50-50.

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

i)4
ii)4
iii)400

Correct.

Originally Posted by JaneBennet

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Gosh how difficult is that? Not!

Originally Posted by Selene

Originally Posted by JaneBennet

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Gosh how difficult is that? Not!

Nope, the answer is not 40.

But don’t worry, you’re not the only one. Lots of other people whose wits are no quicker than yours also think the answer is 40.

There are two lengths of rope.
Each one can burn in exactly one hour.
They are not necessarily of the same length or width as each other.
They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.

By burning the ropes, how do you measure exactly 45 minutes worth of time?

Originally Posted by JaneBennet

Originally Posted by Selene

Originally Posted by JaneBennet

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Gosh how difficult is that? Not!

Nope, the answer is not 40.

Idiot

Originally Posted by Selene

Originally Posted by JaneBennet

Originally Posted by Selene

Originally Posted by JaneBennet

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Gosh how difficult is that? Not!

Nope, the answer is not 40.

Idiot

I take it that means “oh, I’m an idiot, why didn’t I think of that before�? Well, don’t worry. I agree with you.

Originally Posted by KALSTER

There are two lengths of rope.
Each one can burn in exactly one hour.
They are not necessarily of the same length or width as each other.
They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.

By burning the ropes, how do you measure exactly 45 minutes worth of time?

Light one end of one rope and both ends of the other rope simultaneously. The second rope (with both ends lit) will take 30 minutes to burn out. As soon as it burns out, light the unlit end of the first rope immediately. Now when that rope burns out, another 15 minutes will have passed. So you have 30 + 15 = 45 minutes altogether.

You got it.

We need more difficult ones!

There are six eggs in a basket

Six people take one of the eggs each

How is it that one egg can still be left in the basket?

Originally Posted by Selene

There are six eggs in a basket

Six people take one of the eggs each

How is it that one egg can still be left in the basket?

5 people took an egg from the basket and the 6th person took the basket with the egg inside

One of the people is a pregnant woman? :?

Originally Posted by KALSTER

One of the people is a pregnant woman? :?

Hee he he heee

You know I am kind of feeling hormonal tonight

and my tits are killing me

Originally Posted by Swetchy

Originally Posted by Selene

There are six eggs in a basket

Six people take one of the eggs each

How is it that one egg can still be left in the basket?

5 people took an egg from the basket and the 6th person took the basket with the egg inside

Sigh

If you say so

Originally Posted by Selene

Not quite

It's plane plain Jane

You are the idiot

Idiot

Let’s look at this objectively, shall we?

You looked at the question, and you immediately jumped to the conclusion that the answer must be 40. It turns out to be the wrong answer.

There. People make mistakes sometimes. Learn to learn from them!

Originally Posted by JaneBennet

Originally Posted by Selene

Originally Posted by JaneBennet

Originally Posted by Selene

Originally Posted by JaneBennet

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Gosh how difficult is that? Not!

Nope, the answer is not 40.

Idiot

I take it that means “oh, I’m an idiot, why didn’t I think of that before�? Well, don’t worry. I agree with you.

Not quite

It's plane plain Jane

You are the idiot

Idiot

Originally Posted by "Somebody

If 4 readers can read 4 books in 4 days …

(i) how many readers can read 40 books in 40 days?
(ii) in how many days can 40 readers read 40 books?
(iii) how many books can 40 readers read in 40 days?

Is the answer (i) 10
(ii) 4
(iii) 1600

i'm not sure!