To see just how smart some of y'all are, here are some brain teasers. This should be fun!!


I'll post the answers to these tomorrow.

1. Do they have a 4th of July in England?

2. If there are 7 months that have 31 days in them and 11 months that have 30 days in them, how many months have 28 days in them?


3. How many birthdays does the average man have?


4. What is boiled then cooled, sweetened then soured?


5. A woman gives a beggar 50 cents; the woman is the beggar's sister, but the beggar is not the woman's brother. How come?

6. How many outs are there in an inning?


7. What is brought to the table and cut, but never eaten?


8. Is it legal for a man in California to marry his widow's sister? Why?


9. What is neither inside a house nor outside a house, but no house would be complete without it?

10. Divide 30 by 1/2 and add 10. What is the answer?

To see just how smart some of y'all are, here are some brain teasers. This should be fun!!


I'll post the answers to these tomorrow.

1. Do they have a 4th of July in England?

I would hope so.

2. If there are 7 months that have 31 days in them and 11 months that have 30 days in them, how many months have 28 days in them?

All of them.


3. How many birthdays does the average man have?

One a year.



8. Is it legal for a man in California to marry his widow's sister? Why?

Because he is dead.


10. Divide 30 by 1/2 and add 10. What is the answer?

70

Ok, so I am only semi-smart. But I knew that already.

Lisa

To take a guess at a couple of these that Lisa didn't answer:

5. A woman gives a beggar 50 cents; the woman is the
beggar's sister, but the beggar is not the woman's brother. How come?

I'm guessing that the beggar is also a woman which would mean they are sisters and not sister/brother?

6. How many outs are there in an inning?

6, 3 for each team.

9. What is neither inside a house nor outside a house, but no house would be complete without it?

I'm not sure on this one, but it's probably something silly like a roof or a door or something. :D

Keep 'em coming folks. These are fun aren't they. Some of them are kinda tricky. I have the answers and it still took me a min to get a few of them. :D

7. What is brought to the table and cut, but never eaten?
The cheese? ;) :D

The cheese? ;) :D

Ewwwww:o :rolleyes:

1. Do they have a 4th of July in England?
Yes every 4th of July

2. If there are 7 months that have 31 days in them and 11 months that have 30 days in them, how many months have 28 days in them?
Normally 12


3. How many birthdays does the average man have?
Most of the time one.


4. What is boiled then cooled, sweetened then soured?
Rich!!!


5. A woman gives a beggar 50 cents; the woman is the beggar's sister, but the beggar is not the woman's brother. How come?
The beggar is a woman

6. How many outs are there in an inning?

Normally 6 except if the Team that is behind is first in the 9th inning.

7. What is brought to the table and cut, but never eaten?

?????
8. Is it legal for a man in California to marry his widow's sister? Why?
Hummpt, unless it is like “The Bride Corpse”


9. What is neither inside a house nor outside a house, but no house would be complete without it?
????

10. Divide 30 by 1/2 and add 10. What is the answer?
70

7. What is brought to the table and cut, but never eaten?
A deal?

I don't know about the house question.
Nor do I have a clue what is boiled and cooled, sweetened and soured.

9. What is neither inside a house nor outside a house, but no house would be complete without it?

The house.

7. What is brought to the table and cut, but never eaten?

Cards?

To see just how smart some of y'all are, here are some brain teasers. This should be fun!!


I'll post the answers to these tomorrow.

1. Do they have a 4th of July in England?
Yes

2. If there are 7 months that have 31 days in them and 11 months that have 30 days in them, how many months have 28 days in them?
12


3. How many birthdays does the average man have?
One


4. What is boiled then cooled, sweetened then soured?
Sweet Ice Tea with Lemon


5. A woman gives a beggar 50 cents; the woman is the beggar's sister, but the beggar is not the woman's brother. How come?
Her Sister

6. How many outs are there in an inning?
6


7. What is brought to the table and cut, but never eaten?
Cards


8. Is it legal for a man in California to marry his widow's sister? Why?
Illegal to marry a corpse


9. What is neither inside a house nor outside a house, but no house would be complete without it?
Window

10. Divide 30 by 1/2 and add 10. What is the answer?

70


Way to go guys.

If you want more, I have about 20 more I can post. Just say the word.

Have a great weekend guys.

9. What is neither inside a house nor outside a house, but no house would be complete without it?
Window

Could have been a door too. :mad: :D

Could have been a door too. :mad: :D

Actually when a door is open it is either inside the house or outside. The window go straight up, unless you have those pivoting windows.

Unless you were meaning a sliding glass door. :p :D

Actually when a door is open it is either inside the house or outside. The window go straight up, unless you have those pivoting windows.

Unless you were meaning a sliding glass door. :p :D

We used to have those casement windows that swing out before we had our windows done. Thank goodness we got rid of those old things. :rolleyes:

We used to have those casement windows that swing out before we had our windows done. Thank goodness we got rid of those old things. :rolleyes:

Wow, those types are popular here. Says that it is easier to clean and maintain.

I personally like the windows with the blinds build in. Those kind look nifty to me.

Wow, those types are popular here. Says that it is easier to clean and maintain.

There isn't anything wrong with those types of windows per se, but our house was built in 1953 and those were the original windows and so were single pane aluminum and terribly energy inefficient. Just so I know we're talking about the same thing, I'm talking about the windows that have the hand crank you turn and the window swings out. The ones we have now are double hung, which are the ones that you can swing inside the house for easy cleaning. :)

I personally like the windows with the blinds build in. Those kind look nifty to me.

Yes, those are nifty but they're really friggin expensive. :eek:

There isn't anything wrong with those types of windows per se, but our house was built in 1953 and those were the original windows and so were single pane aluminum and terribly energy inefficient. Just so I know we're talking about the same thing, I'm talking about the windows that have the hand crank you turn and the window swings out. The ones we have now are double hung, which are the ones that you can swing inside the house for easy cleaning. :)



Yes, those are nifty but they're really friggin expensive. :eek:


Ohh, OK, I understand what you are saying now. Yes my Grandmothers house had those hand cranked kind. It was three horizontal windows and each cranked out to hope up.

Yes, those are very expensive, but awfully purtty. :D

Actually when a door is open it is either inside the house or outside. The window go straight up, unless you have those pivoting windows.

Unless you were meaning a sliding glass door. :p :D
It could have been the walls then too :cool:

It could have been the walls then too :cool:

I knew you would pipe in and try and complicate this...*giggles*. At least you read it. :p

I suppose you are correct.

Did you get any of the other Rich?? I am curious.

Did you get any of the other Rich?? I am curious.
Most of them had been answered before I saw it, however the answers to this question are all incorrect


How many outs are there in an inning?


There may not be any, it really depends on how good the bowler is :cool: :p

Most of them had been answered before I saw it, however the answers to this question are all incorrect

Ahh, they beatcha to it eh?

I'll give you one just for you honey. You're supper smart you should get this one.

A man builds a house rectangular in shape. All sides have southern exposure. A big bear walks by. What color is the bear? Why?


There may not be any, it really depends on how good the bowler is :cool: :p
Umm, OK, I don't get it!! Bowling doesn't have innings. It has frames. As well as there aren't any outs in bowling? Or are you speaking of Cricket?

White polar bear on North Pole.

Or are you speaking of Cricket?
Of course :cool:

A man builds a house rectangular in shape. All sides have southern exposure. A big bear walks by. What color is the bear? Why?

Well I hate to be pedantic here but it's impossible for a rectangular house to have southern exposure on all sides :confused:

duh Bloody drink!

Well I hate to be pedantic here but it's impossible for a rectangular house to have southern exposure on all sides :confused:


Not if you build it right on top of the North Pole (Thus the white Polar Bear)

Dave

Not if you build it right on top of the North Pole (Thus the white Polar Bear)

Dave
yes, I'd already posted that drink and logic don't mix :mad: :D

6. How many outs are there in an inning?
???

Sorry it is 10.... Has to be...., ;)

Most of them had been answered before I saw it, however the answers to this question are all incorrect




There may not be any, it really depends on how good the bowler is :cool: :p


You are not thinking of an over are you Rich??

Pardon my obsessive need to be acknowledged as having a perfectly acceptable answer, but....

I believe one can bring a deal to the table, and one can cut a deal.
I agree one can also do this with cards, but a deal would also fit the criteria.
:p

Polar Bear is the correct answer to the last one.

As for the 'deal' answer Tess, that is a good answer. I won't argue with you on that one deary!

yes, I'd already posted that drink and logic don't mix :mad: :D

Not so fast...

One night at Cheers, Cliff Clavin explained the "Buffalo Theory" to his buddy, Norm. "Well ya see, Norm, it's like this. A herd of buffalo can only move as fast as the slowest buffalo. And when the herd is hunted, it is the slowest and weakest ones at the back that are killed first. This natural selection is good for the herd as a whole, because the general speed and health of the whole group keeps improving by the regular killing of the weakest members. In much the same way, the human brain can only operate as fast as the slowest brain cells. Excessive intake of alcohol, as we know, kill brain cells. But naturally, it attacks the slowest and weakest brain cells first. In this way, regular consumption of beer eliminates the weaker brain cells, making the brain a faster and more efficient machine! That's why you always feel smarter after a few beers.
:D

This is why we all feel more intelligent after a few drinks. :D
But I was only half way through the second glass :mad: :D

But you are a Pom... Ha Ha

But you are a Pom...
Fortunately :cool: :p

Fortunately

for some, but not for others. ;)

What is it?

The buyer doesn't use it.
The user doesn't see it.
The seller doesn't care.

What is it?

The buyer doesn't use it.
The user doesn't see it.
The seller doesn't care.

I am sorry, it was double posted. You may delete one.

So may you. Hit "Edit", Hit "Delete", check the "Delete Message" option, and hit "Delete" again. :)

What is it?

The buyer doesn't use it.
The user doesn't see it.
The seller doesn't care.

a casket? :confused:

RIDDLE & ANSWER

There is something unusual about these words, see if you can figure it out. The answer is below..but don't peek until you've given it a good shot!



Assess

Banana

Dresser

Grammar

Potato

Revive

Uneven

Voodoo

OK, see if you can figure out what these words have in common........




Are you peeking or have you already given up?




Answer:

In all of the listed words, if you take the first letter, place it at the end of the word, and then spell the word backwards, it will be the same word. Did you figure it out?

At a recent local political caucus everyone shook hands with everyone else (politics being what it is). In total there were 91 handshakes. Can you tell me how many people attended the caucus?

At a recent local political caucus everyone shook hands with everyone else (politics being what it is). In total there were 91 handshakes. Can you tell me how many people attended the caucus?

Hidden Answer:

There were 13 people.

Hidden Answer:

There were 13 people.


Visible response: no, but close

I think it's

14 people

You're right, I did the math wrong. Been too long :p

At a recent local political caucus everyone shook hands with everyone else (politics being what it is). In total there were 91 handshakes. Can you tell me how many people attended the caucus?

Is it 10 people - don't know how to hide / unhide - make text white??

The answer is 14. The first person shakes hands with 13 others. The next with the 12 remaining after the one they already shook hands with and so on.

13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 +1 = 91

really should have paid more attention when at school.
How many babies were kissed ?

Okay, how about this one... and no searching on the net for it either, try to honestly explain your answer logically... Here we go.

You have in front of you, two mugs. Mug A is filled with 1 cup of coffee. Mug B is filled with 1 cup of tea. You take exactly 1 teaspoon of liquid from Mug A and place it into Mug B. You then take exactly 1 teaspoon of liquid from Mug B and place it into Mug A. Which mug contains more of it's original contents? Please explain your answers.

Okay, how about this one... and no searching on the net for it either, try to honestly explain your answer logically... Here we go.

You have in front of you, two mugs. Mug A is filled with 1 cup of coffee. Mug B is filled with 1 cup of tea. You take exactly 1 teaspoon of liquid from Mug A and place it into Mug B. You then take exactly 1 teaspoon of liquid from Mug B and place it into Mug A. Which mug contains more of it's original contents? Please explain your answers.

Mug A & B are 100% of Coffee and Tea

Place 1 teaspoon from A into B and you get a Mix of X Tea and Y Coffee in B.

Remove One teaspoon from B and place it in A you have just placed one Mixed spoon into A.

B rec'd 100% Coffee
A rec'd 99% Tea

assumption is A is Purer then B

ooopps Holds more of the Original Contents - B does.
as A had a teaspoon removed from a cup where as B had a teaspoon removed from a cup and 1 teaspoon. Some of the teaspoon's contents would have been from the teaspoon it rec'd from A

If my math is correct:

They have the same amount (which isn't what I thought it would be when I first read it). Using bigger numbers for ease of math, if both cups have 5 units and I take 1 from A to B and then back. After the first move, cup B has 5 units of tea one 1 coffee. When I take out 1 unit, it is 1 of 6 units or 16.6667%. That means I take out .833333 of tea and .166667 of coffee. 5 minus .833333 is 4.166667 of tea, and 4 plus .166667 is also 4.166667 of coffee. There's probably a better way of explaining that, but I'm hurrying to make dinner so my excuse is hunger-induced brain cramps. :p

Paul is correct.

But it also depends on the temperature of the contents of both mugs and if the surface area of the liquids are equal and remain constant and also assuming both are at the same barometric pressure and at the same ambient temperature.

Other factors could include the content of un-dissolved sugar which would reduce in volume and is likely to absorb heat as it dissolves thereby reducing the overall temperature of the container into which it is being transferred. This could then lead to a contraction of volume. However, if the temperature dropped below 4 C then the water content may start to expand thereby increasing the volume.

The arguments, so far, are based on volume, cup and teaspoon, and not mass.

I think it requires more testing at the local where Irish whisky may be pored into our favourite beverage not withstanding that Irish whisky may be our favourite beverage.
(Under that circumstance who would give a rat’s arse?)

More testing required… :D

The answer is 14. The first person shakes hands with 13 others. The next with the 12 remaining after the one they already shook hands with and so on.

13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 +1 = 91


That is correct. Another way to do it is like this:

With two people (A and B), there is one handshake
(A with B).

With three people (A, B, and C), there are three handshakes
(A with B and C; B with C).

With four people (A, B, C, and D), there are six handshakes
(A with B, C, and D; B with C and D; C with D).

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.

Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 91.
This is the quadratic equation n2+ n - 182 = 0. Solving for n, we obtain 13 as the answer and deduce that there were 14 people at the party.
Since 91 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 91. The last number that you entered (91) is n. n(13) + 1 = 14.

Paul is correct.

But it also depends on the temperature of the contents of both mugs and if the surface area of the liquids are equal and remain constant and also assuming both are at the same barometric pressure and at the same ambient temperature.

Other factors could include the content of un-dissolved sugar which would reduce in volume and is likely to absorb heat as it dissolves thereby reducing the overall temperature of the container into which it is being transferred. This could then lead to a contraction of volume. However, if the temperature dropped below 4 C then the water content may start to expand thereby increasing the volume.

The arguments, so far, are based on volume, cup and teaspoon, and not mass.

I think it requires more testing at the local where Irish whisky may be pored into our favourite beverage not withstanding that Irish whisky may be our favourite beverage.
(Under that circumstance who would give a rat’s arse?)

More testing required… :D

However, the more Irish whiskey that is added to the cup, the less sure the answer becomes. Rather like Schroedinger's Cat, this is known as Paddy's Dilemma.

Ok, next one:

Two guards are guarding safes. One of the safes contains $1,000,000.00 and the other will explode on opening, killing everyone around. You are only allowed to ask one guard one question, which must be answered either yes or no. The problem is that one guard always tells the truth and the other always lies, and you have no idea which is which. What one question would you ask and why?

I think it's

14 people

Bingo. You were the first with the answer.:)

That is correct. Another way to do it is like this:

With two people (A and B), there is one handshake
(A with B).

With three people (A, B, and C), there are three handshakes
(A with B and C; B with C).

With four people (A, B, C, and D), there are six handshakes
(A with B, C, and D; B with C and D; C with D).

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.

Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 91.
This is the quadratic equation n2+ n - 182 = 0. Solving for n, we obtain 13 as the answer and deduce that there were 14 people at the party.
Since 91 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 91. The last number that you entered (91) is n. n(13) + 1 = 14.

Whilst you are correct I am sure that most people would prefer to think of it as with n people the number of handshakes is the sum of the first n-1 numbers, and the sum as n(n-1)/2

Brian

Ok, next one:

Two guards are guarding safes. One of the safes contains $1,000,000.00 and the other will explode on opening, killing everyone around. You are only allowed to ask one guard one question, which must be answered either yes or no. The problem is that one guard always tells the truth and the other always lies, and you have no idea which is which. What one question would you ask and why?

You ask the guard which safe the other guard would say contained the money, you know that that will be a lie.

Brian

Ok, next one:

Two guards are guarding safes. One of the safes contains $1,000,000.00 and the other will explode on opening, killing everyone around. You are only allowed to ask one guard one question, which must be answered either yes or no. The problem is that one guard always tells the truth and the other always lies, and you have no idea which is which. What one question would you ask and why?


Without giving it away to spoil the fun

1 * -1 = - 1 as is -1 * 1

so - 1 * -1 = 1

Without giving it away to spoil the fun

1 * -1 = - 1 as is -1 * 1

so - 1 * -1 = 1


Yes, it quite a boolean problem, isn't it?

Yes, it quite a boolean problem, isn't it?

You guys seem to like to complicate matters.

This question is always cropping up and is easily answered by people who have never heard of boolean but can apply simple logic.

Brian

Another classic - (I and probably others have posted it here before) , even after knowing the answer.


Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats.

You pick a door, say No. 1.

The host then under the rules of the game must open one of the two remaining doors to show you a goat.

The host opens door No. 3, to reveal one of the goats.

He then says to you as you final option , "Do you want to switch from door No 1 and pick door No. 2?" Is it to your advantage to switch your choice?

(For those who know it - I've rewritten this a little - so if there an error in there please tell us)

Another classic - (I and probably others have posted it here before) , even after knowing the answer.


Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats.

You pick a door, say No. 1.

The host then under the rules of the game must open one of the two remaining doors to show you a goat.

The host opens door No. 3, to reveal one of the goats.

He then says to you as you final option , "Do you want to switch from door No 1 and pick door No. 2?" Is it to your advantage to switch your choice?

(For those who know it - I've rewritten this a little - so if there an error in there please tell us)

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it. :D

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it. :D

That can't be right, surely you now have, as you stated a 50% chance, but that applies whichever door is chosen, ie sticking or moving.

Brian

If my math is correct:

They have the same amount (which isn't what I thought it would be when I first read it). Using bigger numbers for ease of math, if both cups have 5 units and I take 1 from A to B and then back. After the first move, cup B has 5 units of tea one 1 coffee. When I take out 1 unit, it is 1 of 6 units or 16.6667%. That means I take out .833333 of tea and .166667 of coffee. 5 minus .833333 is 4.166667 of tea, and 4 plus .166667 is also 4.166667 of coffee. There's probably a better way of explaining that, but I'm hurrying to make dinner so my excuse is hunger-induced brain cramps. :p

You are correct. They are exactly the same (assuming all environmental conditions prevent any affection of said liquids as defined by the people here who over-analyze the problem. :p)

A simpler way to explain it is, you can't possibly have more total volume of coffee or tea than you started with. Since you started with equal amounts of each, you must end with equal amounts of each. You used exactly 1 teaspoon in each transaction, making both cups still contain exactly the same amount of liquid. Mathematically, one cannot contain more tea than the other does coffee or vice-versa.

To make it even easier, find 10 of two small objects, say paper clips and thumbtacks, but any objects around your desk will do. Separate them into different piles. Now, grab any number of paper clips and add them to the thumbtack pile. Next, grab the exact same number of items from the original thumbtack pile, and place them into the original paper clip pile. It doesn't matter what you grab, as long as you grab the same amount, they will be equal of each other's opposite original items. ;)

Brian -

Its even better after a few pints.

Another classic - , even after knowing the answer.

That can't be right, surely you now have, as you stated a 50% chance, but that applies whichever door is chosen, ie sticking or moving.

Brian

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though. :o

It'd naturally be better to switch doors. Just because he opens a door, it doesn't remove it from the mathematical probability of 33.3% of winning, but if you switch doors, you'd have a 50% chance of winning. It's not much, but it's enough to make it worth it. :D

Almost - but I think you may have a % wrong. Its a good one - even knowing the answer, its still makes you think.

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though. :o

I would like to see the proof, tho' I doubt I'd understand it. ;)

I think that you have now got a new choice, door1 or door2, it doesn't matter what you chose before its a new choice. So 50 -50 stick or move.

Brian

That's not true. It's a mathematical phenomenon, but staying where you are still keeps a 33% chance. He didn't really tell you anything about the door that you are on, to raise the chance of winning. But he did tell you that by switching, since there are two mystery doors left, you'll have a 50% chance. By switching, the third door is taken out of the equation.

I remember this riddle from AP Physics, we had to prove it mathematically. I don't remember the formula though. :o

I have heard this one for years and seen the mathematical "proof" but still have a very difficult time following the logic, which seems to say that if you switch and they remove your original door and then he gives you a final chance to choose the door you did not choose the second iteration, and you switch yet again, you have a 100% chance of winning? That cannot be true. The prize could have been behind the door you chose originally or the second door. Or am I just really dense on this one?

By switching, the third door is taken out of the equation.

I guess I'm not following this one, but I never took Physics. The third door is already out of the equation at this point. I may have had a 33% chance when I made my original choice, but with door 3 being revealed, that choice now has a 50% chance of being correct. Switching does not increase that percentage. Either door I choose now has a 50% chance of being correct.

Almost - but I think you may have a % wrong. Its a good one - even knowing the answer, its still makes you think.

I went back and read it again. You are right, I do have the match wrong. You actually have a 66.7% chance of winning upon switching. Let me explain better if I can.

If you stay, you still only have a 1/3 chance of winning. Logically, since there is only one door left, that door must have a 2/3 chance of winning. The third door is taken out of the equation. My original explanation didn't even sound right to me. :p

Does that make more sense Brian?

Let me see if I can break it down... let's assume Door #2 has the prize...

3 scenarios -

You choose Door Number 1, stay, lose
switch, you win...

You choose Door Number 2, stay, win
switch, you lose...

You choose Door Number 3, stay, lose
switch, you win...

These are the only scenarios possible. If you switch, you win 2/3 times. If you stay, you win 1/3 times. Get it?

I went back and read it again. You are right, I do have the match wrong. You actually have a 66.7% chance of winning upon switching. Let me explain better if I can.

If you stay, you still only have a 1/3 chance of winning. Logically, since there is only one door left, that door must have a 2/3 chance of winning. The third door is taken out of the equation. My original explanation didn't even sound right to me. :p

Does that make more sense Brian?

Sorry , no.
Its like tossing a coin, each toss is a new chance what went before doesn't matter. Here the new chance is between 2 doors the fact that you chose 1 before doesn't matter, sticking is just as much a chose as moving and the odds are the same for either door, just as it was before, but changing from 33.3% to 50%.

Brian

Let me see if I can break it down... let's assume Door #2 has the prize...

3 scenarios -

You choose Door Number 1, stay, lose
switch, you win...

You choose Door Number 2, stay, win
switch, you lose...

You choose Door Number 3, stay, lose
switch, you win...

These are the only scenarios possible. If you switch, you win 2/3 times. If you stay, you win 1/3 times. Get it?

But door3 has gone for the final choice. you cannot include it in the 2nd choice scenario.

Brian

Vass

It looks like we are going to have to agree to disagree on this :D

Bri

I would like to see the proof, tho' I doubt I'd understand it. ;)

I think that you have now got a new choice, door1 or door2, it doesn't matter what you chose before its a new choice. So 50 -50 stick or move.

Brian


Where this logic fails is that now BOTH chances could be wrong since Door 1.may have had the prize.

So, you switch to door 2, here are the chances:

Door 2 has the prize, Door 3 does not
Door 2 has no prize, Door 3 has the prize
Door 2 has no prize, Door 3 has no prize

So, your odds of winning are still one in three or 33%

Where this logic fails is that now BOTH chances could be wrong since Door 1.may have had the prize.

So, you switch to door 2, here are the chances:

Door 2 has the prize, Door 3 does not
Door 2 has no prize, Door 3 has the prize
Door 2 has no prize, Door 3 has no prize

So, your odds of winning are still one in three or 33%

:confused:
You have 2 doors left, 1 has the prize, that's the scenario, if none have it you have already lost.

Brian

Chipper, I think Brian's statement was based on the original premise that door 3 has been revealed and does not have the prize. Thus it has to be either behind 1 or 2, thus a 50% chance either way. I'm with Brian on this one. At the point the "switch" decision is made, it's a coin flip.

:confused:
You have 2 doors left, 1 has the prize, that's the scenario, if none have it you have already lost.

Brian


But nothing in the scenario given guarantees that one has the prize. The only stipulation in the scenario was that one of THREE doors had the prize, you choose a door, the host gives you a chance to switch to one of the other two. That does not remove the possibility that the door you orginally chose does not have the prize. That is why I have so much difficulty with this question in all the numerous times I have encountered it.

Chipper, I think Brian's statement was based on the original premise that door 3 has been revealed and does not have the prize. Thus it has to be either behind 1 or 2, thus a 50% chance either way. I'm with Brian on this one. At the point the "switch" decision is made, it's a coin flip.


In that case if the door is revealed BEFORE you make your choice you might as well stay with your original choice since you will still have a 50% chance of being right. If it is revealed AFTER then when you are given the opportunity to switch, you still have a 33% chance of being right WHEN you make the choice to switch or not. Right?

But door3 has gone for the final choice. you cannot include it in the 2nd choice scenario.

Brian

It's an overall scenario... ignoring the original riddle. Honestly, he shouldn't have mentioned a specific door that was opened, only that the host opened one that was not a winner. ;)

Okay, let's look at it like this...

Let's say instead of opening one of the doors, the host tells you you can either stick with the door you chose or take the OTHER TWO doors instead of the one you chose. That's essentially what he is doing. Does that make more sense?

Sure, but the original premise has the switch decision being made after door 3 is revealed to be a loser.

I would like to see the proof, tho' I doubt I'd understand it. ;)

I think that you have now got a new choice, door1 or door2, it doesn't matter what you chose before its a new choice. So 50 -50 stick or move.

Brian

The important point in this puzzle is that the host knows which door the car is behind, so the door with the car will never be eliminated. So the outcomes can be seen in the following table:



Winning Door Door Chosen If you switch do you win?
1 1 No
1 2 Yes
1 3 Yes
2 1 Yes
2 2 No
2 3 Yes
3 1 Yes
3 2 Yes
3 3 No


So with 6 outcomes in your favour if you switch and 3 outcomes that are bad you are more likley to win if you do switch.

Here's a single game simulator...

http://www.theproblemsite.com/games/monty_hall_game.asp

Here's a multigame simulator...

http://dl.dropbox.com/u/548740/www/DoorKeeperGame/index.html

It's hard to explain, but I see it in my head so clearly. Argh! lol

This backs you guys up, but I'm going to have to ponder it:

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

The important point in this puzzle is that the host knows which door the car is behind, so the door with the car will never be eliminated. So the outcomes can be seen in the following table:



Winning Door Door Chosen If you switch do you win?
1 1 No
1 2 Yes
1 3 Yes
2 1 Yes
2 2 No
2 3 Yes
3 1 Yes
3 2 Yes
3 3 No


So with 6 outcomes in your favour if you switch and 3 outcomes that are bad you are more likley to win if you do switch.
I wish I'd never started reading this one :D

The host has already shown you that door # 3 definitely isn't a winner? I can see that, at that point, if you take all three doors into account, he's offering you the option of one door or two (one of the two being a definite loser). However, the fact that the third door exists at all is now irrelevant, as you're never going to pick it.

That would immediately reduce your options to the winning door being 1 or 2. Since you would now only have a choice between door 1 or door 2, surely the odds are 50/50?

I wish I'd never started reading this one :D

The host has already shown you that door # 3 definitely isn't a winner? I can see that, at that point, if you take all three doors into account, he's offering you the option of one door or two (one of the two being a definite loser). However, the fact that the third door exists at all is now irrelevant, as you're never going to pick it.

That would immediately reduce your options to the winning door being 1 or 2. Since you would now only have a choice between door 1 or door 2, surely the odds are 50/50?

The odds of an item being behind one of two objects is 50/50 but thats not the point.

There are nine total scenarios which can occur in the game. If you were to switch in all nine scenarios you would win six of the nine. That is the important part.

I wish I'd never started reading this one :D

The host has already shown you that door # 3 definitely isn't a winner? I can see that, at that point, if you take all three doors into account, he's offering you the option of one door or two (one of the two being a definite loser). However, the fact that the third door exists at all is now irrelevant, as you're never going to pick it.

That would immediately reduce your options to the winning door being 1 or 2. Since you would now only have a choice between door 1 or door 2, surely the odds are 50/50?

The odds are only 50/50 if the host doesn't know where the car is. You must assume he does since he's obviously not going to open the winning door.

If you stay where you are, you're in the exact same scenario you started with. You still have 1/3 of a chance of winning, even one exact door that contains a goat. The reason why is because you already knew one of them would. Knowing which one doesn't change anything. It's like if you were never given an option to switch...

However, if you switch, you'll have the remaining 2/3 chance of winning, because obviously you're not going to choose Door #3. Basically, it's now like this...

Door #1 - 1/3 probability...
Door #2 - 2/3 probability...
Door #3 - 0/3 probability... since you know it's a goat

It still have to come out to 100%. ;)

I have read the wiki and your answers and am going out on a limb to say that it is all smoke and mirrors. Door3 is kept in in all discussions whereas it is now eliminated.
Whether the host knows or not is irrelevent, you now have 2 doors 1 contains the car one doesn't, you have a choice, how does that differ from if that had been the original choice?

Brian

The odds are only 50/50 if the host doesn't know where the car is. You must assume he does since he's obviously not going to open the winning door.

If you stay where you are, you're in the exact same scenario you started with. You still have 1/3 of a chance of winning, even one exact door that contains a goat. The reason why is because you already knew one of them would. Knowing which one doesn't change anything. It's like if you were never given an option to switch...

However, if you switch, you'll have the remaining 2/3 chance of winning, because obviously you're not going to choose Door #3. Basically, it's now like this...

Door #1 - 1/3 probability...
Door #2 - 2/3 probability...
Door #3 - 0/3 probability... since you know it's a goat

It still have to come out to 100%. ;)
I get what you're saying, but it still assumes that the door that everyone knows to be wrong would still be included in the equation. By proving it to be wrong, the host has removed it from any future consideration.

You may as well say that there are 1,000 doors, 998 of which are open and can be seen to have a goat behind them. That doesn't make your odds of winning 1 or 2 in 1000, it's 50/50, since you've only realistically got a choice of two. The open doors are never in the equation.

I get what you're saying, but it still assumes that the door that everyone knows to be wrong would still be included in the equation. By proving it to be wrong, the host has removed it from any future consideration.

You may as well say that there are 1,000 doors, 998 of which are open and can be seen to have a goat behind them. That doesn't make your odds of winning 1 or 2 in 1000, it's 50/50, since you've only realistically got a choice of two. The open doors are never in the equation.

I have read the wiki and your answers and am going out on a limb to say that it is all smoke and mirrors. Door3 is kept in in all discussions whereas it is now eliminated.
Whether the host knows or not is irrelevent, you now have 2 doors 1 contains the car one doesn't, you have a choice, how does that differ from if that had been the original choice?

Brian

It's not out of future consideration though, that's what people don't seem to understand. Just because the host showed you what was behind it, did not remove it from the equation. I don't know how to explain it any easier than the post I made above... :rolleyes: It's still there as an option, but obviously your change of winning with it becomes 0 out of 3.

Okay, how about this. You choose Door number 1, and I DON'T show you what's behind one of the doors. You know for a FACT that at least ONE of the remaining doors will have a goat, correct? I tell you that you can trade your one door for BOTH of the other doors, would you do it? You still know for a FACT one of them is a goat.

Knowing which one is a goat, DOES NOT MATTER! You'd have a 2/3 chance of winning.

Okay, how about this. You choose Door number 1, and I DON'T show you what's behind one of the doors. You know for a FACT that at least ONE of the remaining doors will have a goat, correct? I tell you that you can trade your one door for BOTH of the other doors, would you do it? You still know for a FACT one of them is a goat.

Knowing which one is a goat, DOES NOT MATTER! You'd have a 2/3 chance of winning.
I'm going to just admit here that you're clearly far better at understanding mathematical proofs than I, Gunga Din.

I do, however, have a ticket for last weekend's lottery that I can let you have cheap, if you're interested? It didn't win, but I'm sure the odds can be shown to stil be good. :D

Is the issue here on when the decision is made to switch?
ie if you decide before the game that you will switch your choice when one of the Goat doors is revealed then the chance of that 2nd choice is 50/50 whereas your first choice is 1/3rd chance.

But, if you only make the choice after the door is revealed, then this is a new game with 50/50 chance and you either pick the other door or stay with your current door, which effectively is a choice in itself.

Your original choice 1/3rd automatically becomes a 50/50 choice even by not switching which is a choice. - I think:confused:

Is the issue here on when the decision is made to switch?
ie if you decide before the game that you will switch your choice when one of the Goat doors is revealed then the chance of that 2nd choice is 50/50 whereas your first choice is 1/3rd chance.

But, if you only make the choice after the door is revealed, then this is a new game with 50/50 chance and you either pick the other door or stay with your current door, which effectively is a choice in itself.

Your original choice 1/3rd automatically becomes a 50/50 choice even by not switching which is a choice. - I think:confused:

Nope, it becomes 2/3, not 50/50.

Isn't your 2/3rd using history to support the number. What if another person had to make the 2nd choice?

What if you took Paul the octopus along? how would that effect the probability?

Isn't your 2/3rd using history to support the number. What if another person had to make the 2nd choice?

What if you took Paul the octopus along? how would that effect the probability?

It's not history... the choices haven't changed. He's only shown you what's behind one, that door is still there. Since you can remove the probability that that door is a winner, the remaining 2/3 must automatically revert to the other option.

Vassago,

The choices have changed. These are not INDEPENDENT events.

In an extreme example, consider 1000 doors; one is the winner.

Your choice has 1/1000 of being right.

If the host removes 998 doors, you are faced with two options:

Keep your door ... still a 1/1000 chance.

Switch to the other door ... 1/2 chance.

Even if they remove only one choice your original 1/1000 is still worse
than switching to 1/999 ... It is a "brand-new" scenario.

You should always switch.

It seems counter-intuitive, but it is not like the consecutive coin-flip
scenario. Those are independent events.

Wayne

I understand the 1000 door example - well put.
The 1 other door has the perception of a better chance of being the winning door because the 998 removed doors were false.

To switch to this door is a higher probability 1/2 then your original choice which was 1/1000 and the chance of your original choice Was correct remains at 1/1000.

In effect, for the first choice to be correct it has to perform two criteria.
Be the 1/1000 and now the 1/2 - long shot - against the 2nd door which only has to beat the 1/2 odds.

Still think it will be a hard choice to make at the time.

What entertained me the most was reading from the beginning of this thread... pondering one question that stumped me for a few moments.... and then thought... "A DEAL!" only to scroll down a few posts to find TessB suggest "A deal?"

I looked at the date.... 2006???? WOW!
I scrolled down a bit more, to look for confirmation of my correct answer.
It said "Cards" I frowned..... lol. I thought... But..... a deal fits the logic as well!

Scrolled down more.... watched myself fight for recognition.
Will I EVER change?

Vassago,

The choices have changed. These are not INDEPENDENT events.

In an extreme example, consider 1000 doors; one is the winner.

Your choice has 1/1000 of being right.

If the host removes 998 doors, you are faced with two options:

Keep your door ... still a 1/1000 chance.

Switch to the other door ... 1/2 chance.

Even if they remove only one choice your original 1/1000 is still worse
than switching to 1/999 ... It is a "brand-new" scenario.

You should always switch.

It seems counter-intuitive, but it is not like the consecutive coin-flip
scenario. Those are independent events.

Wayne

Not quite -

Imagine 1000 doors, 1 of which has a car behind it 999 have a goat.

Your chance of choosing the door with the car behind it are 1/1000.

And 999/1000 you were wrong and choose a goat.

Inevitably 998 of the remaining 999 doors at least have a goat behind them, we knew this before the original choice, after the original choice and we still know it after the host reveals the inevitable 998 goats.

So we know nothing new about the original choice. So the odds of it being correct haven't changed.

- with just 2 doors left unopen - the chance we were originally correct is still 1/1000 and still 999/1000 we were wrong.

All the host in effect asks you - is do you want to stick with the 1/1000 chance that you were originally correct or switch to the 999/1000 chance, that you were originally wrong.

Although theres 2 choices, the odds aren't 50/50.

I understand the brain teaser logically, but it is using disingenious means to make the math look a particular way. Assume someone says "select one of these three options", and one of the three options is labeled "Don't pick me". If the person making the pick is guaranteed that the "Don't pick me" label is not a trick, then it really isn't an option anymore.

Or to put it another way, it would be like having three women stand on stage, a blonde, a brunette, and a red head. Each of the women is holding a bag that contains an amount of money between $0 and $1 million. The blonde's bag is labeled "$0". Then, after watching several different people make selections, drawing the conclusion that no one likes blondes. That would be anagalous (and just as faulty) as the brain teaser.

I understand the brain teaser logically, but it is using disingenious means to make the math look a particular way. Assume someone says "select one of these three options", and one of the three options is labeled "Don't pick me". If the person making the pick is guaranteed that the "Don't pick me" label is not a trick, then it really isn't an option anymore.

Or to put it another way, it would be like having three women stand on stage, a blonde, a brunette, and a red head. Each of the women is holding a bag that contains an amount of money between $0 and $1 million. The blonde's bag is labeled "$0". Then, after watching several different people make selections, drawing the conclusion that no one likes blondes. That would be anagalous (and just as faulty) as the brain teaser.

Uh? Theres nothing disingenuous at all about the question is there?

An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian

From a choice of two it is 50-50, however.... in a choice 1 of 1000 it is 1 of 1000 and stays 1 of 1000 that your pick is right.

So if -before the commercial break- he picked one is has a 999 of 1000 chance to be wrong. So you could argue that if they take away 998 that he still has a 999/1000 chance to be wrong about his current choice. Hence statisticaly, your first choice would most likely be wrong (999 out of 1000 odds on that your wrong).

However... If out of the 998 this also includes your current choice or if you are just to start your pick at the 2 options, then yes 50-50

As PngBill and I have both said, you guys are using history to make your point, history can play no part in the next odds.

With 3 doors there is a 1 in 3 chance for each door.
With 2 doors there is a 1 in 2 chance for each door.

The odds for every door change as doors are taken out of the choice.

Brian

An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian

To him now it is 50/50 - I think.

Hes forgotten which door , door 1 was "associated with"

the important bit is which remaining door, door 1 was associated with has twice the chance of having the car

Door 3 - had a 1/3 chance.
And together Door 1 and 2 had a 2/3 chance

but he cant remember if it was

Door 2 - had a 1/3 chance
And together Door 1 and 3 had a 2/3 chance

So hes 50/50 on which door, door 1 was associated with. Or hes 50/50 on which door he originally picked.

The audience remembers that it was door 2 - that had twice the chance.

--
Think of it in the 1000 box example - he cant remember which door was the 1/1000 chance or which was the 999/1000 chance.

So he guesses - and its now 50/50 to him.

A new one -




Its the year 2008. Last month was still 2008 and next month is 2008.

If the date of the last Monday of last month is added to the date of the first Thursday of next month, the result is 38.



What is the current month?

A new one -




Its the year 2008. Last month was still 2008 and next month is 2008.

If the date of the last Monday of last month is added to the date of the first Thursday of next month, the result is 38.



What is the current month?

August.

The only way to achieve this is for the monday to be the 31st and the Thursday to be the 7th.

So the current month starts on a Tuesday and must end on a thursday so that the first thursday can be the 7th. This means it must have 31 days also. Only July and August are consecutive months, in the same year, which both have 31 days therefore it must be August.

Here's another.

At the Infinity Hotel there are an infinite number of rooms. A man walks upto reception and asks for a room. He is told there are already an infinite number of guests staying at the hotel so all the rooms are occupied.

How do they get him a room to stay in?

Uh? Theres nothing disingenuous at all about the question is there?

I'm not sure if it is the way you phrased your question, or if I am just slow, but I'm not sure what you're asking. If you didn't understand what I was trying to say, Brian hit on the key point with:


As PngBill and I have both said, you guys are using history to make your point, history can play no part in the next odds.

If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching. That means I have missed 3 possible scenarios from my list. If you can provide the 3 scenarios I will agree that the probability is 50%.

An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian

In this situation the probability is 50/50. It's only if the person knows which was originally chosen and which door was not eliminated by the host that the probability is not 50/50.

If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching.

There is. Here:


Winning Door Door Chosen If you switch do you win?
1 1 No
1 2 Yes
2 1 Yes
2 2 No


The moment door x is revealed to no longer be a viable option, it ceases to exist.

If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching. That means I have missed 3 possible scenarios from my list. If you can provide the 3 scenarios I will agree that the probability is 50%.

I think the difference is, is that it becomes a brand new choice, a brand new game, so to speak. So once the question is asked, "do you want to switch?" It presents a new problem, your scenarios are reduced to two (Edit: the prize is behind 1 or behind 2), therefore 50/50 chance that you are correct.

There is. Here:


Winning Door Door Chosen If you switch do you win?
1 1 No
1 2 Yes
2 1 Yes
2 2 No


The moment door x is revealed to no longer be a viable option, it ceases to exist.

yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.

yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.

But that's just it, at the point that this question becomes viable, you do know that door three is excluded, so it doesn't have to be included in the possible outcomes.

August.

The only way to achieve this is for the monday to be the 31st and the Thursday to be the 7th.

So the current month starts on a Tuesday and must end on a thursday so that the first thursday can be the 7th. This means it must have 31 days also. Only July and August are consecutive months, in the same year, which both have 31 days therefore it must be August.

Spot on - a nice uncontraversial one.

yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.

Technically, you're correct. But what you're trying to determine has no value.

Imagine the same scenario, but the host opens every door. Technically your exact same table still holds. You can look at your original choice, all the possible outcomes of switching or not switching and determine the possibilities.

But to then draw a conclusions such as "You win 2/3's of the time when you switch" is incredibly misleading because its quite obvious that the person selecting will pick the winning door.

The point in time that you can draw meaningful conclusions from the data is when the question is actually posed: "Do you want to switch?" If a door has been eliminated, probability to choosing the correct door changes.

But that's just it, at the point that this question becomes viable, you do know that door three is excluded, so it doesn't have to be included in the possible outcomes.

A door can always be excluded - you know this before you start. So before you start your odds are 1/3 - and since you know nothing more about your original pick now - its odds of being coorect are still 1/3- and the remaining odds 2/3 must be on the other door.


Right now we know who the marks are - lets set the venue - to play for money. :-)

The best way for the unconvinced to be convinced is to play the host in the real world with a contestant. Then its more obvious nothing has really changed since the original pick. And they are better of swithing - the more doors the more obvious it is.

The choice is do you think - you were right first pick. The answer is probably not, I think we are all agreed on that.

1/3 chance
1/4 chance
.....
.....
1/n chances of being right first pick.

Right now we know who the marks are - lets set the venue - to play for money. :-)

Yeah I'll happily play this with anyone arguing about it, the only stipulation is they are not allowed to switch.

A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

Here's another.

At the Infinity Hotel there are an infinite number of rooms. A man walks upto reception and asks for a room. He is told there are already an infinite number of guests staying at the hotel so all the rooms are occupied.

How do they get him a room to stay in?

I dunno -

Since Infinity + 1 is a valid number - So infinity hotel gets one bigger and the guest gets to stay?

As an aside -
Parallel lines meet at infinity - I never understood that one!

Yeah I'll happily play this with anyone arguing about it, the only stipulation is they are not allowed to switch.

If the stakes are high enough we can have a joint venue midway between Newbury and Liverpool.

You could play it with the witnesses next time they knock! It'll convince you they are illogical, but at the same time there is a God. I'm joking, Just joking.

The point in time that you can draw meaningful conclusions from the data is when the question is actually posed: "Do you want to switch?" If a door has been eliminated, probability to choosing the correct door changes.

If the person eliminating the door knows where the prize is and you know the history of the game the odds don't change.

If the host eliminates a door and then shuffles the doors so you don't know which one you originally picked the odds are 50/50 but as long as you know what you're original choice is then the odds don't change.

and since you know nothing more about your original pick now - its odds of being coorect are still 1/3


Wrong. Once a door is eliminated, the odds change. You know one of the choices is no longer valid. Your using what's called junk science.

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.

seeing as only losing doors are "lost" in the equation, the odds turn out to be 50%-50%

Winning Chosen Rmvd Win?
1 1 2 TRUE
1 1 3 TRUE
1 2 3 FALSE
1 3 2 FALSE
2 1 3 FALSE
2 2 1 TRUE
2 2 3 TRUE
2 3 1 FALSE
3 1 2 FALSE
3 2 1 FALSE
3 3 1 TRUE
3 3 2 TRUE
6 wins - 6 loses


If you make it 4 doors, removing two duds.... 12 wins, 11 loses (52% win)
Winning Chosen Rmvd Rmvd 2 Win?
1 1 2 3 TRUE
1 1 3 4 TRUE
1 1 2 4 TRUE
1 2 3 4 FALSE
1 3 2 4 FALSE
1 4 2 3 FALSE
2 1 3 4 FALSE
2 2 1 3 TRUE
2 2 3 4 TRUE
2 2 1 4 TRUE
2 3 1 4 FALSE
2 4 1 3 FALSE
3 1 2 4 FALSE
3 2 1 4 FALSE
3 3 1 2 TRUE
3 3 2 4 TRUE
3 3 1 4 TRUE
4 1 2 3 FALSE
4 2 1 3 FALSE
4 3 1 2 FALSE
4 4 1 2 TRUE
4 4 2 3 TRUE
4 4 1 3 TRUE


However 5 Doors, 15 wins - 20 loses (43% win)
Winning Chosen Rmvd Rmvd 2 Removed 3 Win?
1 1 2 3 4 TRUE
1 1 3 4 5 TRUE
1 1 2 4 5 TRUE
1 2 3 4 5 FALSE
1 3 2 4 5 FALSE
1 4 2 3 5 FALSE
1 5 2 3 4 FALSE
2 2 2 3 4 TRUE
2 2 3 4 5 TRUE
2 2 2 4 5 TRUE
2 1 3 4 5 FALSE
2 3 2 4 5 FALSE
2 4 2 3 5 FALSE
2 5 2 3 4 FALSE
3 3 2 3 4 TRUE
3 3 3 4 5 TRUE
3 3 2 4 5 TRUE
3 2 3 4 5 FALSE
3 1 2 4 5 FALSE
3 4 2 3 5 FALSE
3 5 2 3 4 FALSE
4 4 2 3 4 TRUE
4 4 3 4 5 TRUE
4 4 2 4 5 TRUE
4 2 3 4 5 FALSE
4 3 2 4 5 FALSE
4 1 2 3 5 FALSE
4 5 2 3 4 FALSE
5 5 2 3 4 TRUE
5 5 3 4 5 TRUE
5 5 2 4 5 TRUE
5 2 3 4 5 FALSE
5 3 2 4 5 FALSE
5 4 2 3 5 FALSE
5 1 2 3 4 FALSE

6 doors, 24 wins, 30 loses (44%)
At the logic of your first choice is 5/6 wrong, that should mean that the odds of it being correct/winning is 1/6 or 16%

If above is correctly worked out, it does seem to be slightly favorable when there are multiple doors (> 4)? Though I can offcourse have errors here :eek:
BUT the big differences vs the 50-50 are diffinatly not there I dont think

Wrong. Once a door is eliminated, the odds change. You know one of the choices is no longer valid. Your using what's called junk science.

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.


Sorry adam that's junk science.

You're example should be the answer is 1 or X where X is any number in your range except Y. 99,999/100,000 times the answer will be X.

seeing as only losing doors are "lost" in the equation, the odds turn out to be 50%-50%

Winning Chosen Rmvd Win?
1 1 2 TRUE
1 1 3 TRUE
1 2 3 FALSE
1 3 2 FALSE
2 1 3 FALSE
2 2 1 TRUE
2 2 3 TRUE
2 3 1 FALSE
3 1 2 FALSE
3 2 1 FALSE
3 3 1 TRUE
3 3 2 TRUE
6 wins - 6 loses





The error here is that which door he removes is irrelevant, as he will never remove the winning door. What matters is if you switch or not.

If the winning door is 1 and you have chosen 1 it doesn't matter if he removes door 2 or 3 as if you switch when given the option you will lose if you do.

A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

So 50 * 2 profit for day one assuming they haven't dehydrated.
= 100

Day two -

it was 99 water and 1 strawberry.

Its now 2% strawberry an 98% water.

So has the weight halved (99/100 is 99% 49/50 is 98%) and the profit is 25 * 2 = 50

So he makes 50.

?

Sorry adam that's junk science.


Well, now we agree :p


You're example should be the answer is 1 or X where X is any number in your range except Y. 99,999/100,000 times the answer will be X.

At this stage you're correct. You're still talking hypothetical. The odds of you successfully guessing the correct number are 1/100,000. The odds of you guessing an incorrect number are 99,999/100,000.

As soon as you add in the "Would you like to switch" question, the above no longer applies.

After reading all this, I have to say, the simple explination for the door problem seem to be that your simply Betting that you were wrong on your first guess.
Because your odds are 1/3 initially, and you made your pick with those odds, given that you have the option to switch to another door.. they are basically asking you.. "Do you think you were right when you choose that door with odds of 1/3"
Your best bet is to assume you were wrong and choose the other door.
Which gives you the 66% odds...yes?

Oh, and the farmer made $198

Nam

1 1 2 TRUE
1 1 3 TRUE

Choosing 1 when the winner is one is only one example, because inevitably the host can remove 2 or 3.
Choosing 1 is one possible choice - somehow you are counting it as 2.

The point is - its irrelevent to the odd that the host can reveal a wrong answer - he can reveal a wrong answer for whatever you first pick as you illustrate. He never as you suggest though gives you two wrong answers.

If you choose 1 - he reveals either, 2 or 3 - just one revaeal. Even if he has two choices of what the one reveal would be.

So you have twice the number of correct choices as there really are.

so 3 out of 9 are correct, and 6 out of nine are wrong.

As soon as you add in the "Would you like to switch" question, the above no longer applies.


yes it does. Really think about it. Why does giving you an option to switch change the odds?

Let me give you an example using your logic. I say "Pick a number between 1 and 100,000", you pick 1, then I say "the number is not greater than 2, would you like to change your guess?"

Your original odds of winning were 1 in 100,000. Since you still don't know if your original guess was correct or not, the odds are still 1 in 100,000. The other possible choice, 2, must contain all the rest of the odds. So 2 will win 99,999/100,000 times.


Thats correct. 99,999 times - the right answer is differant from the one I originally picked. All you have done is clarify that 99, 998 of the ones i didn't pick were wrong - which I already knew.

Its still 99,999 times more likely i was originally wrong, and since 2 is the only answer possible left I didnt go for - its 99,999 times more likely that 2 is correct.



Look at it the other way - play the game over a few times in your head - you are saying my chances of getting the right answer out of 100,000 is 50/50 each time.


So 2 will be right 99,999 out of 100,000
or 3 will correct 99,999 out of 100,00
.......

Or 100,000 will be correct 99,999 out of 100,000.

I doubt thats clarified it but still.

In effect, for the first choice to be correct it has to perform two criteria.
Be the 1/1000 and now the 1/2 - long shot - against the 2nd door which only has to beat the 1/2 odds.



This is the crux of it for me. The showing of the third door allows the first door to conform to the first criterion you described.

There is now an equal chance that door 1 and 2 will confirm to the second criterion unless the first door is somewhat fatigued from the first event.

The first door doesn't have to conform to the first criterion on both occasions.

So 50 * 2 profit for day one assuming they haven't dehydrated.
= 100

Day two -

it was 99 water and 1 strawberry.

Its now 2% strawberry an 98% water.

So has the weight halved (99/100 is 99% 49/50 is 98%) and the profit is 25 * 2 = 50

So he makes 50.

?

Sorry, your answer is not correct.

This is the crux of it for me. The showing of the third door allows the first door to conform to the first criterion you described.

There is now an equal chance that door 1 and 2 will confirm to the second criterion unless the first door is somewhat fatigued from the first event.

The first door doesn't have to conform to the first criterion on both occasions.

Are you saying its 50/50?

Thats correct.

The problem is, is that both door 1 and 2 have conformed to the same criterion but you want to apply the consequences of this conformation differently to each door.

A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

Cost of Strawberries = $200

Day 1 Sales: 50 * $4 = $200

So 50 pounds of strawberries with 1% strawberry content =

1% * 50 Pounds = 0.5 pound strawberry content


0.5 pounds/2% = New Weight/100%

New Weight = (0.5 * 100)/2

New Weight = 25 Pounds

25 pounds * $4 = $100

Total Revenue = $300

Total Expenditure = $200

Profite = $100

Are you saying its 50/50?

You can't have it both ways.

If door 1 remains 1 in 3, then so does door 2.

There is no benefit in switching because both doors still contain the same probability.

Sorry, your answer is not correct.

Maybe Ishould have done sales and added it up (definately adding day 1 and 2 would be good) to get

50 * 4 + 25* 4 = 300 = 200 profit?

Oh no - thats not right either - I worked out the expenditure wrong too.

I dunno -

Since Infinity + 1 is a valid number - So infinity hotel gets one bigger and the guest gets to stay?



No but your thinking is in the right place.

No but your thinking is in the right place.


If the number of rooms is infinite then there will always be room for one more since you cannot fill up an infinite (non-ending) number of rooms.

Infinite number of guests + 1 = infinite number of guests

Originally Posted by ChipperT
A local entrepeneur bought 100 pounds of strawberries for $2.00 per pound and expected to double his investment by selling the strawberries for $4.00 per pound on a convenient street corner. The seller only managed to sell 50 pounds of strawberries the first day and he sold the remainder on the second day. The fresh strawberries had a content of 99% water, but because of the hot weather, the strawberries dehydrated and contained only 98% water on the second day. How much profit did the seller make?

Farm purcahsed 100 pounds at $2. = $200 expense.
First day, sold 50 pounds fresh strawberries at $4.00 = $200 (expenses = 0 now)

Second day..
Loss of 1%.
50 pounds x 0.01 = 0.5 pounds.
50 - 0.5 = 49.5 pounds.
49.5 pounds x $4 = 198 (profit)

You can't have it both ways.

If door 1 remains 1 in 3, then so does door 2.

There is no benefit in switching because both doors still contain the same probability.

The probability must add up to 100/100.

It cant be explained better than already tried.

All I can say is try it on a friend and it becomes obvious its not 50/50

then look back at the explanations to see why.

Do it with 5 doors or something.

Farm purcahsed 100 pounds at $2. = $200 expense.
First day, sold 50 pounds fresh strawberries at $4.00 = $200 (expenses = 0 now)

Second day..
Loss of 1%.
50 pounds x 0.01 = 0.5 pounds.
50 - 0.5 = 49.5 pounds.
49.5 pounds x $4 = 198 (profit)


No, wrong answer.

Chergh: You are correct.

My 3rd and final guess - is Chergh is correct.

You mean Chergh.

No, wrong answer.


I think if there was a 1% loss in weight then there would be a 1% loss in profit, so if the first day he made $100 then the second day he would make the $100 - 1% of $100, so $100-$1 = $199

No but your thinking is in the right place.

I would only paraphrase my original answer if I tried again.

Time for a white answer!

I think if there was a 1% loss in weight then there would be a 1% loss in profit, so if the first day he made $100 then the second day he would make the $100 - 1% of $100, so $100-$1 = $199


Sorry, no prize behind door #2 for you... yet...

Sorry, no prize behind door #2 for you... yet...

Ha! I went up and read Cherghs answer, which you said was correct. So I know I was wrong.... But you can't blame a girl for trying. :p

The probability must add up to 100/100.



Here is where I don't follow.


A door can always be excluded - you know this before you start. So before you start your odds are 1/3 - and since you know nothing more about your original pick now - its odds of being coorect are still 1/3- and the remaining odds 2/3 must be on the other door.

You know nothing more about either door but you apply different rules to each door. Why can the odds on door2 change but not for door1 when they both conform to the rule of knowing nothing more about them?

How does guessing at door1 force door2 to become immune to your rule?

Sorry, your answer is not correct.

My mistake was day 2 he doesn't make a profit! Isn't it? Arggh!

Keep them coming.

Here is where I don't follow.



You know nothing more about either door but you apply different rules to each door. Why can the odds on door2 change but not for door1 when they both conform to the rule of knowing nothing more about them?

How does guessing at door1 force door2 to become immune to your rule?

As I said - its been explained as well as seemingly able to - so try it for real - with money.

But

You do know more about door 2 after the reveal , cos door 2 was part of the group that was 2/3 chance of being right. But you know the rest of that set ie door 3 is definatley wrong. So the 2/3 chance is all concentrated in door 2.

The choice is really were you likely to be right first pick or not.
We are all agreed I think, that the chances of that was only 1/3 - still is. So switch.

Chergh: You are correct.

How is it that a 1% reduction in water is a 50% reduction in weight.... this answer can't be correct based on the info.

If the first day, the water content was 99% and that equates to 50 pounds.
Strawberry weight is 0.5 and water weight is 49.5

Looking at Just the water, we loose 1% of our water due to evap.
1% being 0.495 pounds of water.
Day two would have a water weight of 49.5 - 0.495 = 49.005

The weight of the strawberries stays the same at 0.5 which means we add them
49.005 + .5 = 49.505

Loss of 50% weight doesn't make any sense.

1 part strawberry - 99 parts water = 99% water.
1 part strawberry (the strawberry doesnt diminish) and 49 parts water (the water diminishes) = 98% water.

So the weight is now 50 was 100.

1 part strawberry - 99 parts water = 99% water.
1 part strawberry (the strawberry doesnt diminish) and 49 parts water (the water diminishes) = 98% water.

So the weight is now 50 was 100.

i c.... interesting. lol.
Thanks.

You do know more about door 2 after the reveal , cos door 2 was part of the group that was 2/3 chance of being right. But you know the rest of that set ie door 3 is definatley wrong. So the 2/3 chance is all concentrated in door 2.

Where has this group of '2/3 chance of being right' come from and why isn't door1 a part of it? Before the reveal each door stands on its own. That is from where your 33% for each door derives.

An opinion on the state of a particular door is completely arbitary when defining the probability of its true state.

In no way throughout the entire process is the state of door1 and door2 different. Yet door2 is allowed a 33.33% increase in probability whereas door1 is not.

Door 2 was part of the group that was 2/3 chance of being right whereas Door 1 was not? How, when and where? :p

My mistake was day 2 he doesn't make a profit! Isn't it? Arggh!

Keep them coming.

No, he broke even on day 1, he made $100.00 on day 2 so the total profit was $100.00.

No, he broke even on day 1, he made $100.00 on day 2 so the total profit was $100.00.



Yep - we are agreeing I think. In isolation day 1 was profitable based on what he paid for what he sold that day and the difference between.

Day two - he made nothing cos although it was 2 times the price it had diminished over night to half the weight.
So in isolation day two he made no profit.

Where has this group of '2/3 chance of being right' come from and why isn't door1 a part of it? Before the reveal each door stands on its own. That is from where your 33% for each door derives.

An opinion on the state of a particular door is completely arbitary when defining the probability of its true state.

In no way throughout the entire process is the state of door1 and door2 different. Yet door2 is allowed a 33.33% increase in probability whereas door1 is not.

Door 2 was part of the group that was 2/3 chance of being right whereas Door 1 was not? How, when and where? :p

Perhaps it would help people to think of it another way.... the opposite way.
When you have the option of all 3 doors, only one door is a winner. That we know. So you have a 66% chance of picking the wrong door.
So by that, you always have higher odds of being wrong. If given the option to switch doors, you should take it (according to the fact that you had a 66% chance being wrong on your first guess) and since you only get to choose one door because the host revealed the other one, then you increase your odds to make that OTHER door 66% chance of being the winning door.

Perhaps it would help people to think of it another way.... the opposite way.
When you have the option of all 3 doors, only one door is a winner. That we know. So you have a 66% chance of picking the wrong door.
So by that, you always have higher odds of being wrong. If given the option to switch doors, you should take it (according to the fact that you had a 66% chance being wrong on your first guess) and since you only get to choose one door because the host revealed the other one, then you increase your odds to make that OTHER door 66% chance of being the winning door.


Well, THAT sure cleared it up... NOT! :D:confused::confused::confused::confused:

Here is a simple one. How many errors are in the following?

"Their are four misteaks in this sentence"

Here is a simple one. How many errors are in the following?

"Their are four misteaks in this sentence"

Is it just me or is this meant to trip us up? There are only 4 errors if you count the fact that there are only 3 errors, thus making the count of "four" an error intelf bringing the total to 4 errors, thus negating the falsehood of it, and reducing the count to three.... and round and round it goes.... Or am I completely off base?

Is it just me or is this meant to trip us up? There are only 4 errors if you count the fact that there are only 3 errors, thus making the count of "four" an error intelf bringing the total to 4 errors, thus negating the falsehood of it, and reducing the count to three.... and round and round it goes.... Or am I completely off base?


I like the way you think. It is wrong, but I like the way you think.

Here is a simple one. How many errors are in the following?

"Their are four misteaks in this sentence"

None, all the errors were intentional and therefore aren't really errors.

I like the way you think. It is wrong, but I like the way you think.

I thought of my answer on my own, but with your response googled it and came up with This (http://www.cut-the-knot.org/selfreference/index.shtml#answers)

This confirms my suspicions, but given the comments by others there is debate..... So in one way I am right, it just depends on how you interpret it.

I also like the reasoning behind Chergh's answer.... I wish I had thought of it first. :P

I thought of my answer on my own, but with your response googled it and came up with This (http://www.cut-the-knot.org/selfreference/index.shtml#answers)

This confirms my suspicions, but given the comments by others there is debate..... So in one way I am right, it just depends on how you interpret it.


Actually This (http://www.cut-the-knot.org/selfreference/index.shtml#answers) is wrong, but I would not say that emphatically or empirically.

I was going to say:

3: misuse of "their", spelling error on "mistakes", and no period at the end

I disagree with that site; it doesn't seem like you should count the same word as both a spelling and grammatical problem. It's one or the other.

None, all the errors were intentional and therefore aren't really errors.


Good try and could possibly be correct in an alternate universe other than the one I exist in. :cool:

I was going to say:

3: misuse of "their", spelling error on "mistakes", and no period at the end

I disagree with that site; it doesn't seem like you should count the same word as both a spelling and grammatical problem. It's one or the other.
pbaldy, you pointed out an error that I had inadvertantly introduced - that of the missing punctuation. Therefore I must eat crow. Kryst51 is correct and I must humbly beg her forgiveness.

pbaldy, you pointed out an error that I had inadvertantly introduced - that of the missing punctuation. Therefore I must eat crow. Kryst51 is correct and I must humbly beg her forgiveness.

Can I change my answer to 1:p

pbaldy, you pointed out an error that I had inadvertantly introduced - that of the missing punctuation. Therefore I must eat crow. Kryst51 is correct and I must humbly beg her forgiveness.

Consider yourself forgiven <said haughtily and regally, and if I had a sceptor I might just use it to tap your shoulders in a pardoning gesture. :p>

Okay, here's one I remembered from my youth. It's a visual, attached (forgive my non-existent art skills). The lines represent 4 toothpicks, arranged in the shape of a martini glass. The circle is an olive. You are to move two of the toothpicks in such a way as to redraw the glass but leaving the olive outside the glass.

Okay, here's one I remembered from my youth. It's a visual, attached (forgive my non-existent art skills). The lines represent 4 toothpicks, arranged in the shape of a martini glass. The circle is an olive. You are to move two of the toothpicks in such a way as to redraw the glass but leaving the olive outside the glass.


An easy one. You move the "stem" and the right side so that the glass is upside down and the olive is beside the stem. Wish I could draw it!

That damn matchstick one i have no idea... lol.
how about this:

A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment.
Can you explain why?

An easy one. You move the "stem" and the right side so that the glass is upside down and the olive is beside the stem. Wish I could draw it!


You got to it before me! :D

That's just it guys, if you have 1,000,000 doors and the host reveals that 999,998 of them are goats, you really do have a 999,999 chance of winning if you switch! It's simple math... The fact that you know which ones are goats is IRRELEVANT and thrown in there to make you second guess... What the host is essentially doing is trading your one door for all of the rest of the doors, but letting you know what's behind 999,998 of them. If you stay where you are, it doesn't matter what's behind the rest of the doors, you might as well not even know because your chances of winning will not change.

Check the attachment... I know the code isn't perfect and I didn't put in any error control to make sure you pick a door in the range of doors you enter, but you should get the idea. You can enter the number of games to play, number of doors to choose from, and the door number you pick. Then just click Play, it'll show you the total number of wins in either choice, give you two kept doors, the winning door, and a percentage of wins on staying or switching. With 3 doors, it was roughly 66% on 100 games win on switching. With 5 doors, it was roughly 80% on 100 games win on switching. If you select the winning door, it keeps a random losing door. You really can't argue with the math here guys!

The Goats behind the Doors is like the God threads some believe and some don't.

I still don't see why the odds of the opened doors are given to the door you haven't chosen, not that it matters since we now know them to be zilch.

You can produce all of the maths that you like but a choice from two is still 1/2 even if the original was 1/infinite.

Brian

A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment.
Can you explain why?

I was stumped but my wife proffered this explanation.

The guy is very short so he cannot reach the 10th floor button, only the 7th floor. When anyone else is in the elevator he can ask them to press it for him and when it is raining he can use his umbrella to press the button. Otherwise he has to walk the last three floors.

I was stumped but my wife proffered this explanation.

The guy is very short so he cannot reach the 10th floor button, only the 7th floor. When anyone else is in the elevator he can ask them to press it for him and when it is raining he can use his umbrella to press the button. Otherwise he has to walk the last three floors.

If I were him I would carry the umbrella always. Hee Hee...

The Goats behind the Doors is like the God threads some believe and some don't.

I still don't see why the odds of the opened doors are given to the door you haven't chosen, not that it matters since we now know them to be zilch.

You can produce all of the maths that you like but a choice from two is still 1/2 even if the original was 1/infinite.

Brian

But the math proves what your brain cannot see. I can't explain it better than that. You're behaving like someone religious who can't see science proving something he believes wrong! :D :p

I was stumped but my wife proffered this explanation.

The guy is very short so he cannot reach the 10th floor button, only the 7th floor. When anyone else is in the elevator he can ask them to press it for him and when it is raining he can use his umbrella to press the button. Otherwise he has to walk the last three floors.

Correct!!! ok.. how about this one, it's pretty common, i'm sure it wont take long for someone to get.

Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29.
Where is the remaining dollar?

While i'm at it... What can you put in a bucket, that you CAN see! that makes it lighter.

Correct!!! ok.. how about this one, it's pretty common, i'm sure it wont take long for someone to get.

Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29.
Where is the remaining dollar?

Ah, trick question:

They paid 25 for the room, not 30. They paid out 27, LESS the 2 kept by the bellboy is 25.

That is the unrefundable portion of the HST so I would say that Dalton McGuinty has the $1 :rolleyes:

Correct!!! ok.. how about this one, it's pretty common, i'm sure it wont take long for someone to get.

Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29.
Where is the remaining dollar?

Ah, trick question:

They paid 25 for the room, not 30. They paid out 27, LESS the 2 kept by the bellboy is 25.

Good stuff.... but i did like Fifty2one's answer!

ok...

Brothers and sisters, I have none, but this man's father is my fathers son. Who am I?

Good stuff.... but i did like Fifty2one's answer!

ok...

Brothers and sisters, I have none, but this man's father is my fathers son. Who am I?

It works better if you include the whole riddle:

A man is holding a photograph of a man and says "brothers and sisters i

have none but this mans father is my fathers son". who is the man in the photo?

Answer: Himself

It works better if you include the whole riddle:

A man is holding a photograph of a man and says "brothers and sisters i

have none but this mans father is my fathers son". who is the man in the photo?

Answer: Himself

Yes.. when i pasted it in, i didn't copy the "top line" my mistake, thank you..
and that is not correct.

Good stuff.... but i did like Fifty2one's answer!

ok...

Brothers and sisters, I have none, but this man's father is my fathers son. Who am I?

Is it

the speaker's son?

Is it

the speaker's son?

yep!

Anyone got the bucket teaser on the previous page?

yep!

Anyone got the bucket teaser on the previous page?

I don't believe you.... I challenge you to prove it! :p

While i'm at it... What can you put in a bucket, that you CAN see! that makes it lighter.

That should be...

a hole?

That should be...

a hole?

Or.... a candle/lantern/lamp

Or.... a candle/lantern/lamp

Clever... ;)

yep!

Anyone got the bucket teaser on the previous page?


Not to be a stickler but I think the teaser should be:

"Put me in your bucket and I will lighten your load, and a poorer man ye'll be."

This teaser is attributed to Benjamin Franklin.

Okay, I read over the entire Monty Hall Problem article. I still didn't buy it, so I broke out the playing cards. Now I get it. You guys were right, I was wrong. Hard to wrap the brain around it, but I finally get it. :o

While i'm at it... What can you put in a bucket, that you CAN see! that makes it lighter.

Courtney Love's Band

Two buddies were driving from their house to Sheffield, Karen drove the first 90km, and Dave took over the rest of the trip. On the way back, Karen drove first, and Dave took over for the last 100km. Who drove furthest?

Two buddies were driving from their house to Sheffield, Karen drove the first 90km, and Dave took over the rest of the trip. On the way back, Karen drove first, and Dave took over for the last 100km. Who drove furthest?

Dave drove furthest.

Karens Total journey = 90 + y km (y being what she drove on the way back)

Dave's Total Journey = x + 100 km (x being what he drove on the way there)

Therefore we can say the following:

90 + x = 100 + y

Therefore

x = y + 10

So karen drove:

90 + y

Dave Drove

100 + x

Substituting x gives

100 + y + 10

so he drove 110 + y

So dave drove 20km more than Karen

Another puzzle:

You are playing a game with an other person. There is a perfectly circular table, and an
unlimited supply of 10p pieces. You take it in turns to place a coin on the table, taking care that no coin overlaps another (touching is fine), and no coin can stick out over the edge of the table. The first person to break either of these rules loses. What strategy must you adopt in order to be guaranteed to win every time?

Another puzzle:

You are playing a game with an other person. There is a perfectly circular table, and an
unlimited supply of 10p pieces. You take it in turns to place a coin on the table, taking care that no coin overlaps another (touching is fine), and no coin can stick out over the edge of the table. The first person to break either of these rules loses. What strategy must you adopt in order to be guaranteed to win every time?

super glue?

I don't believe you.... I challenge you to prove it! :p

I accept your challenge. :p

"Brothers and sisters I have none" -Key piece of information.
"My fathers son" is the end of the sentance and is really just a wierd way of saying "myself" because we know his father had no other children, because of the first line.

So no we have.

"This man's father is me"

I accept your challenge. :p

"Brothers and sisters I have none" -Key piece of information.
"My fathers son" is the end of the sentance and is really just a wierd way of saying "myself" because we know his father had no other children, because of the first line.

So no we have.

"This man's father is me"
Does that make sense? :confused:

"This man's father is my father's son."
Which would mean the man speaking is the father of the one in the picture.

Not

"This man is my father's son."
Which would make the man speaking the one in the picture.

Another puzzle:

You are playing a game with an other person. There is a perfectly circular table, and an
unlimited supply of 10p pieces. You take it in turns to place a coin on the table, taking care that no coin overlaps another (touching is fine), and no coin can stick out over the edge of the table. The first person to break either of these rules loses. What strategy must you adopt in order to be guaranteed to win every time?
Go first and use a table the size of a 10p coin.

Does that make sense? :confused:

"This man's father is my father's son."
Which would mean the man speaking is the father of the one in the picture.

Not

"This man is my father's son."
Which would make the man speaking the one in the picture.

Your first statement is correct.

I was simplifiying the original statement.

Brothers and sisters I have none, but this man's Father, is my Fathers son.

and as i said:
"My fathers son" is the end of the sentance and is really just a wierd way of saying "myself" because we know his father had no other children, because of the first line.

So no we have.

"This man's father is me"

Which would mean he is looking at a picture of his son.
Not sure where your second statement originated from.

I accept your challenge. :p

"Brothers and sisters I have none" -Key piece of information.
"My fathers son" is the end of the sentance and is really just a wierd way of saying "myself" because we know his father had no other children, because of the first line.

So no we have.

"This man's father is me"

Ah Ha! I see! Thank you for accepting my challenge. ;)

Go first and use a table the size of a 10p coin.


The solution would work with a table of any size

Your first statement is correct.

I was simplifiying the original statement.

Brothers and sisters I have none, but this man's Father, is my Fathers son.

and as i said:
"My fathers son" is the end of the sentance and is really just a wierd way of saying "myself" because we know his father had no other children, because of the first line.

So no we have.

"This man's father is me"

Which would mean he is looking at a picture of his son.
Not sure where your second statement originated from.
I can explain that:
1. My son wanted to get up at 3:00 this morning
2. I'm still not fully awake
3. I misread your original answer :o

I can explain that:
1. My son wanted to get up at 3:00 this morning
2. I'm still not fully awake
3. I misread your original answer :o

oh, well... Being that my body doesn't even recognize that there IS a 3 in the morning... I'll let that one slide. On a side note, you are making me really happy that I don't have kids, and really dreading the day I do.

33087

Okay, my attempt at a puzzle.

There are three neighbours (Mr Jones, Mr Brown and Mr Smith) who live in three houses (see attached image).
Each wants to be connected to the gas supply, the water supply and the electrical supply. This is shown on the diagram as a simple line between a house and a supplier (Jones' water supply is used as an example).
The different utility companies all have the following rules in common - and here's where I hope I don't miss one out:
1. No two lines can cross (either from one supplier or multiple).
2. No line can split and supply two or more houses.
3. No line can pass through a house or another supplier on its way to its target.

How would you draw the lines on the diagram to ensure that each house receives a supply from all three suppliers?

oh, well... Being that my body doesn't even recognize that there IS a 3 in the morning... I'll let that one slide. On a side note, you are making me really happy that I don't have kids, and really dreading the day I do.
It was certainly a lot different back when my only concern at that time was getting a cab home, but I wouldn't swap. :D

A ship anchored in a port has a ladder (beginning and ending with a tave), where the bottom tave touches water. The distance between taves is 20 cm and length of the ladder is 180 cm. Tide is raising water at 15 cm each hour. When will the water be on the third tave from the top?

A ship anchored in a port has a ladder (beginning and ending with a tave), where the bottom tave touches water. The distance between taves is 20 cm and length of the ladder is 180 cm. Tide is raising water at 15 cm each hour. When will the water be on the third tave from the top?
Unless it already is, a large wave hits, or the cargo is increased to change the waterline, it will never be, since the ship will rise with the tide.

A ship anchored in a port has a ladder (beginning and ending with a tave), where the bottom tave touches water. The distance between taves is 20 cm and length of the ladder is 180 cm. Tide is raising water at 15 cm each hour. When will the water be on the third tave from the top?

When it loads.

A ship anchored in a port has a ladder (beginning and ending with a tave), where the bottom tave touches water. The distance between taves is 20 cm and length of the ladder is 180 cm. Tide is raising water at 15 cm each hour. When will the water be on the third tave from the top?


Well unless the ship is sinking or a really big wave comes in, I would think never.

Well played you three, Well played.


Two girls are born to the same mother, on the same day, at the same time, in the same year and yet they're not twins. How can this be?

Well played you three, Well played.


Two girls are born to the same mother, on the same day, at the same time, in the same year and yet they're not twins. How can this be?
Two of triplets, or quads, etc.

Two girls are born to the same mother, on the same day, at the same time, in the same year and yet they're not twins. How can this be?

They were more than two of them born so they are triplets or greater.

It looks like i'm going to have to look deeper through emails for harder ones....

Four members of a band are walking to a night concert. They decide to take a shortcut, but must cross a bridge. Luckily they have one flashlight. Because of the varying size of their instruments, it takes each member a different amount of time to cross the bridge - it takes the first person one minute, the second person two minutes, the third person five minutes and the fourth person ten minutes. They must cross the bridge in pairs, traveling at the slower speed so if the one minute person went with the ten minute person, it would take a total of ten minutes. Since there is only one flashlight, one person must come back across the bridge, then another pair can cross. They only have 17 minutes to cross the bridge and still get to the concert on time. What order should they cross to get everyone across and get to the concert?

Paul’s height is six feet, he’s an assistant at a butcher’s shop, and wears size 9 shoes. What does he weigh?

Paul’s height is six feet, he’s an assistant at a butcher’s shop, and wears size 9 shoes. What does he weigh?

Meat?.....

Paul’s height is six feet, he’s an assistant at a butcher’s shop, and wears size 9 shoes. What does he weigh?

He weighs meat!

Paul’s height is six feet, he’s an assistant at a butcher’s shop, and wears size 9 shoes. What does he weigh?

Geez, you got this all wrong. I'm 6'1", wear size 10.5 shoes and I'm an IT guy, not a butcher. :p

But if I was a butcher, I'd weigh

meat

but as an IT guy it's more like 175.

Apparently I type slowly too. :o

You guys are great...
Anyone figure out the band members?


100 Politicians are attending a political convention. You know that the politicians are either corrupt or honest, nothing in between.
There is at least one honest politician.
If you randomly choose 2 politicians at least one of them will be corrupt.
How many politicians are corrupt and how many are honest?

Four members of a band are walking to a night concert. They decide to take a shortcut, but must cross a bridge. Luckily they have one flashlight. Because of the varying size of their instruments, it takes each member a different amount of time to cross the bridge - it takes the first person one minute, the second person two minutes, the third person five minutes and the fourth person ten minutes. They must cross the bridge in pairs, traveling at the slower speed so if the one minute person went with the ten minute person, it would take a total of ten minutes. Since there is only one flashlight, one person must come back across the bridge, then another pair can cross. They only have 17 minutes to cross the bridge and still get to the concert on time. What order should they cross to get everyone across and get to the concert?

p1 = 1 min
p2 = 2 min
p3 = 5 min
p4 = 10 min

p1 & p2 cross = 2 mins across and 1 min back for p1 = 3 mins
p3 & p4 cross = 10 mins and 2 mins back for p2 = 12 mins
p1 & p2 cross = 2 mins

Total time 17 mins

100 Politicians are attending a political convention. You know that the politicians are either corrupt or honest, nothing in between.
There is at least one honest politician.
If you randomly choose 2 politicians at least one of them will be corrupt.
How many politicians are corrupt and how many are honest?

99 corrupt, 1 honest.

If you choose 2 and 1 must be corrupt then there is a max of 1 honest politician

Paul’s height is six feet, he’s an assistant at a butcher’s shop, and wears size 9 shoes. What does he weigh?

He weighs meat.

DAMN!!! lol,
ok...


A mother is 21 years older than her son.
In 6 years time the son will be 5 times younger than the mother.
Where is the father?

DAMN!!! lol,
ok...


A mother is 21 years older than her son.
In 6 years time the son will be 5 times younger than the mother.
Where is the father?

21 + 6 = 27

27 / 5 = 5.4

5.4 - 6 = -0.6

So the father is probably in the mother.

21 + 6 = 27

27 / 5 = 5.4

5.4 - 6 = -0.6

So the father is probably in the mother.

LOL, ya, I love that one.. Ok well i'm fresh out for the time being. Have a good weekend everyone! (I'm leaving at lunch!! YAY! Gonna go to Niagara falls try and win me some money at the BlackJack tables!)

Another puzzle:

You are playing a game with an other person. There is a perfectly circular table, and an
unlimited supply of 10p pieces. You take it in turns to place a coin on the table, taking care that no coin overlaps another (touching is fine), and no coin can stick out over the edge of the table. The first person to break either of these rules loses. What strategy must you adopt in order to be guaranteed to win every time?

You must go first and place your coin in the center of the table? I'm thinking in this sense, no matter the size of the table, you'll have the advantage mathematically.

p1 = 1 min
p2 = 2 min
p3 = 5 min
p4 = 10 min

p1 & p2 cross = 2 mins across and 1 min back for p1 = 3 mins
p3 & p4 cross = 10 mins and 2 mins back for p2 = 12 mins
p1 & p2 cross = 2 mins

Total time 17 mins






p3 is 5 mins not 2 mins, so the 2nd line should read 5 mins back for a total of15 mins

You must go first and place your coin in the center of the table? I'm thinking in this sense, no matter the size of the table, you'll have the advantage mathematically.

Yep you have the first part but there is a further step to the strategy.

p3 is 5 mins not 2 mins, so the 2nd line should read 5 mins back for a total of15 mins


Not so, p3 is not the person coming back over p2 is.

Not so, p3 is not the person coming back over p2 is.

I was just about reply with a post full of bold text when I realised your were right.
:o

Two fathers and 2 sons go fishing. Between them they catch 3 fish. Each takes one fish home.

How?

Answer...

Father, Son, Grandfather

Answer...

Father, Son, Grandfather

Your location will be the kennel not the shed if you don't give people time to reply. :D

Brian

Well played you three, Well played.


Two girls are born to the same mother, on the same day, at the same time, in the same year and yet they're not twins. How can this be?

Superfetation!

Yep you have the first part but there is a further step to the strategy.

You need to MIRROR each of your opponent's moves?

You need to MIRROR each of your opponent's moves?

Yep thats it.