todays riddles

[Camuldonum 0]

[quote user="WeAreYellows49"]Ohhhhhh that''s a hard one, said the nun to the vicar [;)][/quote]

Being Irish, we have to keep them simple.[8-|]

[Fuglestad 0]

[quote user="Yellow Rages"]

Three men book a hotel room. The room costs 30 pounds and the men pay ten pounds each.

After arriving in the room, the men are dissapointed by the size and complain to reception. The manager agrees to refund them five pounds and sends the bell boy up with five pound coins.

The bell boy can work out how to divide the five pounds by three people so he gives them a pound back each and keeps two for himself.

The men have now paid 9 pounds each, which is 27 pounds in total. The bell boy has 2 pounds. 27+2=29. Where is the missing pound?

[/quote]

Theif! I posted this one the other day!

[yellow hammer 0]

heres one for ya

I have no legs or arms but I still eat with a fork everyday, What am I?

You, sir, are a knife.

There is a town in Texas where 5% of all the people living there have unlisted phone numbers. If you selected 100 names at random from the town''s phone directory, on average, how many of these people would have unlisted phone numbers?

If they''re unlisted they wouldn''t be in the phone book

Here''s an Irish one:

My face was washed with water that was neither rain nor run Dew?

My face was dried by something that was neither woven nor spun The  wind?

 

You, are a prisoner alone in a cell with two doors,

At midnight you are allowed to go through one of those doors.

One is to freedom,the other is to the gallows,

Your prison guards are Identical twin brothers, one is a born lier, the other can only tell the truth,

Now at midnight you can ask one of these gaurds 1 question, then you go through a door,

What would your Question be.

Which door would the other guard says leads to freedom? (You have to establish which guard is the liar)

 

OK here''s one,

According to army regulations, only the regimental barber is allowed to cut men''s hair, or face punishment. Since the barber cannot practically cut his own hair, how does he get a haircut without facing punishment?

 

[Camuldonum 0]

I''ll give you the Irish one although in the original the sun does the drying but the wind would have the same effect.

[Syteanric 1]

you are the pilot of an Air craft travelling daily from Heathrow to NYC.

on Monday you fly 2000 people

On tuesday you fly 1800 people.

on Wednesday you fly 400 people

on thursday you fly 1200 people

On Friday you fly 900 people.

How olds the pilot?

jas :)

[yellow hammer 0]

[quote user="jas the barclay king"]

you are the pilot of an Air craft travelling daily from Heathrow to NYC.

on Monday you fly 2000 people

On tuesday you fly 1800 people.

on Wednesday you fly 400 people

on thursday you fly 1200 people

On Friday you fly 900 people.

How olds the pilot?

jas :)

you are the pilot might provide a clue?

[/quote]

[pete_norw 0]

Ok then a much easier one, What was the prime minister''s name in 1979.?

[YankeeCanary 0]

[quote user="pete_norw"]

Have a go at this one,

You, are a prisoner alone in a cell with two doors,

At midnight you are allowed to go through one of those doors.

One is to freedom,the other is to the gallows,

Your prison guards are Identical twin brothers, one is a born lier, the other can only tell the truth,

Now at midnight you can ask one of these gaurds 1 question, then you go through a door,

What would your Question be.

(only one guard  will appear at midnight)

[/quote]

"I have to walk through one of these doors but, regardless of what happens to me, do you happen to know if Peter Cullum is buying Norwich City?"

[pete_norw 0]

[quote user="yellow hammer"]

heres one for ya

I have no legs or arms but I still eat with a fork everyday, What am I?

You, sir, are a knife.

There is a town in Texas where 5% of all the people living there have unlisted phone numbers. If you selected 100 names at random from the town''s phone directory, on average, how many of these people would have unlisted phone numbers?

If they''re unlisted they wouldn''t be in the phone book

Here''s an Irish one:

My face was washed with water that was neither rain nor run Dew?

My face was dried by something that was neither woven nor spun The  wind?

 

You, are a prisoner alone in a cell with two doors,

At midnight you are allowed to go through one of those doors.

One is to freedom,the other is to the gallows,

Your prison guards are Identical twin brothers, one is a born lier, the other can only tell the truth,

Now at midnight you can ask one of these gaurds 1 question, then you go through a door,

What would your Question be.

Which door would the other guard says leads to freedom? (You have to establish which guard is the liar)

The question is, Which door would your brother tell me to go through, then you take the opposite door.

Which ever gaurd you speak to will tell you the opposite door,

 

 

OK here''s one,

According to army regulations, only the regimental barber is allowed to cut men''s hair, or face punishment. Since the barber cannot practically cut his own hair, how does he get a haircut without facing punishment?

 

[/quote]

Haven''t got a clue on this one

[yellow hammer 0]

[quote user="pete_norw"][quote user="yellow hammer"]

heres one for ya

I have no legs or arms but I still eat with a fork everyday, What am I?

You, sir, are a knife.

There is a town in Texas where 5% of all the people living there have unlisted phone numbers. If you selected 100 names at random from the town''s phone directory, on average, how many of these people would have unlisted phone numbers?

If they''re unlisted they wouldn''t be in the phone book

Here''s an Irish one:

My face was washed with water that was neither rain nor run Dew?

My face was dried by something that was neither woven nor spun The  wind?

 

You, are a prisoner alone in a cell with two doors,

At midnight you are allowed to go through one of those doors.

One is to freedom,the other is to the gallows,

Your prison guards are Identical twin brothers, one is a born lier, the other can only tell the truth,

Now at midnight you can ask one of these gaurds 1 question, then you go through a door,

What would your Question be.

Which door would the other guard says leads to freedom? (You have to establish which guard is the liar)

The question is, Which door would your brother tell me to go through, then you take the opposite door.

Which ever gaurd you speak to will tell you the opposite door,

 

 

OK here''s one,

According to army regulations, only the regimental barber is allowed to cut men''s hair, or face punishment. Since the barber cannot practically cut his own hair, how does he get a haircut without facing punishment?

 

[/quote]

Haven''t got a clue on this one

[/quote]

It''s unsolvable and is an example of logic called Russell''s Paradox. From wiki:

 

 Informal presentation

Let us call a set "abnormal" if it is a member of itself, and "normal" otherwise. For example, take the set of all squares. That set is not itself a square, and therefore is not a member of the set of all squares. So it is "normal". On the other hand, if we take the complementary set of all non-squares, that set is itself not a square and so should be one of its own members. It is "abnormal".

Now we consider the set of all normal sets – let us give it a name: R. If R were abnormal, that is, if R were a member of itself, then R would be normal since all its members are. So, R cannot be abnormal, which means R is normal. Further, since every normal set is a member of R, R itself must be a member of R, making R abnormal. Paradoxically, we are led to the contradiction that R is both normal and abnormal.

A longer argument often given for this contradiction, by Russell himself, for example, proceeds by cases. Is R a normal set? If it is normal, then it is a member of R, since R contains all normal sets. But if that is the case, then R contains itself as a member, and therefore is abnormal. On the other hand, if R is abnormal, then it is not a member of R, since R contains only normal sets. But if that is the case, then R does not contain itself as a member, and therefore is normal. Clearly, this is a paradox: if we suppose R is normal we can prove it is abnormal, and if we suppose R is abnormal we can prove it is normal. Hence, R is both normal and abnormal, which is a contradiction.

Formal derivation

Let R be "the set of all sets that do not contain themselves as members". Formally: A is an element of R if and only if A is not an element of A. In set-builder notation:

Nothing in the system of Frege''s Grundgesetze der Arithmetik rules out R being a well-defined set. The problem arises when it is considered whether R is an element of itself. If R is an element of R, then according to the definition R is not an element of R. If R is not an element of R, then R has to be an element of R, again by its very definition. The statements "R is an element of R" and "R is not an element of R" cannot both be true, thus the contradiction.

The following fully formal yet elementary derivation of Russell''s paradox^[1] makes plain that the paradox requires nothing more than first-order logic with the unrestricted use of set abstraction. The proof is given in terms of collections (all sets are collections, but not conversely). It invokes neither set theory axioms nor the law of excluded middle explicitly or tacitly.

Definition. The collection , in which is any predicate of first-order logic in which is a free variable, denotes the individual satisfying .

Theorem. The collection is contradictory.

Proof. Replace in the definition of collection by , so that the implicit definition of becomes Instantiating by then yields the contradiction

Remark. The above definition and theorem are the first theorem and definition in Potter (2004), consistent with the fact that Russell''s paradox requires no set theory whatsoever. Incidentally, the force of this argument cannot be evaded by simply proscribing the substitution of for . In fact, there are denumerably many formulae giving rise to the paradox.^[2] For some examples, see reciprocation below.

The paradox holds in intuitionistic logic

Often derivations of Russell''s paradox employ the law of the excluded middle (Russell''s own derivation did). Thus it may be tempting to conclude that the paradox is avoided if the law of excluded middle is disallowed, as with intuitionistic logic. However, as is clear from the first of the two informal arguments presented above, the law of excluded middle is not needed for the paradoxical argument. Here we show that the paradox can still be generated formally by means of the intuitionistically valid law of non-contradiction.

Theorem. The collection is contradictory even if the background logic is intuitionistic.

Proof. From the definition of , we have that . Then (biconditional elimination). But also (the law of identity), so . But by the law of non-contradiction we know that . By modus tollens we conclude .

But since , we also have that , and so we also conclude by modus ponens. Hence we have deduced both and its negation using only intuitionistically valid methods.

More simply, it is intuitionistically impossible for a proposition to be equivalent to its negation. Assume P ⇔ ¬P. Then P ⇒ ¬P. Hence ¬P. Symmetrically, we can derive ¬¬P, using ¬P ⇒ P. So we have inferred both ¬P and its negation from our assumption, with no use of excluded middle.

 Reciprocation

Russell''s paradox arises from the supposition that one can meaningfully define a class in terms of any well-defined property Φ(x); that is, that we can form the set P = {x | Φ(x) is true }. When we take , we get Russell''s paradox. This is only the simplest of many possible variations of this theme.

For example, if one takes , one gets a similar paradox; there is no set P of all x with this property. For convenience, let us agree to call a set S reciprocated if there is a set T with ; then P, the set of all non-reciprocated sets, does not exist. If , we would immediately have a contradiction, since P is reciprocated (by itself) and so should not belong to P. But if , then P is reciprocated by some set Q, so that we have , and then Q is also a reciprocated set, and so , another contradiction.

Any of the variations of Russell''s paradox described above can be reformulated to use this new paradoxical property. For example, the reformulation of the Grelling paradox is as follows. Let us agree to call an adjective P "nonreciprocated" if and only if there is no adjective Q such that both P describes Q and Q describes P. Then one obtains a paradox when one asks if the adjective "nonreciprocated" is itself nonreciprocated.

This can also be extended to longer chains of mutual inclusion. We may call sets A₁,A₂,...,A_n a chain of set A₁ if for i=1,2,...,n-1. A chain can be infinite (in which case each A_i has an infinite chain). Then we take the set P of all sets which have no infinite chain, from which it follows that P itself has no infinite chain. But then , so in fact P has the infinite chain P,P,P,... which is a contradiction. This is known as Mirimanoff''s paradox.