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        riddles >> general problem-solving / chatting / whatever >> paradoxes
        
(Message started by: srn347 on Aug 28^th, 2007, 9:46pm)

Title: paradoxes
Post by srn347 on Aug 28^th, 2007, 9:46pm

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

Try solving the paradoxes.

Title: Re: paradoxes
Post by mikedagr8 on Aug 29^th, 2007, 12:19am

Quote:

Buridan's ass: How can a rational choice be made between two outcomes of equal value?

Lucky dip, blind choice etc ;D

Quote:

Boy or Girl: A two-child family has at least one boy. What is the probability that it has a girl?

Quote:

An ambiguous real-life version
Two old classmates, Anna and Brian, meet in the street, not having seen each other since they left school.

Anna asks Brian: "Have you got any children?"
Brian answers: "Yes, I've got two."
Anna: "Do you have a boy?"
Brian: "Yes, I do!"
Here, for some reason, the conversation is cut short.

Formally, this corresponds to the second version as Brian only has told Anna that at least one child is a boy. Accordingly, the probability that Brian has a girl should be 2/3. However, in real conversation, if Brian had two boys, he would be more likely to answer e.g. "Yes, they are both boys!" The fact he does not answer like that could reasonably be taken by Anna as a clue increasing her posterior probability of one child being a girl above 2/3. This highlights the need for precision when stating such problems in probability.

It seems to me that the answer is that the other child MUST be a girl, because
Anna: "Do you have a boy?"
Brian: "Yes, I do!"

Anna asks if he has 1, no more , no less, so the other must be a girl. But that is my reasoning, so bring on arguments, I am open minded.

Title: Re: paradoxes
Post by srn347 on Aug 29^th, 2007, 9:51am

Good answer, but no. The other child doesn't have to be a girl.

Title: Re: paradoxes
Post by ThudanBlunder on Aug 29^th, 2007, 10:40am

on 08/28/07 at 21:46:18, srn347 wrote:

http://en.wikipedia.org/wiki/List_of_paradoxes
Try solving the paradoxes.

Try taking your Ritalin.

Title: Re: paradoxes
Post by srn347 on Aug 29^th, 2007, 10:43am

Ritalin? What is that, a species of archaebacteria?

Title: Re: paradoxes
Post by mikedagr8 on Aug 29^th, 2007, 5:31pm

Quote:

But that is my reasoning, so bring on arguments, I am open minded.

Quote:

Good answer, but no. The other child doesn't have to be a girl.

You didn't argue, you just said what is on the site. I mean, find why my argument is flawed. It is logical and makes sense, so it is correct.

Title: Re: paradoxes
Post by srn347 on Aug 29^th, 2007, 5:39pm

If a person has two children and at least one of them is a boy, what is the probability the other one is a girl. What if the older one is a boy.

Title: Re: paradoxes
Post by mikedagr8 on Aug 29^th, 2007, 5:41pm

Then he would answer to "Do you have a boy?" he would say no, I have none or 2.

Title: Re: paradoxes
Post by srn347 on Aug 29^th, 2007, 5:42pm

If you refuse to take it seriously, try another one.

Title: Re: paradoxes
Post by mikedagr8 on Aug 29^th, 2007, 5:43pm

on 08/29/07 at 17:42:35, srn347 wrote:

If you refuse to take it seriously, try another one.

You should do the same.

Title: Re: paradoxes
Post by srn347 on Aug 29^th, 2007, 5:47pm

Which one am I not taking seriously?

Title: Re: paradoxes
Post by pex on Aug 29^th, 2007, 11:32pm

on 08/29/07 at 17:47:28, srn347 wrote:

Which one am I not taking seriously?

Like, the rest of the world?

Title: Re: paradoxes
Post by srn347 on Aug 30^th, 2007, 10:11am

Yet you have no specific examples.

Title: Re: paradoxes
Post by Barukh on Aug 31^st, 2007, 5:07am

I think moderators should take care of this.

Title: Re: paradoxes
Post by mikedagr8 on Aug 31^st, 2007, 5:12am

Sorry srn347, I apologise for intimidating you. I'll be deleting the offensive post(s).

Title: Re: paradoxes
Post by ThudanBlunder on Aug 31^st, 2007, 6:21am

on 08/31/07 at 05:07:20, Barukh wrote:

I think moderators should take care of this.

I am THUD, Janissary at the Gates of Hard. 8)
Barukh, why have you put the frighteners on wee mikedagr8? ;D
All his wimpy little insults have now been lost to posterity for ever.

Title: Re: paradoxes
Post by mikedagr8 on Aug 31^st, 2007, 4:16pm

Lol, Not really, just I am trying not to piss everyone off. If I can leave it at one person in the real world, that's enough for me. So I'll try not to post B.S. to often.

Title: Re: paradoxes
Post by srn347 on Aug 31^st, 2007, 9:38pm

Let the riddle solving continue.

Title: Re: paradoxes
Post by Barukh on Sep 1^st, 2007, 5:01am

on 08/31/07 at 21:38:22, srn347 wrote:

Let the riddle solving continue.

Could you please restate explicitly what riddle are we trying to solve?

Title: Re: paradoxes
Post by Archae on Sep 1^st, 2007, 7:45am

on 08/29/07 at 00:19:44, mikedagr8 wrote:

Anna asks if he has 1, no more , no less, so the other must be a girl. But that is my reasoning, so bring on arguments, I am open minded.

I do not believe that Anna specifically asks if Brian has 1 (no more, no less) boy; she only asks if he has a boy. If Brian does indeed have two boys, he could (although it would be strange) respond with a 'yes' and not a modification such as 'Both of my children are boys.'

Title: Re: paradoxes
Post by ThudanBlunder on Sep 1^st, 2007, 1:50pm

on 08/28/07 at 21:46:18, srn347 wrote:

http://en.wikipedia.org/wiki/List_of_paradoxes
Try solving the paradoxes.

As your post is merely an invitation to view an interesting page, it does not belong in Hard.
I have therefore moved it.

Title: Re: paradoxes
Post by mikedagr8 on Sep 1^st, 2007, 4:39pm

on 08/29/07 at 17:31:56, mikedagr8 wrote:

You didn't argue, you just said what is on the site. I mean, find why my argument is flawed. It is logical and makes sense, so it is correct.

on 09/01/07 at 07:45:51, Archae wrote:

That, is what I was looking for.

Title: Re: paradoxes
Post by mfirmata on Sep 5^th, 2007, 9:16am

Quote:

"Is the answer to this question no?"

I asked a guy I work with and he answered, "Know?"

At least, that's what it sounded like he said.

Title: Re: paradoxes
Post by srn347 on Sep 8^th, 2007, 8:10pm

on 09/01/07 at 13:50:43, ThudanBlunder wrote:

As your post is merely an invitation to view an interesting page, it does not belong in Hard.
I have therefore moved it.

Understood.

Title: Re: paradoxes
Post by temporary on Feb 12^th, 2008, 11:01pm

http://en.wikipedia.org/wiki/Necktie_Paradox
You could also say it is in both of their disadvantage since they each either win the other person's tie, or lose a tie more valuable than the other person's tie.

Title: Re: paradoxes
Post by Icarus on Feb 13^th, 2008, 8:09pm

Which still leaves you with the apparent paradox. What is wrong with either the original, or your version, of the reasoning? Clearly something is, as each version says that both people have the advantage (in your case, each thinks the other has the advantage).

There are four ways to deal with paradoxes:
1) Note that the "paradox" is not really a paradox, because there is no actual contradiction. Buridan's Ass is an example of this. The poor critter may be left to starve, but there is no contraction in this. The Banach-Tarski paradox is another example. The only thing it contradicts is our intuition.
2) Show that the "paradox" is a result of misinterpreting the situation. This is the case in the "two children" paradox, or Monty Hall paradox. One calculation of the probability makes a false assumption.

If the paradox is real, you have the remaining two means of resolving it:
3) Redefine the situation so that the paradox is prevented from arising. Such is the way Russell's paradox was handled. Set theory was redefined so that either not all relations determined sets (Zermelo-Frankel), or else so that paradoxical relations could not be formed in the first place (Russell-Whitehead).
4) Accept that the language in question allows paradoxical statements. As such, the full language is not an appropriate place for deductive logic, though subsets of the language may be. This is the resolution of many natural language paradoxes. They just exist, so get over it. Just don't trust anything that comes out of them.

For the Necktie Paradox, (2) is the appropriate means of resolution. The men's calculation of the value of their bet is flawed. To resolve the paradox means finding the mistake they made.

So far, all you've done is restate the flawed argument from the other side.

Title: Re: paradoxes
Post by mikedagr8 on Feb 13^th, 2008, 11:31pm

on 02/13/08 at 20:09:55, Icarus wrote:

So far, all you've done is restate the flawed argument from the other side.

Temporary could play a ping-pong like that with himself.

Title: Re: paradoxes
Post by temporary on Feb 14^th, 2008, 8:53pm

I've tried tic-tac toe against myself, but I just can't win, and I know I didn't let myself win. Anyway, there are two perspectives and two people. One perspective gives them both the advantage, but the other gives them both a disadvantage, this is no coincidence. One has an advantage, one has a disadvantage, but you don't know which is which, so it's fair.

Title: Re: paradoxes
Post by Icarus on Feb 15^th, 2008, 3:24pm

That the situation is fair, and neither has an advantage is clear, since their situations are symmetric. The "paradox" is that the reasoning shows them both to have an advantage (or disadvantage from the view you stated). This is clearly impossible. Therefore, either there is something wrong with their reasoning, or this is a true paradox.

Given that this isn't a true paradox, the problem becomes: what is the flaw in their reasoning? All you have shown is that they can also by the same flaw, reason in the opposite direction. But you have not spotted the flaw itself.

Title: Re: paradoxes
Post by BenVitale on Mar 7^th, 2008, 10:11am

I found this paradox with solution. But in the solution there's a sentence i'm not sure about.

Imagine a pack of cards, each of which has one number on one side and the number directly above on the other. There is one card with 1 on one side and 2 on the other, two cards with 2 on one side and 3 on the other, four cards with 3 and 4, eight cards with 4 and 5, and so on, ad infinitum, so that there are 2^{n+1} cards with n on one side and n+1 on the other.

Let us now use that pack to play a game of chance.

Two people, Ann and Brian, stand facing each other,while the host draws one card from the pack and puts it between them, so that each can see the side facing him or her, but not the other side. The winner is the one who sees the lowest number.

What is Ann’s probability of winning? If she sees a 1, the other side must be a 2, and she has won. If she sees a number n>1, the hidden number, on the other side of the card, is either n-1 or n+1. In the first case she loses, in the other case she wins. Since there are twice as many cards of the second type as there are cards of the first, she has a probability 2/3 of winning.

Unfortunately, the same argument holds for Brian: he also has a probability 2/3 of winning. Since one must win, but not both, the two probabilities should sum to 1, so that 2/3 + 2/3 =1, a remarkable equality!!!

This is a true paradox. The argument is perfectly correct. Still, there is a problem with this argument.

Solution:
The argument is perfectly correct. The only problem with it is that there is no such pack of cards. One cannot physically construct a pack with infinitely many cards. Even if one could, what this argument shows is that one could not draw at random a card from it, otherwise one would end up with a contradiction. In other words, even in mathematics, there is no such thing as drawing a card at random from an infinite pack.

It is the last sentence of the solution,"In other words, even in mathematics, there is no such thing as drawing a card at random from an infinite pack."

Could we say that we can draw cards at random from an infinite pack, but not with equal probability to each card. If we assign a discrete probability measure on an infinite deck the probability of different cards will be different and the argument breaks.

Title: Re: paradoxes
Post by Eigenray on Mar 7^th, 2008, 12:59pm

Yes, that's right. We can pick card {n,n+1} with probability p_n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif [0,1], and there is no problem as long as http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif p_n = 1.

However, in the 'paradox', they are picking the card at random in such a way that {n,n+1} is always twice as likely as {n-1,n}, i.e., p_n = 2p_n-1. This is absurd; there is no way to have http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif p_n = 1.

The probability of seeing the number n is (p_n+p_n-1)/2. The probability of winning, given that you see the number n, is p_n/(p_n+p_n-1). So the probability that you win is

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif (p_n+p_n-1)/2 * p_n/(p_n+p_n-1) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif p_n/2 = 1/2,

as it should be.

For a non-absurd example, we can take p₀=0, and p_n = 1/2ⁿ, n>0. Then the probability of winning given that you see a 1 is 1, while the probability of winning given that you see any other number is only 1/3. But since you see a 1 with probability 1/4, your overall probability of winning is 1/4*1+3/4*1/3 = 1/2.

Title: Re: paradoxes
Post by Christine on Mar 7^th, 2008, 2:05pm

I wonder whether or not Cantor's diagonal and Russell's self-reference argment are linked or connected.
Could anyone explain the relationship between the diagonal argument and self-reference?

Title: Re: paradoxes
Post by Icarus on Mar 7^th, 2008, 4:33pm

Russell's demonstration can be thought of as a sort of super-diagonal argument. Let U be the universe. To every set S in U we can assign a function c_S : U --> {0,1} defined by c_S(x) = 1 if x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif S, = 0 if x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif S. Note that S = c_S^-1(1).

The functions c_S serve in the place of Cantor's exhaustive sequence. Now we go along the diagonal to construct a function c : U --> {0,1} that cannot be c_S for any S.

Define c(x) = 0 if x is not a set or if x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif x, c(x) = 1 if x is a set, but x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gifx. We see that for any set S, if c_S(S) = 1, then c(S) = 0. But if c_S(S) = 0, then c(S) = 1. Therefore, for no set S is c = c_S.

Up to this point, there is a fairly direct connection between the two arguments. Here, they diverge. Cantor's adiagonal number doesn't lie in the sequence, so the sequence cannot be all real numbers. Well and good.

Russell's adiagonal c cannot be any characteristic function c_S, so the characteristic functions cannot be all the functions from U --> {0,1}. Alas, this is not so good, as c clearly IS the characteristic function of the set c^-1(1).

This is why Cantor's argument simply shows that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif is bigger than http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbn.gif, while Russell's argument is a true paradox that required re-inventing mathematics to not allow the shenanigans I pulled off above.

Title: Re: paradoxes
Post by Christine on Mar 8^th, 2008, 10:54am

i have a question about Russell's paradox:
The set M is the set of all sets that do not contain themselves as members. Does M contain itself?
Are the theory of types or Zermelo's special axiom of comprehention or VonNeumanns proper classes, required to show that the Russell class, {x:~(x e x)} does not exist?

could we say...
~EyAx(xRy <-> ~(xRx)) is sufficient. ie. ~EyAx(x e y <-> ~(x e x)) is a theorem. The Russell class does not exist.

Title: Re: paradoxes
Post by Icarus on Mar 8^th, 2008, 3:01pm

Zermelo's axiom of comprehension was not created to solve Russell's paradox. Zermelo's solution to Russell's paradox is simply this: drop the axiomatic schema "for all unary relations R, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/exists.gifyhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/forall.gifx(R(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigleftrightarrow.gifx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gify) is an axiom". This schema of naive set theory is responsible for the paradox. Zermelo resolved the paradox by dropping it, but this meant he needed to give some other tools of set creation in it's place. That is why he introduced the axiom of comprehension. The axiom itself has nothing to do with the paradox. Similar remarks also apply to the theory of types and VonNeumann's proper classes. Both were introduced as alternatives to give us back as much as possible of what we lost when the naive schema above was revoked, without re-introducing the paradoxes.

And "paradoxes" is why your approach doesn't work. It is not one paradox we must avoid. Russell's is just the simplest. But there are infinitely many alternate versions. We do not have to simply avoid his paradoxical set, but the entire situation that allows paradoxes to arise.

Another paradox - not directly related to Russell's - is this one (in many different variants): "the least integer not describable in nine English words." Note that the sentence describes the number in 9 words. This paradox was resolved by removing relational variables from the mathematical theory itself into metamathematics.

Title: Re: paradoxes
Post by Eigenray on Mar 8^th, 2008, 6:23pm

Cantor's diagonal argument goes like this: Let S be a set, and suppose f : S http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/scrp.gif(S) is any function (so for all x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif S, f(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif S). Then A={x : x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif f(x)} is not in the image of f. For, if f(x) = A, then x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif f(x) iff x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif f(x), a contradiction.

For example, if we have a set of people S, and f(x) is the set of people that x shaves, then A is the set of people who don't shave themselves. So there can be no x such that x shaves all those who don't shave themselves. But there's no paradox here. We just have Cantor's theorem that for any set S, there is no surjection from S to the power set of S.

But there's a problem if we let U be the universe. For any x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U, x = { y : y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif x} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif U, so the identity maps U to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/scrp.gif(U). But now Cantor's argument tells us that A = { x : x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif x } is not in the image of the identity function, i.e., it doesn't exist! But this is only a paradox if you assume that A does exist (i.e., if you assume A is a set).

If the universe were a set, then we would have by the axiom of comprehension that { x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U : x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif x} is also a set, and this would be a contradiction. So the way out (in ZFC) is just to accept the fact that the universe is not a set.

As for the Berry paradox, it is interesting to try to formalize it. Fix a universal Turing machine U, and write U(x)=y if given program x, U halts with output y. Let K(y) be the length of the shortest string x such that U(x)=y. Then the Berry paradox becomes:

Let x_n be a program which computes the smallest positive integer y such that K(y) > n, so U(x_n)=y, but for all strings x of length |x|http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif n, U(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif y.

The problem now is that if x_n exists, then surely |x_n| = O(log n), so |x_n|<n for sufficiently large n, and therefore U(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif U(x_n).

The conclusion is therefore that x_n does not exist. If K(y) were computable, then we could simply write a program to compute K(1), K(2), ..., until we find y with K(y) > n. So therefore K(y) is not computable. And if the halting problem were computable, then K(y) would be computable, since we could simply run through all programs in order until we find x with U(x)=y. Therefore the halting problem is not computable.

Of course, the easier explanation of the Berry paradox is that if F is a function from strings to integers, then F("the smallest integer which is not F(s) for some string s of length < n") need not actually be the smallest integer which is not F(s) for some string s of length < n.

Title: Re: paradoxes
Post by BenVitale on Mar 9^th, 2008, 12:07pm

How do you get the paradox from ""for all unary relations R, Ey, for all x R(x) <=> Ey) is an axiom". ?

It seems to me that "Ey, for all x R(x) <=> Ey)" is not well formed.

Did you mean; .. for some y, for all x (R(x) <=> (x is a member of y))?

It is easy to prove that the naive comprehention axiom is invalid...
From ie. ~EyAx(x e y <-> ~(x e x)) is a theorem, we can assert EF(~EyAx(x e y <-> Fx)),
That is ~AF(EyAx(x e y <-> Fx)) is a theorem.

Title: Re: paradoxes
Post by Icarus on Mar 9^th, 2008, 2:18pm

on 03/09/08 at 12:07:50, BenVitale wrote:

Yes, that is what I meant - and if you go back and look at my post, you will see that that is also what I said. You've misread it.

Quote:

It is easy to prove that the naive comprehention axiom is invalid...
From ie. ~EyAx(x e y <-> ~(x e x)) is a theorem, we can assert EF(~EyAx(x e y <-> Fx)),
That is ~AF(EyAx(x e y <-> Fx)) is a theorem.

Yes, it is easy to show it NOW, more than a century after mathematicians first formalized things enough to come across the paradox. Remember, you were trained in the formalism, and you have benefitted from a century of discussion of it. To the mathematicians of that day, a rigorous approach to set theory was a new concept, and the discovery of these paradoxes was something entirely unexpected. It took careful study for them to figure out where the problem was, and what were the best ways of fixing it.

Title: Re: paradoxes
Post by BenVitale on Mar 9^th, 2008, 4:21pm

No, I don't think I've misread it, I may be wrong, but, perhaps your claim is not correct.
Could you please enlighten me?
"for all unary relations R, Ey, for all x R(x) <=> Ey)", what is that? what does it mean?

Title: Re: paradoxes
Post by Icarus on Mar 9^th, 2008, 7:03pm

I don't know what it means because YOU and YOU ONLY have made that statement.

My statement does not drop the x and turn the element sign into some meaningless big E.

This is why I asked you to re-read it. I had hoped you would actually do so, and carefully this time.

Title: Re: paradoxes
Post by rmsgrey on Mar 10^th, 2008, 11:59am

on 03/09/08 at 16:21:39, BenVitale wrote:

No, I don't think I've misread it[...]
"for all unary relations R, Ey, for all x R(x) <=> Ey)"

on 03/08/08 at 15:01:13, Icarus wrote:

"for all unary relations R, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/exists.gifyhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/forall.gifx(R(x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bigleftrightarrow.gifx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gify) is an axiom"

x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gify
is not equivalent to
Ey

Title: Re: paradoxes
Post by BenVitale on Mar 11^th, 2008, 12:31pm

Sorry guys. I've made an error. Mea Culpa!

Title: Re: paradoxes
Post by temporary on Mar 11^th, 2008, 10:39pm

Are you all still discussing the tie paradox, or solving random math problems?

Title: Re: paradoxes
Post by BenVitale on Apr 21^st, 2008, 3:48pm

Take two round coins of equal size. Hold one still so that it does not move and then roll the other coin around it. Make sure the rims touch at all times. So my question is, how many times will the moving coin have rotated after it has completed one revolution of the stationary coin?

Title: Re: paradoxes
Post by ThudanBlunder on Apr 21^st, 2008, 4:23pm

on 04/21/08 at 15:48:30, BenVitale wrote:

So my question is, how many times will the moving coin have rotated after it has completed one revolution of the stationary coin?

Twice (http://mathworld.wolfram.com/CoinParadox.html).

Title: Re: paradoxes
Post by BenVitale on Apr 21^st, 2008, 5:03pm

But, we could argue that relative to a fixed point on the stationary coin, the moving coin only rotates once.

Title: Re: paradoxes
Post by ThudanBlunder on Apr 21^st, 2008, 5:28pm

on 04/21/08 at 17:03:38, BenVitale wrote:

But, we could argue that relative to a fixed point on the stationary coin, the moving coin only rotates once.

I was aware of the ambiguity in your question but presumed you were looking for the more interesting answer.

The same phenomenon explains the difference between solar time and sidereal time.
That is, why the constellations appear roughly four minutes earlier per day.

Title: Re: paradoxes
Post by temporary on Apr 22^nd, 2008, 5:20pm

Not knowing what it is will give you the false result of gaining x or losing x/2. The problem is you change the sum of the envelopes from 3x to 3/2 x when you do that. If you consider that x can be in either one, you either gain x or lose x. If you know how much your envelope has, you consider that switching is best. the fallacy is that you change how much the sum is between the 2 outcomes, which influences a variable that shouldn't change.

Title: Re: paradoxes
Post by BenVitale on May 5^th, 2008, 11:27pm

Voting paradox :

There are 3 candidates: Sen. John McCaine, Sen. Obama and Sen. Hillary Clinton.

Let Sen. John McCaine be A, Sen. Obama be B, and Sen. Hillary Clinton be C.

Normally, in a voting system, if A beats B and B beats C we might reasonably expect A to beat C.

But, in the following example, it doesn't work this way :

Consider the case where 3 voters cast the following votes: ABC, BCA and CAB:

A beats B by 2 choices to 1.
B beats C by 2 choices to 1
but A loses to C, again by 2 choices to 1.

Find the probability of collective choice arising by considering all the possible permutations of the votes that could be cast by 3 voters ranking 3 candidates by order of preference in this way.

Title: Re: paradoxes
Post by towr on May 6^th, 2008, 7:19am

on 05/05/08 at 23:27:14, BenVitale wrote:

Voting paradox

It's no more paradoxical than rock-paper-scissors, to be honest.

Quote:

Find the probability of collective choice arising by considering all the possible permutations of the votes that could be cast by 3 voters ranking 3 candidates by order of preference in this way.

Well, there's only 36 combination to consider, so you can easily write it out.
But I'm not sure what the criterion is for a collective choice. Suppose we have ABC ABC BCA, then we get
A beats B 2 to 1
A beats C 2 to 1
B beats C 3 to 0
We have 4 times A is preferred to someone, 4 times B is preferred to someone, only once C is preferred. A is preferred to everyone by 2 out of 3 voters. Should we go with A, or is it tied between A and B?

Title: Re: paradoxes
Post by BenVitale on May 6^th, 2008, 10:15am

Take 3 people : #1, #2 and #3

Suppose these 3 people have these preferences:
#1 : A > B > C
#2 : C > A > B
#3 : B > C > A

If we ask these 3 people to make a group choice, that is majority vote, between A and B, they would choose A, and a choice between B and C, then they would choose B, and between C and A, then they would choose C.

I expected this relation to be transitive, but it is not!

Title: Re: paradoxes
Post by towr on May 6^th, 2008, 10:50am

Yes, I get that; that's exactly what you said in the previous post. Unfortunately it doesn't address my question.
You wanted the probability that there is a collective choice; I want to know what the condition is for a collective choice.

Suppose if we have a ranking ABC, that we thengive 2 points to A, 1 to B, and 0 to C.
Then if three people vote ABC, ABC and BCA respectively, A and B both get 4 points. Which would be a draw in this scheme.

Now consider another scheme, where given ABC, we gave A 3 points, B 1 and C 0.
Then if three people vote ABC, ABC and BCA (as before) respectively, A gets 6 points and B gets 5. So with this scheme A wins.
So the same votes give different results depending on the scheme for aggregating the votes.

Other schemes (not even necessarily using a point system) are possible.
So what I want to know, who wins in this case, if anyone. What is the criterion for a collective choice? Otherwise I won't be able to answer your question.
If you pick the first scheme, then the answer is a probability of 31/36.

Title: Re: paradoxes
Post by BenVitale on May 6^th, 2008, 11:45am

Sorry, you're right. I got stuck with the transitivity/non-transitivity aspect. I did not frame it properly to resolve it.
Are you suggesting that if we reduce the number of cases to analyze in order to consider what happens if two voters agree on their first choice?

If we frame it differently, then we don't have a paradox.

How about in real life?
would you change your vote if your favorite candidate is falling behind? If your vote isn't likely to matter a lot, in the sense of breaking a tie,
would you still vote for your favorite candidate? Many perceive that it would be wasting your vote to vote for a candidate who has a diminished chance of winning. It was reported on the news that people who vote for the democrat candidates are divided. If, for example, Sen. Obama gets the nomination, a number of Sen. Hillary Clinton's supporters would switch side and vote for the republican candidate.
I just find this type of dynamics fascinating.

Title: Re: paradoxes
Post by towr on May 6^th, 2008, 12:16pm

on 05/06/08 at 11:45:28, BenVitale wrote:

Are you suggesting that if we reduce the number of cases to analyze in order to consider what happens if two voters agree on their first choice?

I'm not quite sure what to make of this sentence.
I just want to know how you propose combining the results from the three voters, such that you get a result that either is or is not decisive.
For ABC, BCA, CAB, no scheme will give a decisive result; but for other votes, for example ABC, ABC, BCA, some schemes give a decisive result while others don't.

So just tell me; if the three voters vote ABC, ABC, BCA, who, if anyone, wins.

Quote:

If we frame it differently, then we don't have a paradox.

There isn't a paradox either way; non-transitivity is not a paradox.

Title: Re: paradoxes
Post by BenVitale on May 6^th, 2008, 1:39pm

At this point i cannot answer, i'm looking at these 2 articles:

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

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

Title: Re: paradoxes
Post by BenVitale on Jul 4^th, 2008, 5:10pm

Murphy's Law says that if anything can go wrong, it will. But, this stupid law applies to itself: itself can go wrong, that is, there must be a situation where something can go wrong and it won't go wrong. So, Murphy Law is paradoxal.

Your thoughts, please.

Title: Re: paradoxes
Post by towr on Jul 5^th, 2008, 3:28am

on 07/04/08 at 17:10:04, BenVitale wrote:

It's not an actual law; it's a statement of pessimism.
In day to day life, obviously not everything that can go wrong goes wrong; you would have failed posting that message. In fact you'd have been long dead from an eating or breathing mishap.
It's insane to even consider this in a literal sense. However that doesn't mean it's stupid.

As for your claim that it's paradoxical; Murphy's law is not something that "goes", so "going wrong" doesn't apply to it. And it also doesn't apply to things that can't go wrong. (Which is kind of funny, because what it typically tries to say is that everything can go wrong; but in a literal sense it says nothign of the sort).

Also, again, you just plagarized from another site. If you're not going to use your own words, attribute them to their rightfull source: http://uncyclopedia.org/wiki/Murphy%27s_Law#Murphy.27s_Paradox

Title: Re: paradoxes
Post by BenVitale on Jul 5^th, 2008, 10:54am

I'm on the computer everyday, I play chess online, I'm on social network websites discussing politics, and someone sent me the Murphy's thing,

Quote:

and I thought it was cute, so i decided to post it here. It was not my intention to plagiarize. I know, I could have asked my virtual friend the source of this thing, or i could have checked the source myself, or i could mentioned on this thread how i got this thing. I did not do any of those things. I got sidetracked by other things, that's no excuse

I'm sorry. I respect you, guys, and I appreciate greatly for your courtesy and taking the time to answer to my posts. Please accept my apologies.

Title: Re: paradoxes
Post by ThudanBlunder on Jul 5^th, 2008, 5:46pm

on 07/05/08 at 10:54:26, BenVitale wrote:

I'm sorry. I respect you, guys, and I appreciate greatly for your courtesy and taking the time to answer to my posts. Please accept my apologies.

Fair enough. No harm done. It's just that towr likes to know exactly whose opinions he is grinding into the dust. LOL

But the other (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_microsoft;action=display;num=1215204891) example required some surgical editing.

Title: Re: paradoxes
Post by towr on Jul 6^th, 2008, 7:33am

on 07/05/08 at 17:46:03, ThudanBlunder wrote:

Fair enough. No harm done. It's just that towr likes to know exactly who he is grinding into the dust. LOL

Hah, well, it's also because of that other thread, and I was starting to wonder if Ben was just a collection of quotes gathered from around the internet.
I suppose I also don't really care specifically which source it is, just that there is another source. Then I can always try google if I want to check it out. So anything from putting quotes around a quote, or adding "I read somewhere that" or "My friend send me this" is fine by me.

Of course, in this case, the fact it's from uncyclopedia gives a big hint on how seriously the "paradox" should be considered. Uncyclopedia is a parody version of wikipedia; so typically they aren't entirely serious.

Title: Re: paradoxes
Post by BenVitale on Jul 6^th, 2008, 2:09pm

Thanks Towr and ThudanBlunder. I will be more thorough next time. After all, since I am aiming for a Master's degree in math, I need to be thorough in every thing. I noticed after re-visiting my own posts my English is not that great. I need to pay more attention to my syntaxes, grammar and spelling. Is there a spelling check button here?

I didn't know an Uncyclopedia existed.

Is there a reliable site for Theoretical Physics?

I really don't want to be labelled "reality-challenged" again.

Title: Re: paradoxes
Post by towr on Jul 6^th, 2008, 3:15pm

on 07/06/08 at 14:09:14, BenVitale wrote:

Is there a spelling check button here?

No, there isn't. But if you use firefox (2 or 3), it has a build-in spell check. If you travel around a lot, you could use firefox-portable and put it on a USB-stick. (Also convenient for having your own bookmarks and history etc wherever you go)

Quote:

Is there a reliable site for Theoretical Physics?

Possibly you can find a few things at http://ctp.lns.mit.edu/links.html, like the journal section (If it's good enough for MIT to link, it can't be all bad)

Quote:

I really don't want to be labeled "reality-challenged" again.

Well, if that is the only argument someone can give as to why he thinks you're wrong (or worse, simply to stop you asking questions), then at the very least the problem doesn't lie entirely with you. Challenging preconceived notions of what constitutes reality isn't necessarily the same as being reality-challenged. (Mind you, this is not a good argument to use against people, because it will just upset them.)
I suppose in a basic sense, being challenged by reality is what being a scientist is about: standing up to that challenge. Rather than letting it bully you around ;)

On the other hand there's a time and place for certain types of questions (and a way to ask them). So don't take this as an advice to interrupt class with questions your professor is unable and/or unwilling to answer or discuss. If it can't be answered in 5 minutes and is outside the curriculum, it's probably best asked after class or via email. (I figure this is about your physics professor and that question about QM interpretations, but of course I don't know the situation at all. So consider this a general shot in the dark type of advice.)

Title: Re: paradoxes
Post by BenVitale on Jul 6^th, 2008, 6:40pm

Well, before the cat incident in class, the physics professor always welcomed our questions and encouraged us to think and discuss. He encouraged us to think outside the box. We usually have 5 minutes for discussions before we start with his lectures. So, the other day, I raised my hand and expressed my fascination with theoretical physics, I said that I was contemplating on the eleven dimensions, parallel universes, and a world made out of strings, and that the string theory that might hold the key to unifying the four forces of nature according to Brian Greene, the supersymmetry, and beyond the standard model.

Then I ask, "What's your take on the Schroedinger's cat experiment?"

His mood changed all of sudden. He got upset. I didn't understand it. Theoretical physicists have been spending time thinking about it. People like Penrose, Hawkings, Dr. Michio Kaku.

Title: Re: paradoxes
Post by BenVitale on Jul 6^th, 2008, 8:21pm

Imagine that there is a scientific law that says "All crows are black." If only three or four crows are observed, then the law is weakly confirmed. If millions of crows are seen to be black, then it is strongly confirmed.

Professor Carl Hempel, who invented this paradox, believes that a purple cow actually does slightly increase the probability that all crows are black.

Source: http://www.geocities.com/CapitolHill/Lobby/3022/hempel.html

I am not totally satisfied with the explanation offered in the linked document.

Proposition 1: All crows are black

Double-negation gives the same proposition: All !(crows) are !(black) <--> All things that are not black are not crows

Observation 1: There exists a cow which is purple

Thus, Proposition 1 is consistent with Observation 1.

But, it is also consistent with Proposition 2: All crows are blue (Through the same method); thus Observation 1 supports two propositions which cannot be true at the same time. Thus, Observation 1 is inadmissible.

I asked myself: what do I make of this purple cow?

Then I get that there exists a cow which is purple (which is consistent with Proposition 1)

Doesn't a purple cow provides evidence for the hypothesis that all things that are not black are not crows, but it also provides evidence for, for example, all things that are not pink are not crows - or, All crows are pink. Thus this evidence is inadmissible.

Title: Re: paradoxes
Post by ThudanBlunder on Jul 6^th, 2008, 9:41pm

If we consider multiple hypotheses simultaneously (eg. all ravens are black or all ravens are blue or all ravens are pink), it is true that any observation of a non-black/blue/pink object that is not a raven increases the probability that all ravens are black or all ravens are blue or all ravens are pink by a very, very small amount corresponding to the ratio of ravens to non-ravens. Also, the information that an object is not a raven removes the possibility of this object being a counterexample to the rule. And then there is the small matter that no blue or pink ravens have ever been observed, whereas black ravens are known to exist.

There is a 'little' (http://en.wikipedia.org/wiki/Raven_paradox) more to it than that, but that's my take on the matter. :P

Cf. (http://en.wikipedia.org/wiki/Grue_and_Bleen)

Title: Re: paradoxes
Post by towr on Jul 7^th, 2008, 12:59am

on 07/06/08 at 20:21:53, BenVitale wrote:

The general argument is that this is why "confirmation" is not a good approach to science.
You can "confirm" your theory by observing anything that it doesn't apply to. You study crows, then end up looking for cows.
It's one of the reasons why Karl Popper suggested that instead you should try to aim at falsifying theories; look for non-black crows. You can't prove a universal, except by examining every object in the domain; but proving it false takes finding just one counter-example. So a search directed at the latter makes more sense.

Title: Re: paradoxes
Post by rmsgrey on Jul 7^th, 2008, 7:06am

Your sampling technique is significant when it comes to whether a purple cow supports the "all crows are black" hypothesis - if you go and collect non-crow objects and then look at their colours, that tells you nothing any more than collecting black objects and then checking whether they're crows does (though the latter does give you some evidence for/against the existence of black crows...)

If you go around looking at cows, that tells you nothing about whether all crows are black. If you go around looking at purple objects, then finding a bunch of cows, but no crows is weak evidence that all crows are black. Similarly, looking at a bunch of crows and seeing that they're all black is (relatively) strong evidence that all crows are black.

And "all crows are black" and "all crows are pink" are not necessarily mutually exclusive propositions - they only become mutually exclusive once it is known that some crows exist - at which point, checking the colour of any known crow will disprove at least one of the two.

Title: Re: paradoxes
Post by Grimbal on Jul 7^th, 2008, 8:04am

The more purple cows you see, the more it confirms that there is no crow. And that supports that all crows are black.

Title: Re: paradoxes
Post by rmsgrey on Jul 8^th, 2008, 8:31am

on 07/07/08 at 08:04:19, Grimbal wrote:

The more purple cows you see, the more it confirms that there is no crow. And that supports that all crows are black.

It still depends on how you're finding the purple cows - if you send a PhD student out to find you a thousand cows, and they don't bring back any crows, that doesn't tell you anything about the abundance of crows (regardless of colour) - there was no chance (assuming a competent PhD student) of getting any crows back anyway...

Title: Re: paradoxes
Post by Grimbal on Jul 8^th, 2008, 10:05am

I assumed you wander idly and check what you see.

Title: Re: paradoxes
Post by Qaster Qof Qeverything Q42 on Jul 9^th, 2008, 9:58am

Wait....
All cows ARE black.

Title: Re: paradoxes
Post by BenVitale on Jul 19^th, 2008, 10:38am

I have a question about the Ross-Littlewood paradox, see link

http://www.suitcaseofdreams.net/Paradox_Infinity.htm

According to this document, the answer is none.
I don't get the same answer.

P₁ + P₂ + ... + P₁₀) - P₁ +
P₁₁ + P₁₂ + ... + P₂₀) - P₂ +
P₂₁ + P₂₂ + ... + P₃₀) - P₃ +

....

P_10j-9 + P_10j-8 + P_10j-7 + ... + P_10j - P_j+
...........

= SUM[i=1,i=10]P_i - P₁
+ SUM[i=11,i=20]P_i - P₂
+ SUM[i=21,i=30]P_i - P₃ +
...
+ SUM[i=10j-9,i=10j]P_i - P_j + ...

= lim_{j--> oo} (SUM[i=1,i=10j] P_i - SUM[i=1, i=j] P_i

= lim_{j--> oo} (SUM[ i=j+1,i=10j] P_i = oo

I'm getting infinite elements.

did i go wrong somewhere?

Title: Re: paradoxes
Post by towr on Jul 19^th, 2008, 11:12am

on 07/19/08 at 10:38:37, BenVitale wrote:

did i go wrong somewhere?

Yes. Just name one single term that is still in the sum, and I'll tell you exactly at what step it was removed.
There is no i, such that P_i was added but not later subtracted.

You could look up the "impish pixie" thread, or one of the similar ones.

Title: Re: paradoxes
Post by ThudanBlunder on Jul 19^th, 2008, 11:13am

on 07/19/08 at 10:38:37, BenVitale wrote:

I have a question about the Ross-Littlewood paradox...

Ah, so that's what it (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1117490596;)is called.

Title: Re: paradoxes
Post by BenVitale on Jul 19^th, 2008, 11:49am

Thanks. I didn't know about "impish pixie" thread. But before I go and read impish pixie's thread, I just need to show you the things I wrote:

Things with different rate of growth can't be subtracted to give zero.

What's the limit of a² - a as a approaches infinity? We can sure see that both a² and a approaches infinity as a approaches infinity. So is it that:

lim (a² - a) = lim (a²) - lim (a) = 0 ?
a --> oo

Clearly not. There are two flaws. One: Things with different growth rates cannot have their limits spread over them in a linear fashion.

lim (SUM [i=1, i=10j]P_i - SUM [i=1,i=j]P_i is difinitely not SUM [i=1,oo]P_i - SUM [i=1,oo]P_i.
j --> oo

We have to do the the brackets first.

We cannot do oo - oo.

Title: Re: paradoxes
Post by towr on Jul 19^th, 2008, 12:07pm

on 07/19/08 at 11:49:02, BenVitale wrote:

Things with different rate of growth can't be subtracted to give zero.

Sure they can.
Besides which, you never even wondered if that sum you wrote even converges. As a sum it is utterly meaningless.

Quote:

What's the limit of a² - a as a approaches infinity?

But we're dealing with something completely different here.

Quote:

We cannot do oo - oo.

And we're not, we're subtracting infinite sets. If you remove every element from the set of integers, what you're left with is an empty set. The cardinality of the set you start with and the set you subtract may both be infinite, and the set you end up with has a cardinality of 0, but at no point were we subtracting infinity from infinity to get 0.
So that simply isn't an objection.

{P[10 i] + .. + P[10i+9] | integer i >=0} - {P[i] | integer i >=0} = {}
SUM ({P[10 i] + .. + P[10i+9] | integer i >=0} - {P[i] | integer i >=0}) = SUM {} = 0

Title: Re: paradoxes
Post by BenVitale on Jul 19^th, 2008, 8:21pm

Towr, you're right. Thanks. I see it clearly now. I got carried away with my summations and screwed up.

And, I looked at the "Paradox of Enchantress and Witch " two schemes, see link

http://www.suitcaseofdreams.net/Enchantress_Witch.htm

Title: Re: paradoxes
Post by LeoYard on Jul 22^nd, 2008, 11:34am

Sorry for interrupting. I'm not ready to celebrate your realization yet.

Is there a simple way to verify this?

As the time reaches 12, the number of cards increases without limits.

What about the sums ben posted? why don't they make sense?

What about infinite sums?

Title: Re: paradoxes
Post by ThudanBlunder on Jul 22^nd, 2008, 12:07pm

on 07/22/08 at 11:34:31, LeoYard wrote:

Is there a simple way to verify this?

Don't think so. It is not a real-world problem.

on 07/22/08 at 11:34:31, LeoYard wrote:

As the time reaches 12, the number of cards increases without limits.

When a function is not continuous, as here,
LIM f(x) = b does not imply f(a) = b
x -> a

on 07/22/08 at 11:34:31, LeoYard wrote:

What about the sums ben posted? why don't they make sense?

Ben's sums might make sense before 12 o'clock. But, as with Cinderalla, when the clock strikes 12 there will be no more ball.

Title: Re: paradoxes
Post by towr on Jul 22^nd, 2008, 12:21pm

on 07/22/08 at 11:34:31, LeoYard wrote:

As the time reaches 12, the number of cards increases without limits.

But at the strike of twelve, each card that was added was also removed. There is a discontinuity there, which causes a result that may be counter-intuitive.
A card added in step i will be removed between step 10*i-9 and step 10*i; for any card we can precisely say when it was added and removed.

Quote:

What about the sums ben posted? why don't they make sense?

Well, maybe if we knew something about the values of P_i. But there's not much to say a priori about a sum from infinity to infinity (which is what the limit goes to).
If you take, e.g. P_i = 1/2ⁱ, the sum would nicely converge to 0
If you take P_i = 1/i, then we might get lim i-> inf ln(10i) - ln(i) = lim i-> inf ln(10) = ln(10); and unlike the previous one, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif1/i goes to infinity.

In any case it's a bad model when we're dealing with sets of distinguishable objects.

Quote:

What about infinite sums?

What about them?
I prefer them unconditionally convergent.

Title: Re: paradoxes
Post by BenVitale on Jul 23^rd, 2008, 11:22am

Allais paradox :

http://www.daviddarling.info/encyclopedia/A/Allais_paradox.html

Quote:

A paradox that stems from questions asked in 1951 by the French economist Maurice Allais (1911-).1 Which of these would you choose: (A) an 89% chance of receiving an unknown amount and 11% chance of $1 million; or (B) an 89% chance of an unknown amount (the same amount as in A), a 10% chance of $2.5 million, and a 1% chance of nothing? Would your choice be the same if the unknown amount were $1 million, or if it were nothing?

Quote:

Would your choice be the same if the unknown amount were $1 million, or if it were nothing?

The answer in the linked document is not clear to me, especially when the "unknown" amount is nothing.

Would anyone like to clarify?

Title: Re: paradoxes
Post by towr on Jul 23^rd, 2008, 12:10pm

Given a 100% probability of winning a million, and a 1% of not winning anything or winning slightly more on average, I'd hate to not win anything. You risk a sure million by going for B.

But if chances are stacked against me, with only an 11% chance of winning a million, I may as well push my luck and go for a 10% chance of winning 2.5 million. It's marginal increase in risk for a huge extra payoff.

It may not be rational according to game theory standards, but those are poor standards anyway. What is lacking is a proper mathematical expression for risk. So it's hard to say for what unknown value we should switch from A to B as a risk-averse agent. At what point is a chance of more worth the risk of getting nothing.

Title: Re: paradoxes
Post by BenVitale on Jul 23^rd, 2008, 1:27pm

Thanks towr,

I'm gonna need time to digest this concept.

First, I need to read again about the expected utility theory and see how the Allais paradox is inconsistent with the expected utility theory.

And I'm going to read more about risk aversion.

Title: Re: paradoxes
Post by towr on Jul 23^rd, 2008, 3:17pm

If we use a simple model of valuing a marginal increase with the square of the probability we get it, then option A is better than B if the unknown value is greater than about 725 thousand.

(for 0 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif u http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 1)
A gives a subjective value of 1²uM+0.11²*(1-u)M and
B gives a subjective value of 1²*0M+0.99²uM+0.10²*(2.5-u)M

Might not necessarily work that way in reality (and of course some might be more risk averse than others), but at least it shows that for some models it's a perfectly rational choice to prefer A for high u and B for low u.
Instead of squaring you could use an arbitrary power as parameter for averseness, I suppose.

Title: Re: paradoxes
Post by BenVitale on Jul 23^rd, 2008, 4:01pm

Thanks for your insights, towr.

I find Game Theory the most exciting thing in math.