Another AC anomaly?
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From: WM on 26 Nov 2009 11:24 On 26 Nov., 05:33, Bill Taylor <w.tay...(a)math.canterbury.ac.nz> wrote: > It is an interesting problem in probability generating functions > to work out the individual proabilities for each ball, > and the expected number, left in the pot at the end! Here is another interesting task: Use balls representing the positive rationals. The first time fill in one ball. Then fill in always 100 balls and remove 100 balls, leaving inside the ball representing the smallest of the 101 rationals. If you get practical experience, you will accomplish every Centuria in half time. So after a short while you will have found the smallest positive rational. Instead of 100 balls you can use only 80 balls as well. The Romans also used to have 80 men in a Centuria- --- That pointed already to the final period of the empire. Of course the game is only possible, if there is an actual infinity waiting to be finished. --- And that points to the final period of mathematics. Regards, WM
From: William Hughes on 26 Nov 2009 13:22 On Nov 26, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > Here is another interesting task: Use balls representing the positive > rationals. The first time fill in one ball. Then fill in always 100 > balls and remove 100 balls, leaving inside the ball representing the > smallest of the 101 rationals. [at random with any measure that gives a positive probability to each rational] > If you get practical experience, you > will accomplish every Centuria in half time. So after a short while > you will have found the smallest positive rational. Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim you will have an empty set. - William Hughes
From: WM on 26 Nov 2009 15:51 On 26 Nov., 19:22, William Hughes <wpihug...(a)hotmail.com> wrote: > On Nov 26, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > Here is another interesting task: Use balls representing the positive > > rationals. The first time fill in one ball. Then fill in always 100 > > balls and remove 100 balls, leaving inside the ball representing the > > smallest of the 101 rationals. > > [at random with any measure that gives a positive probability > to each rational] Simply take the first, seconde, third ... Centuria according to Cantor's well-ordering of the positive rationals. Then there is no need for considering any probabilities. > > > If you get practical experience, you > > will accomplish every Centuria in half time. So after a short while > > you will have found the smallest positive rational. > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > you will have an empty set. Besides your assertion, you have arguments too, don't you? In particular you can explain, how the empty set will emerge while throughout the whole time the minimum contents of the vase is 1 ball? Regards, WM
From: LauLuna on 26 Nov 2009 18:32 On Nov 25, 10:07 pm, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "LauLuna" <laureanol...(a)yahoo.es> wrote in message > > news:36cb4819-04fd-4dbe-b204-7168ae550e23(a)p8g2000yqb.googlegroups.com... > > > > > > > On Nov 23, 12:22 pm, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > > > Has anyone seen this before? > > > >http://possiblyphilosophy.wordpress.com/2008/09/22/guessing-the-resul.... > > > > I'm not sure yet what to conclude from it; that AC is horribly wrong, or > > > that WM is horribly right, or something else altogether. > > > > In short the story goes like this: > > > > A game is played, in which infinitely many coins are tossed, and there's > > > one player, who makes infinitely many guesses. Both are done over a > > > finite period of time. The tosses and guesses are not made faster and > > > faster, however, but slower and slower: at t = 1/n. There's no 'first' > > > move. > > > > Claim: > > > There exists a strategy with which you're certain to guess all entries > > > correctly except for at most finitely many mistakes. Not 'certain' as in > > > 'probability is 100%', but absolutely certain. > > > > Reasoning: > > > On 2^w, consider the equivalence relation that makes x equivalent to y > > > when x(n) =/= y(n) for at most finitely many n. Next, using AC, create a > > > set S that contains precisely one element from every equivalence class. > > > Strategy: at every move, you already know the results of the previous > > > tosses, which is an infinite tail of some sequence in 2^w. Now take the > > > unique element from S associated to that tail, take the n'th element of > > > that sequence from S, and deliver that as your move. > > > After some thinking, you will see that with this strategy, you're indeed > > > certain to guess wrong at most finitely many times. > > > > Thanks, AC! Another nice mess you've gotten us into. > > > > -- > > > Cheers, > > > Herman Jurjus > > > In fact, you don't need the axiom of choice. > > > At any 1/n hour past 12pm it is already determinate what equivalence > > class the eventual sequence is in. Since you know what the previous > > results are and there are only finitely many outstanding results, you > > can complete the sequence at random and take it as your > > representative. Which means that from any 1/n hr past 12pm on you can > > guess at random. > > ..but then how would you prove that there have been only a finite number of > incorrect guesses? If the guesses always follow the representative sequence > (whose existence follows from AC) it's this that allows us to deduce at the > end that we've only made finitely many mistakes. > > > > > That is, you can make all your guesses at random and still be certain > > to guess wrong only finitely many times. > > Your proof for this doesn't work... > > Mike.- Hide quoted text - > > - Show quoted text - At any 1/n point you are guessing according a representative of the equivalence class of the eventual sequence, if you just complete the sequence of the previous results with random guesses for the finite number of the future ones.
From: LauLuna on 26 Nov 2009 19:20
On Nov 26, 1:24 am, Tim Little <t...(a)little-possums.net> wrote: > On 2009-11-25, LauLuna <laureanol...(a)yahoo.es> wrote: > > > At any 1/n hour past 12pm it is already determinate what equivalence > > class the eventual sequence is in. Since you know what the previous > > results are and there are only finitely many outstanding results, > > you can complete the sequence at random and take it as your > > representative. Which means that from any 1/n hr past 12pm on you > > can guess at random. > > That strategy fails for the same reason as the previous non-choice > strategy. The choice sequence works only because you provably used > the same sequence in the *past*, and hence at the time of your > decision had already made only finitely many errors. > > > That is, you can make all your guesses at random and still be > > certain to guess wrong only finitely many times. > > No, in your case you can't prove anything at all about how many of > your past guesses were correct. In order to reveal the similarity between this and Benardete's paradox, put it this way: First note that what the relevant equivalence class is, is already determinate at all 1/n points. Then assume that at any 1/n point I construct it -by guessing at random the future results- iff I haven't constructed it at any 1/m with m>n. Certainly, I possess a representative at all 1/n points and certainly I construct or choose it at no 1/n point: so I inherit it, just as Benardete's hearer inherits his deafened ears. So, having no need to construct or choose the representative at any 1/ n point, I don't need AC. As I see it, the essential paradox here is that the relevant equivalence class is already fixed at each 1/n point, so that no toss contributes to its determination. Regards. |