Another AC anomaly?
in [Logic]
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From: Rotwang on 23 Nov 2009 09:19 On 23 Nov, 14:03, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > David C. Ullrich wrote: > > On Mon, 23 Nov 2009 12:22:17 +0100, Herman Jurjus <hjm...(a)hetnet.nl> > > wrote: > > [...] > > > I think the moral is not that AC leads to the weirdness but that > > this is a highly weird situation to begin with. At the time when > > we make any given guess we've already been told the result of > > infinitely many coin tosses... > > Yup - it's a game without a first move. So 'weird' is an accurate > qualification. Yet, it's not particularly difficult to give a > mathematical description of the situation, reason about it, and convince > ourselves that, mathematically, there's no problem with it. Ideally, it > should be much harder to make mathematical sense of a game like this. > That's why I included "WM is horribly right" as a possible moral. > > Apparently there's something wrong with backward supertasks (and not > with ordinary, 'forward' supertasks). But why should that be? I'm not sure what qualifies as a "backward" or "forward" supertask, but this paradox is very similar to the AC solution to the infinite prisoners-and-hats puzzle, which avoids any weirdness involving non- well-ordered games: http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle#Countably_Infinite-Hat_Solution
From: Jesse F. Hughes on 23 Nov 2009 09:16 Herman Jurjus <hjmotz(a)hetnet.nl> writes: > Apparently there's something wrong with backward supertasks (and not > with ordinary, 'forward' supertasks). But why should that be? Well, I'm not at all sure that there's no problem with forward supertasks. Surely, it is not difficult to come up with a problematic case. For instance, take our favorite example: at each time t - 1/n, place balls 10(n-1) to 10n - 1 in a vase and then remove ball n. At the end the vase is empty. Now alter the situation slightly. At each step, again place 10 balls into the vase and then remove one ball, but remove the ball *randomly*. At the end, the vase may contain any number of balls. This strikes me as suitably counterintuitive to say that the forward supertask has something wrong with it. Or, perhaps, with my intuitions. -- "To solve this problem, we define a security flag, known as the 'evil' bit, in the IPv4 [RFC791] header. Benign packets have this bit set to 0; those that are used for an attack will have the bit set to 1." -- RFC 3514
From: Herman Jurjus on 23 Nov 2009 09:36 Rotwang wrote: > On 23 Nov, 14:03, Herman Jurjus <hjm...(a)hetnet.nl> wrote: >> David C. Ullrich wrote: >>> On Mon, 23 Nov 2009 12:22:17 +0100, Herman Jurjus <hjm...(a)hetnet.nl> >>> wrote: >> [...] >> >>> I think the moral is not that AC leads to the weirdness but that >>> this is a highly weird situation to begin with. At the time when >>> we make any given guess we've already been told the result of >>> infinitely many coin tosses... >> Yup - it's a game without a first move. So 'weird' is an accurate >> qualification. Yet, it's not particularly difficult to give a >> mathematical description of the situation, reason about it, and convince >> ourselves that, mathematically, there's no problem with it. Ideally, it >> should be much harder to make mathematical sense of a game like this. >> That's why I included "WM is horribly right" as a possible moral. >> >> Apparently there's something wrong with backward supertasks (and not >> with ordinary, 'forward' supertasks). But why should that be? > > I'm not sure what qualifies as a "backward" or "forward" supertask, > but this paradox is very similar to the AC solution to the infinite > prisoners-and-hats puzzle, which avoids any weirdness involving non- > well-ordered games: > > http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle#Countably_Infinite-Hat_Solution Thanks for the reference. (One of the comments on the OP's blog was "the Unexpected Hanging on steroids" <g>) -- Cheers, Herman Jurjus
From: Herman Jurjus on 23 Nov 2009 09:48 Jesse F. Hughes wrote: > Herman Jurjus <hjmotz(a)hetnet.nl> writes: > >> Apparently there's something wrong with backward supertasks (and not >> with ordinary, 'forward' supertasks). But why should that be? > > Well, I'm not at all sure that there's no problem with forward > supertasks. Surely, it is not difficult to come up with a > problematic case. > > For instance, take our favorite example: at each time t - 1/n, place > balls 10(n-1) to 10n - 1 in a vase and then remove ball n. At the end > the vase is empty. > > Now alter the situation slightly. At each step, again place 10 balls > into the vase and then remove one ball, but remove the ball > *randomly*. At the end, the vase may contain any number of balls. > This strikes me as suitably counterintuitive to say that the forward > supertask has something wrong with it. Or, perhaps, with my > intuitions. More conclusive (at least for me): you switch a light bulb on and off; after infinitely many steps, is the light on or off? (Or: you put one and the same ball in the vase, out of the vase, in the vase, out of the vase, etc. What's in the vase/where's the ball after infinitely many steps?) If you can't trust supertask-reasoning in this case, why should you trust it in other, seemingly less problematic cases? For the sake of clarity: I'm not endorsing or advocating the backward-supertask paradox as extremely important. Just wanted to share it here, because I had never heard of it, and thought perhaps others also hadn't. -- Cheers, Herman Jurjus
From: William Hughes on 23 Nov 2009 10:38
On Nov 23, 10:09 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote: > William Hughes wrote: > > On Nov 23, 7:22 am, 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. > > > Note you can avoid AC, by defining > > a_n(m)= s(m) m>n, a_n(k) heads otherwise. > > What's s(m) ? > > As a matter of fact, what's a_n(m) ? > The game involves tosses and guesses, i.e. two sequences with one index: > toss(n), guess(n). > > -- > Cheers, > Herman Jurjus s is the sequence of tosses, so s(n) = toss(n) As above, a_n(m) is equal to the toss if m>n, otherwise is a head. Note that a_n is equal to s except perhaps on a finite set. For any n, a_n(n) is heads, so guess(n)=a_n(n) is heads. I don't think the fact that time is reversed has any real bearing on this. If the tosses are independent, perfect knowledge of the future is of no help for the present. From the quoted page Now, since the representative sequence and the actual sequence of heads and tails that occurs are in the same equivalence class, they must only differ at finitely many places. Correct. Note this is true of sequence a_n. So, if you guessed according to the representative sequence, you have only made finitely many mistakes. Correct. However, you did not guess acconding to the representative sequence. You only used a_n for guess(n). For other guesses you used other sequences. - William Hughes |