Returning to the thread: On Sat, May 1, 2010 at 10:23 PM, Brent Meeker <meeke...@dslextreme.com> wrote: > On 5/1/2010 7:10 PM, Rex Allen wrote: >> >>> On Sat, May 1, 2010 at 10:01 PM, Brent Meeker <meeke...@dslextreme.com> >>> wrote: >>> >>> It's invalid simply because your conclusion depends taking the cardinality >>> to be the measure. The cardinality of infinite sets doesn't satisfy >>> Kolomogorov's axioms for a probability measure. For example one of the >>> axioms is: If A and B are disjoint then P(A) + P(B) = P(A union B). Let >>> the measure of the integers be 1. The let A be the evens and B be the odds. >>> You get 2=1. If you're going to talk about probabilities of infinite sets >>> you have to introduce some measure other than cardinality. >> >> >> Isn't that exactly what I said here: >> >> >>> SO, I think we have zero information that we can use to base our >>> probability calculation on. Because of the counting issues introduced >>> by the infinity combined with the lack of pattern. There is no usable >>> information. >> > > No that's not the same. If you based the order on a die toss there would be > no pattern, but there would still be a measure even when the cardinality was > infinite. Your use of "information" is ambiguous. On the one hand you use > it to mean "no pattern" and then you assume that must be the same as "we > don't know the measure". > >> >> I claim vindication. >> >> But having done so, what measure are you suggesting for this >> particular example instead? For this particular example. Not for >> general cases involving telephone surveys. > > In my example, a die toss, measure is based on the symmetry of the die.
In that case it seems to me that we are ignoring the *actual* infinite set of randomly generated results and only talking about the measure. Effectively we're saying, "We have no useful information about the random infinite set - because it's random...and infinite. So let's go back to the measure and ask what would we get if we generated another number according to that definition." So returning to my infinite row of numbered squares, let's say we take the 1-squares and put them into a one-to-one correspondence with the not-1-squares. Now, let's put a sticker on each 1-square that says "A". And another sticker on each not-1-square that says "B". Now, let's put them back into an infinite row. What is the probability of hitting an B-square with my randomly thrown dart? What is the probability of hitting a 1-square? It seems to me that we can't say anything about the actual infinite set. We can only talk about various measures on it. Which is what you said. But I think is still consistent with my original example. I'd think that if you have an actual randomly generated infinite set, then you can't draw any probabilistic conclusions about that infinite set, even if you know how it was generated (e.g., dice rolling). You can only draw conclusions about the measure that describes the generating process. Right? Or wrong? But, now returning to the Boltzmann brain problem, Carroll says: "This version of the multiverse will feature both isolated Boltzmann brains lurking in the empty de Sitter regions, and ordinary observers found in the aftermath of the low-entropy beginnings of the baby universes. Indeed, there should be an infinite number of both types. So which infinity wins? The kinds of fluctuations that create freak observers in an equilibrium background are certainly rare, but the kinds of fluctuations that create baby universes are also very rare." Well, since these are physically existing infinities of the same size, then neither infinity wins. So, given eternal recurrence, there are an infinite number of Rexs. And an infinite number of not-Rexs. Let's pair the Rexs off in a one-to-one correspondence with the not-Rexs. Then, let's go down the list and put an "A" sticker on the Rexs. And a "B" sticker on the not-Rexs. Then lets randomly arrange them in an infinitely long row and select one at random. What's the probability of selecting a Rex? What's the probability of selecting an "A" sticker? -- You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-l...@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
