tag:blogger.com,1999:blog-3891434218564545511.post7095878496894844757..comments2024-03-28T19:56:42.305-05:00Comments on Alexander Pruss's Blog: Arbitrariness, regularity and comparative probabilitiesAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-3891434218564545511.post-9759775362715224942020-07-29T03:50:51.535-05:002020-07-29T03:50:51.535-05:00Here are some half-baked thoughts. You may make so...Here are some half-baked thoughts. You may make something of them. I’m still thinking about it.<br /><br />Yes, the approach I have suggested would say that no L can be compared with any L’. This seems reasonable. For example, L1 implies L0, and, granted the non-classical intuition given earlier, P(L1|L0) = 1/2. So L1 and L0 can be compared. But no such relation applies between any L and any L’. Note also, there is no natural correspondence between the indexing of the flips in the two experiments. You could shift it to make L0 correspond the L’1, or any other L’.<br /><br />One moral I draw is that no totally ordered probabilities, standard or not, can reflect the structure implied by the intuitive description given above.<br /><br />That said, classical probabilities seem adequate in practice (if the infinite can ever be practical…). All the Ls and Rs and L’s and R’s have classical probability zero. They are all bad bets at any odds. What more do you need to know in practice? The comparisons that are not captured by classical probabilities are certainly of conceptual interest, but relevant only in cases that happen with classical probability zero.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-90843463744749221672020-07-28T14:11:16.128-05:002020-07-28T14:11:16.128-05:00Ian:
One more thought. If you deny comparability ...Ian:<br /><br />One more thought. If you deny comparability between L and R, then for the same reason you have to say that if the experiment is repeated, and for the repeat you have events L' and R', then you have to deny comparability between L and L', as well as between R and R' (for by symmetry there is no more reason to say that L and L' are comparable than that L and R' are). But this seems to commit one to the view that in infinitary cases one cannot repeat an experiment with the same chances as before. And the repeatability of experiments with the same chances seems foundational to our concept of chance.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-58830963493126246862020-07-27T09:02:31.903-05:002020-07-27T09:02:31.903-05:00Yeah. I wonder if there is a non-arbitrary way to ...Yeah. I wonder if there is a non-arbitrary way to work out when we have lack comparability between two events here.<br /><br />When the classical probabilities are different, we should have comparability.<br />When one event is a subset of the other, we should have comparability.<br />When the two events depend on only finitely many flips each, we should have comparability.<br /><br />Here's another interesting test case. Let A be the event that the strong law of large numbers holds in the leftward direction (i.e., that the frequencies of heads converge to 1/2 leftward) and let B be the event that it holds in the rightward direction. Note that it doesn't matter where we start when we define A and B. Can we say that A and B are comparable? I don't think so. Let A' and B' be the events that the strong laws of large numbers hold in the leftward and rightward directions when we restrict to the even-numbered coins. Then A and B are, respectively, proper subsets of A' and B', so they should be respectively less probable. But there is no more reason to think that A is equiprobable with B' than that A is equiprobable with B. <br /><br />This suggests the hypothesis that when A and B depend on disjoint infinite sets of coin tosses, and their classical probabilities are equal, then they should be incomparable. There may be a counterexample to that hypothesis.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-56524489241700785202020-07-26T03:38:45.848-05:002020-07-26T03:38:45.848-05:00One might give up totality and say the Ls and Rs a...One might give up totality and say the <i>L</i>s and <i>R</i>s are not comparable.<br /><br />This is not merely ad hoc. Intuitively, the coin flips could be described like this: the probability of any particular outcome of any finite set of N of the coins is 2^(-N), regardless of any condition on the outcomes of the other coins (even a condition with classical probability zero).<br /><br />As in the post, it follows easily that the <i>L</i>s are strictly decreasing in probability with n and the <i>R</i>s are strictly increasing. (Use the comparison in your Popper Function post, or my modification of it.) But there seems no natural way to compare any <i>L</i> with any <i>R</i>. For simplicity, take a disjoint <i>L</i> and <i>R</i>. What is P(L | L ∪ R)? As far as I can see, it cannot be deduced from the intuitive description of the flips.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.com