https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice edited Dec 9 '13 at 16:32 asked Dec 9 '13 at 16:16 Denis にも類似の話しがあります しかし、ここの3 Answers 中 下記 Alexander Prussさんと、Tony Huynhさんはこのriddle成立には否定的ですよ 確率を定義する測度が、きちんと決められないという趣旨のことを理由にしていますね なお、Alexander Prussさんは、”The probabilistic reasoning depends on a conglomerability assumption,”も理由に挙げていますね
(引用開始) Alexander Pruss edited Dec 12 '13 at 16:16 answered Dec 11 '13 at 21:07
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice edited Dec 9 '13 at 16:32 asked Dec 9 '13 at 16:16 Denis 3 Answers 11 edited Dec 12 '13 answered Dec 11 '13 Alexander Pruss (抜粋) But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate. A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion). Assume CH. Let < be a well-order of [0,1]. Suppose X and Y are i.i.d. uniformly distributed on [0,1]. Consider the question of which variable is bigger. Fix a value y∈[0,1] for Y. Then P(X<=y)=0, since there are only countably many points <- prior to y. By a conglomerability assumption, we could then conclude that P(X<=Y)=0, which would be absurd as the same reasoning would also show that P(Y<=X)=0. The argument fallaciously assumes conglomerability. We are neither justified in concluding that P(X<=Y)=0, nor that {X<=Y} is measurable (though for each fixed y, {X<=y} is measurable). And indeed it's not measurable: for were it measurable, we could use Fubini to conclude that it has null probability. Note that one can repeat the argument without CH but instead using an extension of Lebesgue measure that assigns null probability to every subset of cardinality <c, so clearly there is no refutation of CH here. (引用終り) 0043現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/06/14(金) 20:30:41.10ID:/k5aIfYN つづき スレ64 https://rio2016.5ch.net/test/read.cgi/math/1556253966/839 839 自分返信：現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 投稿日：2019/05/05(日) 16:53:47.49 ID:1ZCM8Sju [17/24] >>838 つづき
参考(see here for a discussion) https://www.mdpi.com/2073-8994/3/3/636 https://www.mdpi.com/2073-8994/3/3/636/pdf PDF Symmetry 2011, 3(3), 636-652; https://doi.org/10.3390/sym3030636 Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis Paul Bartha Department of Philosophy, University of British Columbia, Vancouver, BC V6T 1Z1, Canada Received: 25 July 2011 / Revised: 27 August 2011 / Accepted: 1 September 2011 / Published: 6 September 2011 (This article belongs to the Special Issue Symmetry in Probability and Inference) Abstract Freiling [1] and Brown [2] have put forward a probabilistic reductio argument intended to refute the Continuum Hypothesis. The argument relies heavily upon intuitions about symmetry in a particular scenario. This paper argues that the argument fails, but is still of interest for two reasons. First, the failure is unusual in that the symmetry intuitions are demonstrably coherent, even though other constraints make it impossible to find a probability model for the scenario. Second, the best probability models have properties analogous to non-conglomerability, motivating a proposed extension of that concept (and corresponding limits on Bayesian conditionalization). (引用終り)
Alexander Prussさん、ちょっと大物かも（＾＾ ”Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals”です で、mathoverflowの”Probabilities in a riddle involving axiom of choice”では、否定的見解を述べていますね～！ｗ（＾＾ そして、mathoverflowにおける”conglomerability ”の詳しい説明が、 「Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018」のP76-77にあり、下記googleブックで読めますね（＾＾ 勝負あり～！！ｗ（＾＾
https://en.wikipedia.org/wiki/Alexander_Pruss Alexander Pruss Alexander Robert Pruss (born January 5, 1973) is a Canadian mathematician, philosopher, Professor of Philosophy and the Co-Director of Graduate Studies in Philosophy at Baylor University in Waco, Texas. Biography Pruss graduated from the University of Western Ontario in 1991 with a Bachelor of Science degree in Mathematics and Physics. After earning a Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals,[4] he began graduate work in philosophy at the University of Pittsburgh.
http://alexanderpruss.com/cv.html Curriculum Vitae Alexander R. Pruss December, 2018 Education Ph.D., Mathematics, University of British Columbia, Spring, 1996 Dissertation title: Symmetrization, Green’s Functions, Harmonic Measures and Difference Equations, advised by John J. F. Fournier B.Sc. (hon.), Mathematics and Physics, University of Western Ontario, Spring, 1991 Books Infinity, Causation and Paradox, Oxford University Press, 2018
https://books.google.co.jp/books?id=RXBoDwAAQBAJ&;pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018 以上 ｗ（＾＾
