Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf P2 Remark. When the number of boxes is finite Player 1 can guarantee a win with probability 1 in game1, and with probability 9/10 in game2, by choosing the xi independently and uniformly on [0, 1] and {0, 1, ・・・, 9}, respectively. ”independently and uniformly”が、独立同分布（IID)を含意
仮定つまり与件は、当たり前だが、数学的な推論をいくら並べても、これを覆すことはできない。もし、矛盾が生じるなら、推論が間違っているか、前提が間違っているかだ ところで、独立同分布（IID)の仮定は、大学の確率過程論で、正しいと認められているので、矛盾が生じるなら、推論が間違っている なお、高校レベルの確率論で、大学レベルの確率論・確率過程論を覆すことはできない。これもまた自明だ これが分からない人は、>>28を実行ください。はよやれ！（＾＾ 0039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) 20:29:13.98ID:/k5aIfYN スレ62 https://rio2016.5ch.net/test/read.cgi/math/1551963737/915 915 名前：現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 投稿日：2019/03/27(水) (抜粋) Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf このP2に Remark. When the number of boxes is finite Player 1 can guarantee a win with probability 1 in game1, and with probability 9/10 in game2, by choosing the xi independently and uniformly on [0, 1] and {0, 1, ・・・, 9}, respectively. とある ここで”independently and uniformly”が、独立同分布（IID)を含意することは、知る人がみればすぐ分かること で、例えば、{0, 1, ・・・, 9}ならば、的中確率は、1/10(for Player 2)(つまり、出題者Player 1は、確率9/10で勝てる) つまり、独立同分布（IID)を仮定すれば、どの箱も同じで、例外はない なお、Sergiu Hart氏 は、時枝先生よりも良く分かっているみたい
game1（選択公理を使う）→game2（選択公理を使わない）→boxes is finite （有限の場合は通常確率論通り） と並べて説明している まあ、落語の落ちですね。最後”boxes is finite （有限の場合は通常確率論通り）”ですから
本気で”通常確率論外し”が成立していることを、読者に説明するなら boxes is finite （有限の場合は通常確率論通り）→しかしgame2（通常確率論外し（選択公理を使わない））→game1（通常確率論外し（選択公理を使う） の並びでしょうからね（＾＾；
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（＾＾
二つの集合 X, Y を用意する。 X の各要素 x に対して Y の要素 f(x) が定められているとき、 この対応の規則を写像 (mapping) という。
写像 f : X → Y でとくに Y が数の集合であるとき、 関数 (function) 、X = N のとき列 (sequence) とい うことが多い。 (引用終り) 0079１３２人目の素数さん2019/06/14(金) 22:27:37.74ID:ebInVaSj>>78 失せろキチガイ詐欺師 0080哀れな素人2019/06/14(金) 22:29:53.24ID:BYpABN67 ID:ebInVaSj
