Andrea Ferretti
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 Mar 19 awarded Nice Answer Dec 15 awarded Enlightened Dec 15 awarded Nice Answer Sep 21 awarded Yearling Jul 27 awarded Yearling Jul 27 awarded Yearling Jul 4 awarded Nice Answer May 24 awarded Good Answer Jul 28 awarded Yearling Dec 11 awarded Quorum Nov 19 awarded Nice Answer Sep 13 comment Proof: Series converges $\implies$ the limit of the sequence is zero You are mistakingly assuming that the sequence actually converges to something. The other proofs also show that the limit exist, AND it is 0. Sep 11 comment Boy Born on a Tuesday - is it just a language trick? @Gadi The formulation you give is indeed ambiguous: we do not know a probability model for who goes to open the door. The formulation in the original question is unambiguous and indeed is a simple problem of conditional probability. Sep 11 comment Boy Born on a Tuesday - is it just a language trick? @T. No, it is not. In the Monty Hall problem there is a random choice, and then the showman has to make a choice about which door to open (he certainly does not want to open the door with the goat). In this case, there is not a random choice, after which the parent has to reveal information (for instance a random question). It is just plain conditional probability: the fact that the parent reveals this piece of information has nothing to do with the fact that he/she could reveal another. Sep 11 comment Boy Born on a Tuesday - is it just a language trick? @Gadi: I do not agree. I think you are making confusion with similar sounding problems like the Monty Hall one. In this case there is a very precise information; there is no need to speculate why the parent gave this piece of information instead of another one. Of course he/she might have said other things, for instance "I have a boy which is not born on Sunday", and you would obtain a different question, but this does not affect our problem. Sep 11 answered Can we define definitions as axioms in logic? Sep 7 comment Induction on Real Numbers It seems to me that ii) is flawed. You could have $[0, x] = [0, y)$ (think of the case of natural numbers where y = x+1). Hence in your argument you cannot deduce that $y \notin S$. Sep 5 awarded Scholar Sep 5 accepted CAS with a standard language Sep 3 awarded Student