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Monty Hell problem - Solution |  | Monty Hell problem - Solution: Encyclopedia II - Monty Hell problem - Solution |  | The solution to the paradox is to observe that the two contradictory arguments are taking different limits. The obvious answer takes the limit of the number of bills in the sack; the less obvious answer takes the limit of the set of bills in the sack first, and counts them up afterwards. It is surprising, but not inconsistent, that these two different computations yield different answers.
Which answer is correct? This depends on how you interpret the problem. If you concentrate on the fate of individual bills, and are comfortable with ...
See also:Monty Hell problem, Monty Hell problem - The paradox, Monty Hell problem - Attacks on the second solution, Monty Hell problem - Everybody dies but that doesn't mean someday no one will be alive, Monty Hell problem - You can't multiply a zero probability by infinitely many elements, Monty Hell problem - What if the Devil pays you out of his heating fee receipts?, Monty Hell problem - Solution, Monty Hell problem - Appendix: Proof that each bill leaves the sack with probability 1, Monty Hell problem - Historical notes |  | | Monty Hell problem, Monty Hell problem - Appendix: Proof that each bill leaves the sack with probability 1, Monty Hell problem - Attacks on the second solution, Monty Hell problem - Everybody dies but that doesn't mean someday no one will be alive, Monty Hell problem - Historical notes, Monty Hell problem - Solution, Monty Hell problem - The paradox, Monty Hell problem - What if the Devil pays you out of his heating fee receipts?, Monty Hell problem - You can't multiply a zero probability by infinitely many elements, Ross-Littlewood Paradox, a problem which is equivalent to this one, but is much more astonishing as it has nothing to do with probability. |  | |
|  |  | Monty Hell problem: Encyclopedia II - Monty Hell problem - Solution
Monty Hell problem - Solution
The solution to the paradox is to observe that the two contradictory arguments are taking different limits. The obvious answer takes the limit of the number of bills in the sack; the less obvious answer takes the limit of the set of bills in the sack first, and counts them up afterwards. It is surprising, but not inconsistent, that these two different computations yield different answers.
Which answer is correct? This depends on how you interpret the problem. If you concentrate on the fate of individual bills, and are comfortable with set-theoretic limits, you are likely to take the set-theoretic limit first and conclude that Monty leaves you with nothing. This leaves you in the embarrassing situation of having to explain how you could have had increasing wealth forever but suddenly lose it all at the end, and the assurance that this outcome is consistent with modern mathematics may not make you feel much better. If you treat the individual bills as mere tokens representing your total wealth, you are likely to prefer the numerical limit and conclude that Monty leaves you with infinite wealth. But now you are in the still more embarrassing situation of having to explain how you came to hold a sack that contains infinitely many bills, every single one of which you previously lost forever.
The rec.puzzles discussion ultimately favored the zero-wealth outcome, but the issue continues to be contentious. For further discussion, see Talk:Monty Hell problem.
Other related archivesBoole's inequality, Devil, Hell, Kolmogorov, Marilyn vos Savant, Monty Hall problem, Parade Magazine, Paradoxes, Ross-Littlewood Paradox, Talk:Monty Hell problem, Transfinite number, countable, countably, diverges, harmonic series, inequality, paradox, probability, probability axioms, probability theory, set theory, set-theoretic limit, usenet
 Adapted from the Wikipedia article "Solution", under the G.N U Free Docmentation License. Please also see http://en.wikipedia.org/wiki |
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