question for MIT probability class

[pzhon]

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So I was a bit bored and decided to solve this problem by hand, but I arrived at a different answer than the above posters. I would appreciate if someone told me where I went awry.

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You used a different set of assumptions. You assumed that all indistinguishable configurations are equally likely. Everyone else assumed that if you can distinguish the balls of the same color, then all configurations should be equally likely.

If you had 100 indistinguishable balls to place randomly in 2 urns, what is the probability that all of the balls end up in the first urn? That's not well-defined until you clarify what is meant by randomly. You assumed that the probability is 1/101, since there could be 0, 1, 2, ..., 100 balls in the first urn. The assumption used by everyone else was that you are essentially tossing a coin independently for each ball, so the probability that they all end up in the first urn is 1/2^100.

Some physicists use the same definition you are using by default, assuming that all indistinguishable cases are equally likely. I don't know whether this is due to laziness (and the fact that it usually doesn't matter in statistical mechanics), or because this is actually a more accurate approximation for how most quantum-mechanical systems work.

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30/84 times the black balls will be in 2 urns, leaving two open. 14 out of these 30 configurations will not overlap with any given placement of the green balls.

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That should be 1/2, or 15/30.

Thanks for taking the time to let me know why my answer was different. Of course, because I didn't fully understand what I was doing, I was unable to see the differing assumptions I and the above posters employed. Also, regarding the 14/30 rather than 15/30 -- you are of course correct. Sloppy work to accompany my sloppy thinking. Cheers.

[BruceZ]

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Some physicists use the same definition you are using by default, assuming that all indistinguishable cases are equally likely. I don't know whether this is due to laziness (and the fact that it usually doesn't matter in statistical mechanics), or because this is actually a more accurate approximation for how most quantum-mechanical systems work.

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The latter, though you mean that all "distinguishable" arrangements are equally likely, and that the particles behave as though they are indistinguishable. Elementary particles obey Bose-Einstein statistics (bosons) where they act as though they are indistinguishable and each distinguishable arrangement is equally probable, or Fermi-Dirac statistics (fermions) in which case they act as though they are indistinguishable, and each distinguishable arrangement has one particle per "urn" and is equally probable. The last I heard, no elementary particles obey Maxwell-Boltzmann statistics, which is the common-sense macroscopic distribution in which each arrangement is equally probable as if each particle were distinguishable.