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tilly
There is no trick.<p>
In the problem, each envelope can contain any number. The requirement of a distribution of numbers that has a continuous probability distribution with non-zero density everywhere is on the one you create for your algorithm. That number is made up out of thin air and has no connection to either of the numbers in the envelopes. The trick is that it gives us a chance of telling the two numbers apart.<p>
The probability density function that you are trying to describe, "uniform over the real numbers", is not a valid probability density function. It is a classic result both in real analysis and probability that it can't be. (The real analysis statement is that no such measure exists, the probability theory statement is that no such probability distribution exists. Those statements are the same.)<p>
This is why you have to be very careful in the wording to even get a well-defined problem. Given any two numbers and the algorithm, there is a well-defined probability that you're right, and that probability is over 0.5. Prior to the numbers and algorithm, the probability of your being right is undefined and undefinable.<p>
Again I'd like to point people to <a href="http://groups.google.com/group/sci.math/tree/browse_frm/thread/c4f7e14f217747d1/7e38eb6b8af5ba08?rnum=11&hl=en&q=probability+group:sci.math+author:benjamin.j.tilly@dartmouth.edu&_done=%2Fgroup%2Fsci.math%2Fbrowse_frm%2Fthread%2Fc4f7e14f217747d1%2F9a7a46cf2051b7fa%3Flnk%3Dst%26q%3Dprobability+group:sci.math+author:benjamin.j.tilly@dartmouth.edu%26rnum%3D1%26hl%3Den%26#doc_14feb6db3679f9da">Bently Preece's excellent explanation</a>. The goal of the question is to come up with a single strategy that will give better than even odds for every game in a particular class of games. And the described strategy does that.<p>
Now I haven't explained why this particular probability distribution can't exist. The technical reason is that the [http://mathworld.wolfram.com/ProbabilityAxioms.html|axioms of probability] require the probability of a countable disjoint union to be the sum of the probabilities, and that the probability of the whole set is 1. But the real line is a countable disjoint union of intervals, each of which has probability 0 (for the reason that you explained), so the probability of the whole set is a countable sum of 0's, which is 0. Contradicting the requirement that it be 1.<p>
Now it is easy to object that one could just define probabilities differently. And it is true, one could. But any way you define it, you'll have a lot of subtleties around infinity, and attempting to think about a uniform probability distribution on the real numbers will be one place that you'll encounter lots of them.<p>
BTW in the USA this fact is generally covered in a real analysis class in advanced undergraduate or beginning graduate school mathematics.
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