note
tmoertel
The trick here – and there <em>is</em> a trick – is a
subtle use of equivocation: You make a claim about one problem but
then explain the claim in terms of another problem that is subtly yet
significantly different.
<p>In the first problem, each envelope can contain <em>any</em>
number. In the second problem, however, you require that the
distribution of numbers have a "continuous probability
distribution with non-zero density everywhere." These problems are
not the same.
<p>To see why, imagine an infinite number line representing all
numbers. If you pick any segment on this line, say that between 0 and
1, the portion of the line that the segment represents is zero. Hence
the probability density function over the segment is also zero, which
contradicts the assumption made in your analysis.
<p>If we repeat your analysis, this time using the distribution of
numbers for your originally stated problem, we find that the
probability of picking a number in between your two numbers is
<em>zero</em>, and thus knowing the first number provides no benefit.
Our intuition turns out to be correct.
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