Is it ALWAYS the case that, for a unimodal probability distribution, the median is between the mode and mean?
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The answer is "No," as the article linked to by leonbloy indicates. Here's an example of the kind of situation in which you would get mean < mode < median. The black rectangle below contains the mean, the blue rectangle the mode, and the red rectangle the median.
The black rectangle is $5 \times 1/5$, the blue $1/4 \times 4$, and the red $1 \times 3$. The median is in the red rectangle because the areas of the three rectangles are 1, 1, and 3. The blue rectangle clearly contains the mode. Placing the origin at the lower left corner of the blue rectangle, we see that the mean is at $(-2.5)(1) + (1/8)(1) + (3/4)(3) = -1/8$, and so the black rectangle contains the mean.
Of course, this can be scaled to produce an actual probability density function or smoothed to obtain a continuous pdf without changing mean < mode < median.
Added: One of our own users, Henry, has written a detailed article on the relationship between the mean, median, mode, and standard deviation in a unimodal distribution. It's definitely worth a look.