Instead of inventing new medians I suggest we take a closer look at the old one. As a prototypical model, consider a set S of n sticks or segments. These make up a PRE-ordered set with a relation >=,"greater than or equal". Namely given sticks s and s', we have s=s' if the sticks are congruent and s<s' if s is congruent to a subset of s'. This is straight out of Euclid's elements which never mentions the notion of length of a segment, and neither will I. What is a median for this set? NOTATION. if s is a stick then s+ ={s'in S; s'>=s} s- = {s' in S: s'<=s} If X is a set #X is the cardinality of X. DEFINITION: A stick s is a MEDIAN of the set S if #(s+)>=n/2 and #(s-)>=n/2. EXAMPLE. If S consists of two sticks s<s' then both s and s' are medians. EXERCISE 1. Prove the "Fundamental Theorem of Median Theory".Every finite preordered set has at least one median. (not completely trivial). A slightly fancier way of doing this is to take the ordered set consisting of the equivalence classes {s} with respect to =. Then #{s} is a given distribution etc. Another example, going back to the thing that started all this, is an ordered set of men M where m'>=m if m' has had at least as many sexual partners as m. Now here is a "real world" example. The pre-ordered set is a set of Gifts ordered "chronologically" as follows; G_1=a partridge in a pear tree < G_2= a turtle dove <G_3= _<. . . .<G_12= a drummer drumming Depending on how one interprets the song there are two possibilities. I. S consists of 66 gifts; 1 partridge, 2 turtle doves,. . ,12 drummers. II. S consists of 364 gifts; 12 partrigdes, 22 turtle doves, . . ,12 drummers. EXERCISE 2; Find the median gift for both interpretations I and II. The link below is provided for people (like me) who don't remember which gifts go with which days. http://www.night.net/christmas/12-Days.html David