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I know that does not sound like a math question, but I believe there should be a really simple solution to this somewhere in combinatorics (I am sorry in advance if I am wrong). There is a group of 6 people, whom we have to divide into two equal groups of 3 according to their stated preferences. How this mapping of preferences to subgroups should like so the groups are always of the same size of 3?

For example, in this famous experiment by Tejfel the participants had to choose between Klee and Kandinsky and were assigned into two different groups based on their choice (Tajfel, Henri. "Experiments in intergroup discrimination." Scientific American 223.5 (1970): 96-103.)

But there is no guarantee that the preferences will split the group into two equal parts.

If for instance, we would offer them to order N pictures from the most to the least preferred, and then assign the group membership based on the preferences, that would allow us to divide them. Unless all 6 of them would choose the same order (unlikely but possible).

Does such ordering exist?

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  • $\begingroup$ In the original Tejfel experiment the boys were allocated between the two groups at random, ignoring their claimed preferences between unlabelled pictures. If you start with an even number of boys, this is usually possible. $\endgroup$ – Henry Sep 21 '17 at 16:38
  • $\begingroup$ This question is a bit broad, but could easily be a good fit for the site with some more specifics about what the "stated preferences" look like. For example, how many preferences? Just binary or multiple options? $\endgroup$ – Joffan Sep 21 '17 at 16:38
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A group of 6 people can be divided into two groups of equal size in 20 different ways. Depening on your definition of similarity all of those 20 ways could be a good division. In order to answer your question, first you have to define what exaclty do you mean by similarity.

For example imagine three different preferences vectors:

$p1: A>B>C>D>E$

$p2: E>A>B>C>D$

$p3: A>E>D>C>B$

And now tell whether $p1$ is more similar to $p2$ or to $p3$? As you see the answer is not so obvious.

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