🚀 go-pugleaf

In article <1992Aug19.090212.1658@gacvx2.gac.edu> logic@gacvx2.gac.edu writes:
>Now for a riddle of a different flavor..  I first saw this in an interview
>article with Paul Halmos in the Mathematical Intelligencer..  I offer it here
>because it is (1) not as widely known as the riddle above, (2) more difficult
>than the riddle above, and (3) its solution offers some surprises..  Here we
>go.  My wife and I go to a party..  There are 4 other couples there for a total
>of 5 couples.  (No funny business here..  a couple is husband and wife..)  As
>is usual at a social gathering a certain amount of handshaking takes place.. 
>After the handshaking has completed, I interview all of the people at the party
>including my wife.  I ask them how many hands they shook.  (note: spouses don't
>shake hands and no one shakes their own hand)..  Now each person, therefore,
>could have shook as few as 0 hands or as many as 8..  Now, in fact, that range
>of answers is exactly what i get for answers!  I.e., the first person I
>interview responds 0, the second 1, the third 2, ..., the ninth 8..  The riddle
>is "How many hands did my wife shake?"..  I'll even give you the answer..  It
>is 4..  The real riddle, of course, is explaining why this is the correct
>answer..

While this isn't particularly difficult, it's much easier if we are allowed
to assume that there is a unique answer.  If you just have each person shake
hands with exactly the people (s)he previously did not shake hands with, you
get the same distribution, but each person shook hands with 8-n people.  The
assumption that you know how many people one person shook hands with means
that n=8-n, so n=4.