Obviously, the 10th person has a 50% chance to survive, as noone
except the monster can see his/her hat's colour. Nonetheless, the person is
able to see all the other 9 hats. As 9 is odd, there must be an odd number
of either white or black hats remaining. So, by "guessing" the color which
appears an odd number of times, the 10th person survives with 50% chance
and gives away all information needed to save the remaining 9.
The (2n+1)th person survives by confirming his/her neighbour's choice, if
that colour appears an even number of times among the remaining 2n hats,
and altering it otherwise.
The 2nth person survives by confirming his/her neighbour's choice, if that
colour appears an odd number of times among the remaining (2n1) hats,
and altering it otherwise.
If there were an odd number of inhabitants, the one who's asked first would
have to "guess" a preagreedupon colour, if his/her neighbour's hat has the
colour that appears an odd number of times among the remaining ones,
excluding his/her neighbour.
