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Suppose each one has a,b,c,d,e,f,g ounce of milk initially. Everyone will pour their milk into 6 other cups equally
so a+b+c+d+e+f=6g
Similarly
b+c+d+e+f+g=6a
Therefore a=g
and similarly all will be equal
a=b=c=d=e=f=g = 42/7 = 6 ounce

initial quantity of milk in order: 12 ,10, 8, 6, 4, 2, 0
total : 42 ounces

11.56
9.84
8.05
6.17
4.20
2.15
0.0

Correct answer by Akshay Kumar. Detailed Solution is as follows:

Since the final state matches the initial state, we can imagine this process going on continuously. Consider the dwarf whose cup contains the smallest amount of milk just before he begins pouring. Call him D, and call this quantity of milk a ounces. Since D's cup contains the least milk before pouring, then each other dwarf has an equal or greater amount in his own cup before pouring, and thus gives D at least a/6 ounces. This would leave D with at least a ounces in his own cup just before pouring; the only way he can receive precisely a ounces, as we know he does, is if each other dwarf gives him precisely a/6 ounces and no more, i.e., that each cup, just before pouring, contains the same quantity of milk. So after any given pouring the cups contain a, 5/6a, 4/6a, 3/6a, 2/6a, 1/6a, and 0 ounces; if there are 42 ounces in total then this works out to 12, 10, 8, 6, 4, 2, and 0 ounces.

The above solution is from Futility Closet

Interesting solutions on Math Stack Exchange

You can also generalize the solution for any number. But, you need to have n*(n-1) ounces of milk in total. Then, for each such value, answer would be in AP with difference = -2.
For example: for 4 dwarfs, if you have 12 ounces of milk in total, then answer would be 6, 4, 2, 0.
For 3 dwarfs and 6 ounces of milk, answer would be 4, 2, 0.
:-)