The traditional method for two children to divide a piece of cake fairly between them is "you cut, I choose". What process accomplishes this is intuitively clear. Is there a similar process which works for $3$ children?

Suppose that three children $(1),(2)$ and $(3)$ have associated measures (ways of evaluating the pieces of cake) $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$. Assume for the simplicity of the notation that the sum of total pieces of the cake is equal to $1$.

Question : Does the cake have three subsets $C_{1},C_{2}$ and $C_{3}$ that are pairwise disjoint and their union is equal to $C$, and that have moreover the property that $$\alpha(C_{1}) \geq \frac{1}{3}, \quad \alpha(C_{2} )\geq \frac{1}{3} , \quad \alpha(C_{3}) \geq \frac{1}{3}?$$

**Solution:** Let one of the three children, say $(1)$, divide the cake into what he regards as three equal pieces $C_{1},C_{2}$ and $C_{3}$: In other words $\alpha(C_{1})= \alpha(C_2)=\alpha(C_{3})=\frac{1}{3}$. Let each of $(2)$ and $(3)$ claim dibs on one of those three pieces, the one he prefers the most, the one he would be quite satisfied to have. It is possible that $(2)$ and $(3)$ will claim different pieces, and it is possible that they will claim the same one, but no matter, atleast one of the three pieces, $C_{1},C_{2},C_{3}$ will be left unclaimed. Let $(1)$ remove that piece (one of those pieces) and eat it. So far everyone is satisfied. Child (1) got what he believes to be exactly a third, and in the judgement of the children $(2)$ and $(3)$ the best piece (the claimed piece) is still available. The problem reduces therefore, to dividing whats fairly left, - and that's the classical problem of dividing a piece of cake between two claimants. Let that problem be solved by "you cut, i choose" and everybody is happy.

**Solution** The Proposed solution is wrong; Whats wrong with it?