Finding one-third of a uniform distribution

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I have a valuation of a cake (the unit interval $[0,1]$) such that $\int_0^1 v(x) dx =1$. This valuation tells me how much a value different pieces of cake (subintervals of [0,1]).

I am cutting the cake against 2 other guys. I know nothing about their valuations so it is just safe to assume they are uniform (I read somewhere that this minimizes entropy). I know the point x such that $\int_0^x v(x) dx =1/3$ and want to make a guess about the points y and z such that the same thing is true for the other two guys: the point at which they value the cake at 1/3. What would be my expectation about y and z? It is kind of obvious that it should be y=z=1/3, but maybe this is wrong? I am just not sure. More particular, I want to make inference about min(z,y). This is surely less than one third?