Optimal strategy in repeated cake-cutting
Let $v(x) = \frac{V(x) - V(0)}{V(1) - V(0)}$ so $v(x)$ nicely goes from $0$ to $1$. Maximizing $v$ also maximizes $V$. To cover my ass for future mistakes regarding $&lt;$ vs $\leq$, I will assume that you will never exactly guess $x = h$.

Now, if $n = 1$, we can find the expected value from cutting at $x$ by integrating over all possible $h$:

$$e_1(x) = E[v(x)] = \int_0^x (1 - v(x))dh + \int_x^1 v(x)dh = (1 - 2x)v(x) + x$$

Which allows you to maximize $e_1$ with differentiation to find the optimal $x$. Note that in the above example "all possible $h$" meant $[0, 1]$. But what if we know that $h \in [a, b]$ due to previous information?

Then we get the following sum of integrals for when also $x \in [a, b]$:

$$\int_a^x (1 - v(x))dh + \int_x^b v(x)dh = (a + b - 2x)v(x) - a + x$$

Which makes the total expected valuation:

$$e_{1,a,b}(x) = E[v(x) | a \leq h \leq b] = \begin{cases}
v(x) &amp; \text{if $x &lt; a$}\\
(a + b - 2x)v(x) - a + x &amp; \text{if $a \leq x \leq b$}\\
1 - v(x) &amp;\text{if $x &gt; b$}
\end{cases}$$

Now let's analyze $n = 2$. Suppose our first cut is at $x$, then there is a $x$ chance $h \leq x$ and $1 - x$ chance $h \geq x$. In both those scenarios we will then choose our best possible $y$. This gives equation:

$$e_2(x) = e_1(x) + x\left( \max_{0 \leq y \leq 1} e_{1,0,x}(y)\right) + (1-x)\left( \max_{0 \leq y \leq 1} e_{1,x,1}(y)\right)$$

$$\max_{0 \leq y \leq 1} e_{1,0,x}(y) = \max\left(\max_{0 \leq y \leq x} ((x - 2y)v(y) + y), \max_{x \leq y \leq 1} (1 - v(y))\right)$$

$$\max_{0 \leq y \leq 1} e_{1,x,1}(y) = \max\left(\max_{0 \leq y \leq x} v(y), \max_{x \leq y \leq 1} ((x + 1 - 2y)v(y) - x + y)\right)$$

Now let's say $v(x) = x$, a simple linear increase in desire for more cake. Now we can actually solve:

$$\max_{0 \leq y \leq 1} e_{1,0,x}(y) = \max\left(\begin{cases}x(1-x)&amp;0\leq x \leq \frac{1}{3}\\\frac{1}{8}(x+1)^2&amp;\frac{1}{3} \leq x \leq 1\end{cases}, 1 - x\right)$$

$$\max_{0 \leq y \leq 1} e_{1,x,1}(y) = \max\left(x, \begin{cases}\frac{1}{8}(x-2)^2&amp;0\leq x \leq \frac{2}{3}\\x(1-x)&amp;\frac{2}{3} \leq x \leq 1\end{cases}\right)$$

Now $e_2$ is complicated, but is essentially a piecewise function which can be solved.