I have another idea for the case where the stick is divided into $n$ pieces: let $X_{i}$ $1 \leq i \leq n$ be the cut marks, which are independently randomly distributed variables between $0$ and $1$, and set $A=max{X_{i}}$. Now, consider the two pieces, one of length $A$ and the other of length $1-A$. The piece of length $A$ is then subdivided into $n-1$ pieces while the one of length $1-A$ remains whole. If $1-A>A$, then the piece of length $1-A$ will be the longest (obviously). If $1-A<A$, then the longest piece is the maximum of the longest subdivision on $A$ and the one of length $1-A$. Let $L_{n}$ denote the length of the longest segment of a stick of unit length divided into $n$ pieces. Since the length of the longest subdivision on the stick of length $A$ is $AL_{n-1}, we obtain the following formula:
$$E(L_{n})=P(1-A>A)E(1-A)+P(1-A<A)E(\max(AL_{n-1}, 1-A))$$
Since the $X_{i}$ are independent, the CDF for $A$ is $F_{A}(y)=y^{n}$ and the PDF is $ny^{n-1}$, so that we obtain:
$$E(A)= \int_{0}^{1} ny^{n} dy = \frac{ny^{n+1}}{n+1}|_{0}^{1}=\frac{n}{n+1}$$.
Therefore, $P(1-A>A)=P(A<\frac{1}{2})=\frac{1}{2^{n}}$, so that $P(1-A<A)=1-\frac{1}{2^{n}}$. Now, $E(1-A)=E(1)-E(A)=1-\frac{n}{n+1}=\frac{n+1-n}{n+1}=\frac{1}{n+1}$. Therefore, the above formula for $E(L_{n})$ reduces to:
$$E(L_{n})=\frac{1}{2^{n}(n+1)}+(1-\frac{1}{2^{n}})(E(\max(AL_{n-1}, 1-A))$$
Now to deal with $E(\max(AL_{n-1}, 1-A)$, let $f_{L_{n-1}}$ denote the PDF for the random variable $L_{n-1}$. I believe we can assume that $A$ and $L_{n-1}$ are independent. If this is the case, let $M=\max(AL_{n-1},1-A)$ and note the following fact: if $a=L_{n-1}$ is fixed, then $P(M \leq y)= P(1-y \leq A \leq \frac{y}{a}) = F_{A}(\frac{y}{a})-F_{A}(1-y)=(\frac{y}{a})^{n}-(1-y)^{n}$. Therefore, we obtain the following CDF for $M$:
$$F_{M}(y)=\int_{\frac{1}{n-1}}^{1} [\frac{y^{n}}{a^{n}}-(1-y)^{n}]f_{L_{n-1}}(a) da$$
Note that the lower bound $\frac{1}{n-1}$ on the integral is justified by the fact that the longest piece on the piece of a rope of unit length cut into $n-1$ pieces must be longer than $\frac{1}{n-1}$. Pulling out factors that do not depend on $a$ from the above integral yields:
$$F_{M}(y) = y^{n}\int_{\frac{1}{n-1}}^{1} \frac{f_{L_{n-1}}(a)}{a^{n}}da-(1-y)^{n}\int_{\frac{1}{n-1}}^{1}f_{L_{n-1}}(a)da$$
It is obvious from the definition of a probability density function that $\int_{\frac{1}{n-1}}^{1}f_{L_{n-1}}(a)da=1$ and we further recognize $\int_{\frac{1}{n-1}}^{1} \frac{f_{L_{n-1}}(a)}{a^{n}}da$ as $E(\frac{1}{L_{n-1}^{n}})$. We thus obtain:
$$F_{M}(y)=y^{n}E(\frac{1}{(L_{n-1})^{n}})-(1-y)^{n}$$
Differentiating $F_{M}(y)$, we get the following PDF, $f_{M}(y)$ for $M$:
$$f_{M}(y)=ny^{n-1}E(\frac{1}{(L_{n-1})^{n}})+n(1-y)^{n-1}$$ so that we can then calculate $E(M)$:
$$E(M)=E(\frac{1}{(L_{n-1})^{n}})\int_{\frac{1}{n}}^{1} ny^{n} dy + \int_{\frac{1}{n}}^{1} ny(1-y)^{n-1} dy$$
We compute
$$\int_{\frac{1}{n}}^{1} ny^{n} dy = \frac{ny^{n+1}}{n+1}|_{\frac{1}{n}}^{1}=
\frac{n}{n+1}-\frac{n(1/n)^{n+1}}{n+1}=\frac{n}{n+1}-\frac{1}{n^{n}(n+1)}$$
Finding a common denominator for the above two fractions, we get
$$\int_{\frac{1}{n}}^{1} ny^{n} dy = \frac{n^{n+1}-1}{n^{n}(n+1)}$$
From wolfram alpha's integrator, I computed
$$\int_{\frac{1}{n}}^{1} ny(1-y)^{n-1} dy = \frac{2(n-1)^{n}}{n^{n}(1+n)}$$
We can then conclude that
$$E(M)=\frac{(n^{n+1}-1)E(\frac{1}{(L_{n-1})^{n}})+2(n-1)^{n}}{n^{n}(n+1)}$$
Plugging this value back into the formula for $E(L_{n})$, we obtain:
$$E(L_{n})=\frac{1}{2^{n}(n+1)}+(1-\frac{1}{2^{n}(n+1)})\frac{(n^{n+1}-1)E(\frac{1}{(L_{n-1})^{n}})+2(n-1)^{n}}{n^{n}(n+1)}$$
Combining the two fractions using the common denominator of $(2n)^{n}(n+1)^{2}$, we obtain
$$E(L_{n})=\frac{n^{n}(n+1)+(2^{n}(n+1)-1)((n^{n+1}-1)(E(\frac{1}{(L_{n-1})^{n}})+2(n-1)^{n})}{(2n)^{n}(n+1)^{2}}$$
In order for this recursion formula to be useful, however, I need to be able to write $E(\frac{1}{(L_{n-1})^{n}}$ in terms of $E(L_{n-1})$. I have calculated that $E(\frac{1}{(L_{2})^{3}})=3$ and that $E(\frac{1}{(L_{2})^{4}})=12$, but I do not know in general how $E(L_{n-1})$ relates to $E(L_{n})$; does anyone have any ideas on how to do this? Please let me know. Thanks!
