For $c>0$, sample repeatedly and independently from $(0, 1)$ until the sum of the samples exceeds $c$. Let $\mu_c$ be the expected size of the final sample.
For which $c$ is $\mu_c$ maximised?
It is clear that as $c$ tends to $0$, $\mu_c$ tends to $\frac{1}{2}$ and this is its minimum value.
I will write $f(c) = \mu_c$ to make it more clearly a function of $c$. For now I will only investigate $c \in [0,1]$. It turns out (if my math is correct) there is a unique max within $c \in [0,1]$, at:
$$c = \ln 2, ~~~~~~~f(c) = \mu_c = \ln 2$$
But frankly I would only trust my own math below with about 80% confidence... :)
First of all, $c$ can be considered as the amount "still to go". I.e., starting with $c$, if the next sample is $x < c$, then you effectively have a new problem with a new threshold of $c-x$, and the expected value becomes $f(c-x)$. We can build a recurrence from this observation.
Let $X \sim Unif(0,1)$. We have:
Law of total expectation: $f(c) = P(X > c) E[X \mid X > c] + P( X < c) E[f(c-X) \mid X < c]$
$P(X>c) = 1-c$
$E[X \mid X > c] = {1+ c \over 2}$ because conditioned on $X > c$ then $X\sim Unif(c,1)$
$P(X < c) = c$
Conditioned on $X < c$, we have $c-X \sim Unif(0, c)$.
So the most trouble term becomes:
$$E[f(c - X) \mid X < c] = \int_0^c \frac1c f(u) ~du$$
And the overall equation is:
$$f(c) = {1 - c^2 \over 2} + c \int_0^c \frac1c f(u) ~du$$
Differentiate w.r.t. $c$:
$$f'(c) = -c + f(c)$$
which is an ODE with this solution (credit: wolfram alpha!): for some integration constant $K$,
$$f(c) = K e^c + c + 1$$
Substitute in $f(0) = 1/2$ (as observed by OP) and solving, we have $K = -1/2$ and so:
$$f(c) = -\frac12 e^c + c + 1$$
Now we just need to find the max:
$$f'(c) = -\frac12 e^c + 1 = 0 \iff e^c = 2 \iff c = \ln 2$$
at which point we have $f(c) = c = \ln 2$ which is just slightly $> 2/3$.
Further thoughts:
(1) I am pretty rusty (and that's a charitable description!) with "continuous" math, so if someone can critique / verify the above, that'd be much appreciated.
(2) This answer does not cover the case of $c > 1$ so far. For $c> 1$, there is no chance the next sample is enough, and the recurrence becomes:
$$f(c) = \int_{c-1}^c f(u) ~du$$
where $f(u) = -\frac12 e^u + u + 1$ whenever $u < 1$. I don't know how to do this integration. However, intuitively, since $f(c)$ is based on averaging of values of $f(u)$ (or average of averages, etc), the max of $f(c)$ cannot $>$ the max of $f(u)$, and in fact, since the max $f(u)$ is unique within $u \in (0,1)$, the max of $f(c)$ cannot even $=$ the max of $f(u)$ within $u \in (0,1)$. This is not a rigorous proof, but rather an intuitive argument why the max I found within $(0,1)$ is also the global max.
For $1\le c\le2$, we have
$$ f(c)=\int_{c-1}^1\left(-\frac12\mathrm e^x+x+1\right)\mathrm dx+\int_1^cf(x)\mathrm dx;; $$
differentiating yields $f'(c)=\frac12\mathrm e^{c-1}-c+f(c)$, and the solution for the initial condition $f(1)=-\frac12\mathrm e+2$ is $f(c)=\frac12\mathrm e^{x-1}(x-1)-\frac12\mathrm e^x+x+1$. Here's a plot. This section also has a local maximum, at $\hat c\approx1.4770$ with $f(\hat c)\approx0.67139$, already quite a bit closer to $\frac23\approx0.66667$ than $\ln2\approx0.69315$.
– joriki Dec 28 '19 at 00:07antkam's approach is correct and elegant. We can also derive the same result with a bit less calculus by approximating the continuous uniform distribution by a discrete uniform distribution.
So we have an $n$-sided die that we roll repeatedly, summing the results, and we want to know the threshold $k$ that maximizes the final result (where reaching $k$ itself is enough to stop).
We can prove by induction that the expected final result for threshold $k$ is $n+k-\frac n2\left(1+\frac1n\right)^k$. For $k=1$ this is $n+1-\frac n2\left(1+\frac1n\right)=\frac{n+1}2$, which is correct. Assuming the result for $k-1$, we obtain the expected final result for $k$ as
\begin{eqnarray*} &&\frac1n\left((n-k+1)\frac{n+k}2+\sum_{j=1}^{k-1}\left(n+j-\frac n2\left(1+\frac1n\right)^j\right)\right) \\ &=& \frac1n\left((n-k+1)\frac{n+k}2+(k-1)n+\frac{k(k-1)}2-\frac n2\left(\frac{1-\left(1+\frac1n\right)^k}{1-\left(1+\frac1n\right)}-1\right)\right) \\ &=& n+k-\frac n2\left(1+\frac 1n\right)^k\;. \end{eqnarray*}
Setting the derivative with respect to $k$ to zero yields
$$ 1-\frac n2\ln\left(1+\frac 1n\right)\left(1+\frac 1n\right)^k=0\;, $$
so the optimal threshold in the discrete case is an integer near
$$ -\frac{\ln\frac n2+\ln\ln\left(1+\frac 1n\right)}{\ln\left(1+\frac 1n\right)}\;. $$
For instance, for $n=6$, this is
$$ \frac{\ln3+\ln\ln\frac76}{\ln\frac76}\approx5.003\;, $$
so the optimal threshold is $k=5$ with an expected final result of
$$ 6+5-\frac62\left(1+\frac16\right)^5=\frac{11705}{2592}\approx4.516\;. $$
For $n\to\infty$, we have $\ln\left(1+\frac1n\right)\sim\frac1n$ and thus
$$ -\frac{\ln\frac n2+\ln\ln\left(1+\frac 1n\right)}{\ln\frac 1n}\sim-\frac{\ln\frac n2+\ln\frac 1n}{\frac 1n}=n\cdot\ln2\;, $$
in agreement with antkam's result.
