One day you stumble across a magic genie, who says that if you play a simple game with him, you could win fabulous riches. You take the genie up on his offer, and he hands you a stick of length 1. He says that behind his back is another stick, with a random length between 0 and 1 (chosen uniformly over that interval).
Next, you must break your stick into two pieces and present one of those pieces to the genie. If that piece is longer than the genie’s hidden stick, then you win a prize of \$1 million times the length of your remaining piece. For example, if you present to the genie a length of 0.4, and that’s longer than the genie’s stick, then you win $1 million times 0.6, or \$600,000. However, if the genie’s stick is longer, then you win nothing.
Being a regular reader of The Riddler, you crunch some numbers and prepare to break your stick in half. But then you have a thought. You ask the genie if you can have more than one turn. For example, if you present the genie with a length of 0.4, but the genie’s stick is longer, can you break off a piece of the remaining length of 0.6 — say, a length of 0.5 — and then present that to the genie? To keep things fair, your winnings will still be proportional to the leftover length. So had the genie’s length indeed been between 0.4 and 0.5, your first and second guesses, then the remaining length would have been 0.1, and you would have won \$100,000.
The genie doesn’t think any of this really matters and says you can have as many turns as you desire. If your goal is to maximize your expected winnings, what will your strategy be? And how much money would you expect to win on average?
I looked at limited number of turns.
The expected value is the maximum of
$$l\cdot(1-l)\cdot1000000$$
$l = 0.5$ is optimal stick length, which yields $\$250,000$.
The value of the first turn is $l_1\cdot(1-l_1)\cdot1000000$.
Let $g$ be the length of the genie's stick, and let $l_2$ be the length broken off on your second turn, where $l_2 > l_1$. Note that $l_1 < g < 1$ and if $g < l_2$, you win. The remaining stick length is $1-l_1-l_2$
The value of the second turn is $\dfrac{l_2 - l_1}{1 - l_1} \cdot(1-l_1-l_2)\cdot 1000000$.
The expected value of two turns is thus the maximum of
$$EV(l_1, l_2) = l_1\cdot(1-l_1)\cdot1000000 + \dfrac{l_2 - l_1}{1 - l_1} \cdot(1-l_1-l_2)\cdot 1000000$$
$$= \dfrac{1000000}{1 - l_1}\left(l_1\cdot(1-l_1)^2 + (l_2 - l_1)\cdot(1-l_1-l_2) \right)$$
$$= \dfrac{1000000}{1 - l_1}(l_1^3-2l_1^2+l_1 + l_2 - l_1l_2 - l_2^2 - l_1 + l_1^2 + l_1l_2 )$$
$$= 1000000\left(\dfrac{l_1^3-l_1^2}{1 - l_1} + \dfrac{l_2 - l_2^2}{1 - l_1}\right)$$
$$= 1000000\left(-l_1^2 + \dfrac{l_2 - l_2^2}{1 - l_1}\right)$$
Taking the derivative,
$$EV'(l_1, l_2) = \dfrac{\delta EV}{\delta l_1} + \dfrac{\delta EV}{\delta l_2}$$
$$\dfrac{\delta EV}{\delta l_2} = 0 = 1000000\left(\dfrac{1-2l_2}{1-l_1}\right)$$
$$l_2 = 0.5$$
$$\dfrac{\delta EV}{\delta l_1} = 0 = 1000000\left(-2l_1 + \dfrac{l_2-l_2^2}{(1-l_1)^2} \right)$$
Substituting $l_2 = 0.5$,
$$2l_1(1-l_1)^2 = \frac{1}{4}$$
$$8l_1^3-16l_1^2 + 8l_1 - 1 = 0$$
$$(2l_1-1)(4l1^2 - 6l_1 + 1) = 0$$
$l_1 \ne 0.5$, the second factor yields
$$l_1 = \frac{3 - \sqrt{5}}{4} \approx 0.19098$$
Let $l_n$ be the length broken off on your second turn, where $l_n > l_{n-1}$. Note that $l_{n-1} < g < 1$ and if $g < l_n$, you win. The remaining stick length is $1-\sum_{k=1}^n l_k$
The value of the nth turn is
$$\dfrac{l_n - l_{n-1}}{1 - l_{n-1}} \cdot\bigg(1-\sum_{k=1}^n l_k\bigg)\cdot 1000000$$
Let $R = 1-\sum_{k=1}^{n-2} l_k$. The value of the nth turn can be expressed as
$$\dfrac{1000000}{1 - l_{n-1}} (l_n - l_{n-1})(R- l_{n-1}-l_n) $$
$$= \dfrac{1000000}{1 - l_{n-1}} (l_nR - l_n^2 - l_{n-1}R + l_{n-1}^2)$$
This is the only term in $EV(l_1, l_2, ...l_n)$ containing $l_n$.
$$\dfrac{\delta EV}{\delta l_n} = 0 = \dfrac{1000000}{1 - l_{n-1}}(R-2l_n)$$
$$l_n= \frac{R}{2}$$
The maximum expected value will have $l_n < 0.5$ for any strategy with $n$ runs, $3 \le n$. I could not find an expected value greater than the maximum for two runs.
With two runs,
$$EV(0.19098, 0.5) = 0.19098 \cdot 0.80902 \cdot 1000000 + \dfrac{0.30902}{0.80902} \cdot 0.30902 \cdot 1000000 \approx \$272,542.49$$