Let’s first look at the simple case of going exactly 2 rounds. This is instructive before venturing into the general case of \(n\) rounds.

Let \(x_1\) be the first non zero stick length presented and \(x_2 \gt x_1\) be the second, if the first one ends up being shorter than the Genie’s. 

Now the expected reward units \(R_2\) (remember we are counting $1M = 1 unit) is calculated as:

\begin{align*}
R_2 &= (\textit{round 1 payoff}) \cdot P(\textit{winning round 1}) \ + \ (\textit{round 2 payoff}) \cdot P(\textit{winning round 2 having lost round 1}) \ + \ 0 \cdot (\textit{losing in round 2}) \\ \\
&=(1-x_1)x_1 + (1-x_1-x_2)(x_2-x_1) \\ \\
&=x_2 – {x_2}^2 \\
\end{align*}

This, as you notice does not depend on \(x_1\) ! Also, it’s simple to determine that the maximum value occurs at \(x_2 = 0.5\) and the max expected reward units are 0.25. This is the same as what we got in the single round game.

So the 2 round game solution is: pick any \(0 \lt x_1 \lt 0.5 \) for round 1 and if you lose, then present Genie with stick \(x_2 = 0.5\) in round 2 for total expected winnings of 0.25 units. Clearly then there are infinite solutions here all resulting in the same expected winnings.

But what if we were to go beyond 2 rounds? Are we still limited to expected winnings of 0.25? If so, will we at least have more choices/solutions? Turns out, the answers are YES and NO, respectively. Here’s why:

As before consider \(n \) rounds with stick lengths \(x_1, x_2,…x_n \) with  \(0 \lt x_1 \lt x_2 \lt…\lt x_n \leq 0.5 \). Now the expected reward \(R_n\) can be written as 

\begin{align*}
R_n &=(1-x_1)x_1 + (1-x_1-x_2)(x_2-x_1) + (1-x_1-x_2 – x_3)(x_3-x_2)) + …+ (1 – x_1-x_2-x_3…-x_n)(x_n-x_{n-1}) \\ \\
&= \sum_{i=1}^{n}(1 – \sum_{j=1}^{i}x_j)(x_i – x_{i-1}) \qquad \tag{1} \label{eq1} \qquad (\textit{more compact notation if you prefer})
\end{align*} 

Let’s start with the case of \(n = 3\) which is illustrative of the general solution. The above equation simplifies to:

\[
R_3 = x_3 – x_1(x_3-x_2) – x_3^{2}
\]

Now here are the implications of this:

• Since \(x_3 \gt x_2\) the second term above is overall less than zero. Which implies \(R_3 \) is maximized when the first cut \(x_1\) is 0 
• This is the same as playing just a 2 round game which with \(x_1 = 0\) can be maximized as before with any \(0 \lt x_2 \lt 0.5 \) and \(x_3 = 0.5\) for total expected winnings of 0.25 units
• If we forced this to be a three round game picking a non zero \(x_1\), we will reduce expected winnings! 

We’ve just learned that we can never play 3 rounds and do better! What if we played more than 3? We can now easily show that we will necessarily reduce our expected winnings if we play any number of rounds greater than 2.

If we take our general \(n\) round winnings \(R_n\) from \(\eqref{eq1}\) above and group all the terms containing \(x_1\), (the first cut stick length presented to the Genie), we get:

\begin{align*}
R_n &= \textit{a bunch of terms not containing} \ x_1 \ + \ … \ \ [-x_1(x_3-x_2) \ – \ x_1(x_4-x_3) \ – \ x_1(x_1-x_5) \ – \ … \ – \ x_1(x_n – x_{n-1})] \\ \\
&= \textit{non} \ x_1 \ \textit{terms} \ + \ … \ \ -x_1(x_n-x_2)
\end{align*}

Which, as before, knowing that \(x_n \gt x_2\) implies that it is maximized by picking the first cut \(x_1 = 0\) ! i.e. reduce it to an \(n-1\) round game. This of course by iterative/same logic results in it further becoming an  \(n-2\) round game, going all the way back to when it is only 2 rounds!

Hence you cannot play more than 2 rounds and do better.