If we end the game as soon as the score exceeds \(\sqrt{2}-1 \simeq 0.4142\), the expected value of the final score will be approximately \(0.6268\). This is the maximum score that any strategy can guarantee in expectation.
Obviously, the optimal strategy is of the following form: We continue playing the game as long as the current sum hasn’t reached a certain threshold. As soon as it reaches that value, we stop. To determine the optimal threshold, let us assume that the current value of \(X\) is \(q\) and we want to decide whether or not to add another random number. If the next random number is between 0 and \(1-q\), then our score in the next step will be higher than now, otherwise it will be zero. So the average score with one move will be:

\[\int_{t=0}^{1-q} (q+t)\cdot d_t = (1-q^2)/2.\]

Therefore, for adding another random number to be profitable, this value must be greater than \(q\). Thus, for the next move to be reasonable we should have: \(\frac{1-q^2}{2} > q\) and thus \(q^2+2q-1 < 0\) which implies \(q < \sqrt{2}-1\).

Thus in the optimal strategy, we end the game immediately after \(X\) reaches \(\sqrt{2}-1\) (for \(X = \sqrt{2}-1\), the expected value of the score is the same whether or not we continue. In the following, we assume that we continue and generate a new random number in this case).

Next we determine the expected value of the final score with the above strategy when we start from \(x \leq \sqrt{2}-1\). We denote this value by \(f(x)\). Since we only consider the cases that the game has not ended yet, we have \(x \leq \sqrt{2}-1\). Starting from \(X = x\) and adding a new random number, \(X\) will turn into a new number \(t\) in the range \([x, 1+x]\). To determine \(f(x)\), we consider three cases:

• If the new value \(t\) is less than or equal to \(\sqrt{2}-1\), according to the above strategy, we continue the game, so in this case, the expected value of our score will be \(f(t)\).

• If the new value \(t\) is between \(\sqrt{2}-1\) and \(1\), according to the above strategy, we will end the game and our score will be \(t\).

• If the new value \(t\) is greater than 1, the game will be over and our score will be zero.

Based on the above, we can determine function \(f(x)\) for \(x \leq \sqrt{2}-1\) as the following recursive formula:

\[f(x) = \int_{t=x}^{\sqrt{2}-1} f(t)\cdot d_t + \int_{t=\sqrt{2}-1}^1 t\cdot d_t + \int_{t=1}^{1+x} 0\cdot d_t.\]

If we take the derivative of both sides, we get the differential equation \(f'(x)+f(x)=0\). Also, for \(x=\sqrt{2}-1\), we have \(f(\sqrt{2}-1)=\sqrt{2}-1\). Solving this differential equation with the boundary condition \(f(\sqrt{2}-1) = \sqrt{2}-1\), we get the following solution for \(f\):

\[f(x) = e^{-x+\log(\sqrt{2}-1)+\sqrt{2}-1}.\]

Thus the expected value of the game starting from a value of zero for \(X\) is \(f(0) = e^{\log(\sqrt{2}-1)+\sqrt{2}-1} \simeq 0.6268\).