removed from $\Omega_{()}$ and the procedure continues. Define $\Omega_Y$ as the set of all thresholds defined by the above pseudo-code. From the construction of the above code, we see that, $${\mathsf C}(Y_{N-1},Y_N) = \arg \max_{y \in \Omega_Y} \{ Pr(\Omega_{y}) : Y_{N-1} \leq y < Y_N\}$$ For this reason, we call OSA as the maximal probability allocation code (MPA). The following theorem from [@cohen-etal-ternaryfeedback] identifies the exact threshold of OSA for a given range $(y_{min},y_{max}]$.

Given $N$ users and thresholds $(y_{min}, y_{max}]$ (i.e., $y_{min} \leq Y_{N-1} \leq Y_N \leq y_{max}$), the $y$ that maximizes the probability of success in the interval $(y_{min}, y_{max}]$ is the unique stationary point of $(y_{max} - y) (y^{N-1} - y_{min}^{N-1})$.

**Remarks**

1.  For any $N$, and with $y_{min} = 0, y_{max} = 1$, the above expression becomes, $(1 - y)(y^{N-1}) = y^{N-1} - y^N$. The expression is maximized at $y = 1 - \frac{1}{N}$. Hence, for any $N$, $y_1 = 1 - \frac{1}{N}$.

2.  As an example, for $N = 2$, repeating the above procedure will yield us $y_1 = 0.5, y_2 = 0.75, y_3 = 0.25, y_4 = 0.875, \cdots$. Note that the above values are in fact the thresholds reported in Table \[tab:osa\_two\_users\].

3.  