with a probability $P$ close to 1.0. Let’s define by $R_{.99}$ the number of repetitions needed to find the optimal solution with a certainty of 99%. $R_{.99}$ clearly depends on the probability $p_s$ of measuring this desired optimal solution after a single repetition, which in turn depends on the the annealing time, $t_a$, specified as input for each repetition or cycle, i.e., $p_s = p_s(t_a)$. Suppose you request a number of readouts $N_r$ from the quantum processor. Then $p_s(t_a)$ can be estimated as the ratio $n_{gs}/N_r$, with $n_{gs}$ denoting the number of occurrences of the ground states (corresponding to the desired solutions), at the specified $t_a$. For example, the success probability $p_s(t_a)$ can be determined from the histograms in Fig. \[fig:cb21f6\]a by dividing the number of occurrences of the optimal solutions with $E=6$ by the total number of runs $N_r$, set to 50,000 in this particular instance in order to obtain at least some solutions.

The number of repetitions needed to reach $R_P$ can be estimated as follows. The probability of not obtaining the lowest configuration after $R_P$ runs is $(1-p_s)^{R_P}$. Therefore, the probability of measuring the ground state at least once in these $R_P$ experimental runs is given by 