One step of depth for the worst total

If depth $b$ is one deeper than depth $a$, then the worst totals satisfy $k \cdot \mathrm{worstTotal}(X,b) \le (k+1)\cdot \mathrm{worstTotal}(X,a)$, where $k+1$ is the size of a competing group at depth $b$.

The argument avoids any appeal to sorting or to binomial counting. Take a group $S$ of $k+1$ states attaining the worst total at depth $b$. Deleting any single member leaves a group of $k$ states, which is a legitimate competitor at depth $a$; hence for every $i \in S$ we get $\mathrm{worstTotal}(X,a) \ge \mathrm{worstTotal}(X,b) + X_i$. Summing these $k+1$ inequalities and using that the total payoff on $S$ is exactly $-\mathrm{worstTotal}(X,b)$ collapses the sum to the stated bound.

As throughout CoherentRisk, the development counts states and refers to no probability measure, so the depth is an integer count rather than a confidence level.