We iterate through the time steps until we find the point where the EV of firing equals the EV of waiting. 3. Implementation Logic (Pseudocode)
A(t)=∫at1P(x)dxcap A open paren t close paren equals integral from a to t of the fraction with numerator 1 and denominator cap P open paren x close paren end-fraction d x The goal is to find the lower bound We iterate through the time steps until we
def solve_silent_duel(accuracy_func, steps=1000): # Backward induction to find the 'tipping point' for t in reversed(range(steps)): prob_hit = accuracy_func(t / steps) # If the risk of the opponent hitting us next # is higher than our current hit chance, we wait. if prob_hit >= calculated_threshold(t): return t / steps Use code with caution. Copied to clipboard 4. Why This Matters if prob_hit >= calculated_threshold(t): return t / steps
In the final part of this series, we will look at , where one player is faster, but the other is more accurate. In a symmetric duel, both players share the
In a symmetric duel, both players share the same accuracy function,