Silent Duelsвђ”constructing The Solution Part 2 Вђ“ Math В€© Programming -
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
such that the total probability of action equals 1. In a simple linear case where , the optimal strategy is to fire at exactly . 2. The Programming Challenge: Discretizing the Continuous A(t)=∫at1P(x)dxcap A open paren t close paren equals
, the probability of hitting is 100%. We use this boundary condition to calculate the "Expected Value" (EV) of firing at tn−1t sub n minus 1 end-sub The Programming Challenge: Discretizing the Continuous , the
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 Copied to clipboard 4
Constructing this solution is a masterclass in . It’s used in:
, which represents the probability of hitting a target at time goes from 0 to 1). To find the optimal time to fire ( t*t raised to the * power
