Crossposted on MathOverflow.
Suppose a quality inspector is inspecting $b$ black gadgets amongst which $d_b$ are known to be defective and $w$ white gadgets amongst which $d_w$ are known to be defective. The gadgets come down along an assembly line one by one. As each gadget passes, the inspector observes its color, and he chooses to let the gadget pass or use a device to detect whether the gadget is defective. But he can only use the device a total of $n$ times. What is the optimal stopping rule to use the inspecting device to maximize the expected number of defective gadgets found.
Suppose at each pass the number of black gadgets already inspected by the device is $i_b$, amongst them $f_b$ are detected to be defective. Then the probability of this current black gadget detected to be defective is $p_b=\frac{d_b-f_b}{b-i_b}$. The symmetrical probability holds for the white gadgets.
I have a conjecture for the explicit solution, which is a greedy algorithm, as follows and am seeking a proof.
At a pass of the gadget, without loss of generality, suppose the current gadget which has not yet been inspected is black. Suppose there are $n_b$ black including the current one and $n_w$ white gadgets left. Suppose $i_b$ black and $i_w$ white gadgets have been inspected, amongst which $f_b$ black and $f_w$ have been found to be defective. If $p_b\ge p_w$, the inspector inspect the current black gadget with the device. Otherwise, the inspector let the current black gadget pass without inspection.
I have set up the dynamic programming formulation but fail to see either the proof or a counterexample to my conjecture.