Consider a machine, which may or may not have a defect, and the probability q that this machine is defective is unknown. In order to determine whether the machine is defective, it is tested. On each test, the defect is found with probability p, if it has not been found yet. Performing n tests costs cn dollars and there is a fine of 1 dollar if there is a defect and it is not found on the tests. When should we stop testing, in order to minimize the cost? This problem is treated in a minimax setting: we try to find a strategy that works well, even for ‘bad’ qs. It turns out that the minimax optimal stop rule can be unexpectedly complicated. For example, if p = 1/2 and cn = cn = 0.25n, then the optimal rule is to start by performing one test. If a defect is found we stop, otherwise we perform a second test. If a defect is found, then again we stop, else toss a coin and stop if this shows heads. If we still have not stopped, a third and last test is performed.
