Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 4H / ii

There is a genuinely different optimal procedure. Exchange the codewords assigned to items three and four. Both have code length four, so each item's length, and hence the expected number of tests $173/55$, remains unchanged. The first tested subset now contains item three instead of item four, so this changes the tested groups rather than merely reversing the names of the two test outcomes. The new prefix code still identifies every item uniquely and retains the minimum Huffman coding expectation.