Sue picks a number from 0 to 3.
Tom asks questions about the number, with yes/no answers. For example, "Is it odd" or "Is it 3?"
If Sue picked X, she is allowed to lie at most X times. For example, if she picked zero, she must always tell the truth. If she chose any other number, she can decide each question whether to lie, until she runs out of lies, after which she must tell the truth.
Tom wants to minimize the average number of questions. Sue wants to maximize it.
I think Sue's best strategy is random. Sometimes choose $0$, sometimes tell the truth when she doesn't have to. What is her strategy, and how many questions does Tom need on average?
Wei Hua Huang posed this in rec.puzzles newsgroup twenty years ago. I think we solved the worst-case number of questions, but not the average.
To compare, suppose Sue can pick 0 or 1, and may lie once if she picks 1.
Tom asks 'Is it 1?'. Suppose Sue says no. If she picked 1, then she has used up her lie. So, whether Sue picked 0 or 1, she has no lies left. Tom asks 'Is it 1?' again, and will get a correct answer.
Suppose Sue says yes to Tom's first question. She can't say yes if she picked $0$, so Tom now knows Sue picked $1$.
So Sue's best strategy is to say no, and it will take Tom two guesses.