Here's the description of the quiz:
- A player has to answer $w$ questions right in a row in order to win
- There is a pool of $N$ questions
- Every round a question is chosen from the pool uniformly randomly (with replacement)
- If the player has already seen that question, he will answer it correctly (lets assume his memory is perfect)
- If he hasn't seen it before, he has probability $p$ of answering it correctly.
- Regardless of how the player answered, he will be given the correct answer after each question, so he will learn that question for the future.
Question: What is the expected length (number of rounds until the player wins) of the game?
My try: For $N=1$ either the player guesses the question correct first time or he makes error on the first. After that all question (the only one) are seen and the player just has to answer additional $w-1$ (in the first case) or $w$ (in the latter) times. Hence we get
$$pw + (1-p)(w+1) = w+1-p.$$
Can we somehow reduce the games with bigger $N$ to smaller ones? Once the player has seen the first question, we only consider the pool to be the other $N-1$ questions and set the player's probability of knowing the answer to $\frac{1}{N} + \frac{N-1}{N} p$ (either the chosen question will be the excluded question that is already considered known to the player or it's one of the others and then player guesses with probability $p$). Then either the player makes a clear victory from that on or makes an error at some point $2,3,\dots, w$ and has to start from the bottom again and the expectation can be calculated in terms of previous ones and itself.
Now that I think of it, I'm not sure the guessing probability adjusting will work since if it comes to the guessing, it is already a fact that the question wasn't seen before.