Expected number of cards drawn to get two consecutive aces

Question

Here is a question from my probability textbook:

A person draws cards one by one from a pack and replaces them till he has drawn two consecutive aces. How many cards may he expect to draw?

I'm not sure how to even begin this problem. Any hints (not complete solutions) in the correct direction would be well-appreciated.

Answer 1

There are many methods to solve this question. You can use Markov Chain or Martingale if you know them. The method that doesn't need any high level math is recursion. Try to figure out the meaning of the following graph.

Suppose the expected number you are seeking is $\mu$. Then

$$\mu = \frac{12}{13}(1+\mu)+\frac{12}{13^2}(2+\mu)+\frac{1}{13^2}\cdot 2$$

Answer 2

Since he replaces the cards after drawing them, each draw is independent of the others and so $P(ace) = \frac{1}{13}$ for every draw. So we have a process with a constant probability of success, and we're trying to find the number of attempts to get to 2 successes. Is this familiar at all?

Answer 3

I am going to assume it is that the card drawn is then replaced randomly

We have the following:

$P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$ and $P(\text{not ace}) = \frac{48}{52} = \frac{12}{13}$

We can condition on the last drawn card. That is, let $X$ be the number of draws and we have

$\displaystyle \mathbb{E}[X] = \frac{1}{13}\mathbb{E}[X \mid \text{last draw was ace}]+ \frac{12}{13}\mathbb{E}[X \mid \text{last draw was not ace}]$

Can you take it from here?