# Draw from a standard 52-card deck until you get four red cards. What is the expected number of draws?

I tried considered an easier case: what is the expected number of draws you need until drawing a single red card? The solution I came up with for this case is similar to this solution to another question. For any black card, drawing that card before any of the 26 red cards is $$\frac{1}{27}$$. Over all black cards, this is an expected value of $$\frac{26}{27}$$ black cards drawn before reaching a red card.

Is there a way that can extend this strategy? I'm trying visualizing 26 red cards and fill in the 26 black cards inbetween them.

EDIT: Wasn't clear enough in the wording, I meant to ask for "4 consecutive red cards". I've made a new question for this: Draw from a standard 52-card deck until you get four red cards in a row. What is the expected number of draws?

• What if there aren't four consecutive red cards in the deck? Do you just want the expected number of draws given that there are? – mjqxxxx Sep 29 at 21:06

Each black card is equally likely ($$1/27$$) to be in any of the 27 gaps between the red cards. In particular, the probability of a black card being in the first four gaps is $$4/27$$. Thus, the expected number of black cards in the first four gaps is $$26 \cdot \frac{4}{27}$$.
Then, add $$4$$ to account for the red cards.