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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?

identicon
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  • 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 '19 at 21:06

2 Answers2

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Visualize the 26 red cards and the 27 "gaps" between them. You want the expected number of black cards in the first four gaps (before the 4th red card), and then you need to add 4 (since you need to count the first 4 red cards).

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.

angryavian
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  • Oops, I meant to write "4 red cards IN A ROW" --- sorry! This is still the right answer for four red cards in total, so I'll mark it as such and make a new question. Thanks! – identicon Sep 29 '19 at 20:58
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If you ignore the the finer details, ie the red cards being depleted you have 50 % chance to draw one red card, 25% to draw two, 12.5 to draw three, ang 6.25 to draw 4 red card in a row.

This means that you need to draw on average 16, in reality around 17 times to get 4 red in a row from the top of the deck

Paul
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