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: I also wanted to add that I found this as a practice interview question, so I don't think the intended solution is meant to be very strenuous, i.e. there is some sort of trick.

Note that this is a fixed version of Draw from a standard 52-card deck until you get four red cards. What is the expected number of draws?, where I wasn't clear that I was looking for four consecutive red crads.

  • 4
    $\begingroup$ What if you don't ever see four red cards in a row? How do you define "number of draws" in that case? $\endgroup$
    – angryavian
    Commented Sep 29, 2019 at 21:03
  • 1
    $\begingroup$ What happens if you never get $4$ red cards in a row? Does that count as $52$ draws or $\infty$ draws? $\endgroup$
    – saulspatz
    Commented Sep 29, 2019 at 21:03
  • $\begingroup$ It wasn't defined outright for me (I found this question online). Is it possible to calculate a conditional expected value given that four consecutive red cards were drawn? $\endgroup$
    – identicon
    Commented Sep 29, 2019 at 21:09
  • $\begingroup$ "in a row" mean that the four cards are taking one after the other, without no other kind of cards in between? $\endgroup$
    – Masacroso
    Commented Sep 29, 2019 at 21:33
  • 1
    $\begingroup$ Are cards being returned to the pack? $\endgroup$
    – user502266
    Commented Sep 29, 2019 at 21:43

1 Answer 1


Let $A_n$ the event that on $n$ draws no one is red. Then

$$ \Pr[A_n]=\begin{cases} \frac{52-26}{52}\cdot \frac{52-26-1}{52-1}\cdots \frac{52-26-(n-1)}{52-(n-1)}, & n\leqslant 52-26\\ 0, & \text{ otherwise } \end{cases} $$

Then, after no drawing red cards in $n-4$ cards then you need to get four consecutive red cards, so if $B_n$ is the event of drawing $n$ cards where just the last four are red then

$$ \Pr[B_n]=\begin{cases} \Pr[A_{n-4}]\frac{26\cdot 25\cdot 24\cdot 23}{(52-(n-4))(51-(n-4))(50-(n-4))(49-(n-4))}, & n-4\leqslant 52-26\\ 0, & \text{ otherwise } \end{cases} $$

Using notation for falling factorials and Iverson brackets we get the compact expression

$$\Pr[B_n]=\frac{(52-26)^{\underline{n-4}}\cdot26^{\underline{4}} }{52^{\underline{n-4}}\cdot(52-(n-4))^{\underline{4}} }[n\leqslant 30]$$

And the expectation will be $\sum_{n>0}n\Pr[B_n]$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .