Ace and King

What is the expected number of draws (without replacement) from a standard 52-card deck needed to see at least one ace and at least one king?

Answer

The $$8$$ special cards (4 aces + 4 kings) split the $$44$$ non-special cards into $$9$$ gaps, so the average number of cards per special card is

\frac{44}{9}+1=\frac{53}{9}.

Thus, if $$n$$ is the number of special cards needed until we stop, then

\mathbb{E}[T]=\frac{53}{9}\,\mathbb{E}[n].

Now we compute $\mathbb{E}[n]$ . We stop when we have seen at least one ace and at least one king. WLOG assume the first special card is an ace (the king case is symmetric). Then among the remaining $$7$$ specials there are $$4$$ kings and $$3$$ aces, and we are looking for the first king.

The $$4$$ kings split the $$3$$ aces into $$4+1=5$$ parts, so the average number of aces before the first king is

\frac{3}{5}.

Hence the expected number of additional special cards after the first ace until we see a king is

1+\frac{3}{5}=\frac{8}{5},

and therefore

\mathbb{E}[n]=1+\frac{8}{5}=\frac{13}{5}.

Finally,

\boxed{\mathbb{E}[T]=\frac{53}{9}\cdot\frac{13}{5}=\frac{689}{45}\approx 15.31.}