Expected flips to get 2 queens All the red cards are removed from a pack of 52 cards. Remaining cards are shuffled well and flipped one by one without replacement. What is the expected number of cards flips we need to do to get both the queens?
 A: For each of the 24 non-queen cards define  $A_i$ to be  the event 
that card $i$ appears after both queens. The position of the second queen is given by $26-\sum_{i=1}^{24}{\bf 1}_{A_i}$, and so its expected position is 
$$\mathbb{E}\left(26-\sum_{i=1}^{24}{\bf 1}_{A_i}\right)=26-\sum_{i=1}^{24}
\mathbb{P}(A_i)=26-24\left({1\over 3}\right)=18. $$
A: Assume all permutations of the $26$ cards have the same probability. Further, for any choice of the positions of the queens in the decks, there are $24!$ equiprobable permutations. So the mean does not change if you consider only the $26\choose 2$ positions of the queens.
And the number of cards to flip to get the two queens is the position of the second queen.
The mean number of card flips needed is thus, with $n=26$:
$$\frac{1}{26\choose 2}\sum_{1\leq i\lt j\leq 26}j=\frac{2}{n(n-1)}\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}j=\frac{2}{n(n-1)}\sum_{i=1}^{n-1}\left(\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\right)\\=n+1-\frac{2}{n(n-1)}\sum_{i=1}^{n-1}{i+1\choose2}=n+1-\frac{2}{n(n-1)}{n+1\choose3}\\=n+1-\frac{2n(n-1)(n+1)}{6n(n-1)}=n+1-\frac{n+1}{3}=27-9=18$$
A: There is a distribution for problems of this type, it is called the Negative Hypergeometric Distribution. I add it for good measure.
https://en.wikipedia.org/wiki/Negative_hypergeometric_distribution
The formula given is:
$\frac{\text{winners} (\text{total} \text{cards}-\text{winners})}{\text{winners}+1}+\text{winners}$
winners = 2
totalcards=26
$\frac{2 (26-2)}{3}+2 = 18$
