A deck of cards

A deck has fifty-two cards: four suits, thirteen cards in each. One suit is hearts, so the hearts form a pile of thirteen. There are also four aces, one in each suit.

A deck: $52=4\times 13$ cards. Hearts $H$ with $|H|=13$, aces $K$ with $|K|=4$, one in each suit.

$H$ hearts, $K$ aces: $52=4\times 13,|H|=13,|K|=4$.

Draw one card without looking. The chance that it is a heart is thirteen in fifty-two, which is one in four. The chance that it is not a heart is thirty-nine in fifty-two, which is three in four.

One card, unseen: $P(H)=\frac{13}{52}=\frac{1}{4}$. Not a heart: $P(\neg H)=\frac{39}{52}=\frac{3}{4}$.

$P(H)=\frac{13}{52}=\frac{1}{4},P(\neg H)=\frac{39}{52}=\frac{3}{4}$.

What is the chance of a heart or an ace? Careful: thirteen plus four is seventeen, but the two piles together have only sixteen cards. The ace of hearts belongs to both piles and must be counted once. So the chance is sixteen in fifty-two, which is four in thirteen.

Heart or ace, $H\cup K$? Careful: $13+4=17$ but $|H\cup K|=16$, because the ace of hearts lies in $H\cap K$ and $|H\cap K|=1$, so $|H\cup K|=13+4-1$. Hence $P=\frac{16}{52}=\frac{4}{13}$.

$|H\cap K|=1\Rightarrow |H\cup K|=13+4-1=16,P(H\cup K)=\frac{16}{52}=\frac{4}{13}$.

Now draw two cards one after the other, keeping the first. Both aces? Four in fifty-two for the first, then three in fifty-one for the second. Together that is one in two hundred and twenty-one. A hand of two aces is rarer than it feels.

Two cards, the first kept: both aces has chance $\frac{4}{52}\times \frac{3}{51}=\frac{1}{221}$.

$\frac{4}{52}\times \frac{3}{51}=\frac{1}{221}$.