A deck of 36 cards is randomly divided into two equal parts.…

A deck of $36$ cards is randomly divided into two equal parts. What is the probability that both parts will contain the same number of red and black cards?

Step 1: The deck

A 36-card deck has $4$ suits of $9$ cards: two red suits (hearts, diamonds) and two black suits (spades, clubs). So there are $18$ red and $18$ black cards.

Step 2: All outcomes

A random split is completely described by the set of $18$ cards that land in the first half (the rest form the second half). All such sets are equally likely:

\[ \binom{36}{18} = 9\,075\,135\,300. \]

Step 3: What does the event mean?

"Both parts have equally many red and black cards." Each half has $18$ cards, so this means each half has $9$ red and $9$ black. It is enough to require it for the first half: if the first half has $9$ red and $9$ black, the second half gets the remaining $18-9=9$ red and $18-9=9$ black.

Step 4: Favourable outcomes

Choose $9$ of the $18$ red cards and $9$ of the $18$ black cards for the first half:

\[ \binom{18}{9}\cdot\binom{18}{9} = 48\,620^2 = 2\,363\,904\,400. \]

Step 5: The probability

\[ \boxed{P = \frac{\binom{18}{9}^2}{\binom{36}{18}} = \frac{2\,363\,904\,400}{9\,075\,135\,300} \approx 0.2605.} \]

So only about one split in four is perfectly balanced — even though $9$–$9$ is the most likely single result.

Step 6: The whole picture

Let $R$ be the number of red cards in the first half. By the hypergeometric formula (Problem 37), $P(R=j) = \binom{18}{j}\binom{18}{18-j}/\binom{36}{18}$:

The value $9$ is the most likely, but values near it together are much more likely.