A deck of 52 cards consists of four suits of 13 cards each.…

A deck of $52$ cards consists of four suits of $13$ cards each. The deck is thoroughly shuffled, so that every card is equally likely to be drawn. Six cards are drawn. Describe the sample space and:
$1^{\circ}$ find the probability that the king of spades is among the drawn cards;
$2^{\circ}$ find the probability that all four suits are represented among the drawn cards;
$3^{\circ}$ what is the minimum number of cards that must be drawn so that the probability that two of them have the same rank is greater than $\tfrac12$?

Step 1: The sample space

An elementary outcome is a set of $6$ different cards (order does not matter). All of them are equally likely:

\[ |\Omega| = \binom{52}{6} = 20\,358\,520. \]

Step 2: $1^{\circ}$ — the king of spades

Hands containing the king of spades: the other $5$ cards are chosen from the remaining $51$:

\[ P_1 = \frac{\binom{51}{5}}{\binom{52}{6}} = \frac{6}{52} = \boxed{\frac{3}{26}\approx 0.115.} \]

Intuition: we take $6$ of the $52$ cards, so each particular card is taken with probability $6/52$.

Step 3: $2^{\circ}$ — all four suits

Let $S_i$ be the event "suit $i$ is missing". A hand missing a given suit uses only the other $39$ cards; a hand missing two given suits uses only $26$ cards; missing three suits, only $13$ cards. By inclusion–exclusion,

\[ \big|S_1+S_2+S_3+S_4\big| = 4\binom{39}{6} - 6\binom{26}{6} + 4\binom{13}{6} = 13\,050\,492 - 1\,381\,380 + 6\,864 = 11\,675\,976. \]

(All four suits cannot be missing, since we hold $6$ cards.) Hands with all four suits:

\[ 20\,358\,520 - 11\,675\,976 = 8\,682\,544, \qquad \boxed{P_2 = \frac{8\,682\,544}{20\,358\,520}\approx 0.4265.} \]

Check by direct counting

With $6$ cards and all $4$ suits present, the suit counts are either $3,1,1,1$ or $2,2,1,1$:

\[ 4\binom{13}{3}13^3 + 6\binom{13}{2}^2 13^2 = 2\,513\,368 + 6\,169\,176 = 8\,682\,544. \ \checkmark \]

Step 4: $3^{\circ}$ — two cards of the same rank

There are $13$ ranks, each with $4$ cards. If $k$ cards are drawn, they all have different ranks when we choose $k$ different ranks and one of the $4$ cards of each:

\[ P(\text{all ranks different}) = \frac{\binom{13}{k}\,4^k}{\binom{52}{k}}, \qquad P(\text{some rank repeated}) = 1 - \frac{\binom{13}{k}\,4^k}{\binom{52}{k}}. \]

With $5$ cards the probability of a repeated rank is still just below $\tfrac12$; with $6$ cards it is about $0.65$.

\[ \boxed{\text{The minimum number of cards is } 6.} \]