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The easiest way to calculate the probability that at least two cards are of the same suit is to subtract the probability that they are all of different suits from $1$. Since there are only four suits, the answer must be $1$ whenever the number of cards drawn is at least five since we would otherwise need more than four suits. For a hand with four cards, the probability that at least two cards are of the same suit is \begin{align*} \Pr(\text{at least two cards of the same suit}) & = 1 - \Pr(\text{all cards are of different suits})\\ & = 1 - \frac{52 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49} \end{align*}

At least two cards of consecutive ranks: In poker, an ace can be treated as the card of highest rank or lowest rank, but not both. The number of two card straights is $$13 \cdot 4^2$$ since there are $13$ ways to select the low card and four ways to select the suit of each rank.

The easiest way to calculate the probability that at least two cards are of the same suit is to subtract the probability that they are all of different suits from $1$. Since there are only four suits, the answer must be $1$ whenever the number of cards drawn is at least five since we would otherwise need more than four suits. For a hand with four cards, the probability that at least two cards are of the same suit is \begin{align*} \Pr(\text{at least two cards of the same suit}) & = 1 - \Pr(\text{all cards are of different suits})\\ & = 1 - \frac{52 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49}. \end{align*}

At least two cards of consecutive ranks:

In poker, an ace can be treated as the card of highest rank or lowest rank, but not both. The number of two card straights is $13 \cdot 4^2$, since there are $13$ ways to select the low card and four ways to select the suit of each rank.

• exactly two cards with consecutive ranks of the same suit
• exactly three cards with consecutive ranks of the same suit
• two disjoint pairs of two cards with consecutive ranks of the same suit (the suits of the pairs may or may not be the same)
• four cards of consecutive ranks of the same suit

• exactly two cards with consecutive ranks of the same suit
• exactly three cards with consecutive ranks of the same suit
• two disjoint pairs of two cards with consecutive ranks of the same suit*
• four cards of consecutive ranks of the same suit

*Note that it is possible for the pairs to have the same ranks ($5\clubsuit, 6\clubsuit, \color{red}{5\diamondsuit}, \color{red}{6\diamondsuit}$), have the same suit ($7\spadesuit, 8\spadesuit, J\spadesuit, Q\spadesuit$), or neither ($\color{red}{9\heartsuit}, \color{red}{10\heartsuit}, A\spadesuit, 2\spadesuit$).

However, we cannot count an Ace as both the lowest and highest card. Such configurations would have three green balls among the nine spaces between successive green balls, so there are $$\binom{9}{3}$$ impermissible ways to select the ranks so that no two of them are consecutive, giving $$\binom{10}{5} - \binom{9}{3}$$ ways to select the ranks of the five cards so that no two of them are consecutive.

For each rank, there are four possible suits. Hence, the probability that at least two cards are of the same rank is \begin{align*} \Pr(\text{at least two consecutive ranks}) & = 1 - \Pr(\text{no consecutive ranks})\\ & = 1 - \frac{\left[\dbinom{10}{5} - \dbinom{9}{3}\right]4^5}{\dbinom{52}{5}} \end{align*}

However, we cannot count an Ace as both the lowest and highest card. Such configurations would have three green balls among the eight spaces between successive blue balls, so there are $$\binom{9}{3}$$ impermissible ways to select the ranks so that no two of them are consecutive, giving $$\binom{10}{5} - \binom{8}{3}$$ ways to select the ranks of the five cards so that no two of them are consecutive.

For each rank, there are four possible suits. Hence, the probability that at least two cards are of the same rank is \begin{align*} \Pr(\text{at least two consecutive ranks}) & = 1 - \Pr(\text{no consecutive ranks})\\ & = 1 - \frac{\left[\dbinom{10}{5} - \dbinom{8}{3}\right]4^5}{\dbinom{52}{5}} \end{align*}