Poker Rank Probabilities in three cards poker game -- single deck vs. multiple decks. Poker is a game that plays with a single deck of cards.  However, for the purpose of learning, I want to know the probability of poker hands in multiple decks.  To simplify the problem, let's take a popular three-card poker game, and calculate probabilities of a pair and three-of-a-kind ranks for a single deck, and 8 decks.
Quasi answered my question and provided formulas for a pair and a three-of-a-kind probabilities.  I then followed Quasi's approach, and calculated 3-card, single deck combinations for other ranks (see below). But when I applied an 8 deck factor to the single deck equations the results doesn't look right. Are the equations below correct for single deck, and what modifications are needed for 8-deck?
royal flush equation: COMBIN(4,1)
straight flush equation: COMBIN(12,1)*COMBIN(4,1) - COMBIN(4,1) 
straight equation: COMBIN(12,1) * (4 ^ 3 - 4) 
flush equation  : COMBIN(4,1)*(COMBIN(13,3) - 12) 
 A: Assume $k$ standard decks.

Probability of a pair:
$$\large{\frac
{
\binom{13}{1}\binom{4k}{2}\binom{48k}{1}
}
{
\binom{52k}{3}
}}
$$
Explanation:


*

*Choose the rank for the pair: $\binom{13}{1}$ choices.

*Choose the two cards for that rank: $\binom{4k}{2}$ choices.

*Choose the non-pair card: $\binom{48k}{1}$ choices.


Probability of three-of-a-kind:
$$\large{\frac
{
\binom{13}{1}\binom{4k}{3}
}
{
\binom{52k}{3}
}}
$$
Explanation:


*

*Choose the rank for the triple: $\binom{13}{1}$ choices.

*Choose the cards for that rank: $\binom{4k}{3}$ choices.


Update:

Answering the additional questions in the OP's edit . . .

Probability of a royal flush:
$$\large{\frac
{\binom{4}{1}\binom{k}{1}^3}
{\binom{52k}{3}}
}
$$
Explanation:


*

*Choose the suit: $\binom{4}{1}$ choices.

*Choose the cards for that suit: $\binom{k}{1}^3$ choices.


Probability of a straight flush (but not royal):
$$\large{\frac
{\binom{4}{1}\binom{11}{1}\binom{k}{1}^3}
{\binom{52k}{3}}
}
$$
Explanation:


*

*Choose the suit: $\binom{4}{1}$ choices.

*Choose the rank for the high card: $\binom{11}{1}$ choices.

*Choose $3$ cards, one for each rank: $\binom{k}{1}^3$ choices.


Probability of a straight (but not a flush):
$$\large
{
\frac
{\binom{12}{1}\left(\binom{4k}{1}^3-\binom{4}{1}\binom{k}{1}^3\right)}
{\binom{52k}{3}}
}
$$
Explanation:


*

*Choose the rank for the high card: $\binom{12}{1}$ choices.

*Choose $3$ cards, one for each rank: $\binom{4k}{1}^3$ choices.

*Subtract the count for the flushes:  $\binom{4}{1}\binom{k}{1}^3$ choices.


Probability of a flush (but not a straight):
$$\large
{
\frac
{\binom{4}{1}\left(\binom{13k}{3}-\binom{12}{1}\binom{k}{1}^3\right)}
{\binom{52k}{3}}
}
$$
Explanation:


*

*Choose the suit: $\binom{4}{1}$ choices.

*Choose $3$ cards from that suit: $\binom{13k}{3}$ choices.

*Subtract the count for the straights:  $\binom{12}{1}\binom{k}{1}^3$ choices.

