What is the probability of getting a King, Queen, Jack and Ace of same colour, and a Single pair of same colour in a hand of six cards?

Question

What is the probability of getting a King, Queen, Jack and Ace of same colour, and a single pair of same colour in a hand of six cards?

I am working on a casino game however to determine house edge and other factors the probability of winning outcomes must be determined. I am really struggling with this problem as there are so many factors to consider.

I started by considering that there are $16$ royal cards, thus $\binom{16}{1}$. Say a red king is drawn then a red queen, jack or ace can be drawn so $$\binom{6}{1}\binom{4}{1}\binom{2}{1}$$ Then for the pairs it is $\binom{48}{1}$ and the last card must be the same as the previous card so one option is available. This is giving the equation $$\frac{\dbinom{16}{1}\dbinom{6}{1}\dbinom{4}{1}\dbinom{2}{1}\dbinom{48}{1}}{\dbinom{52}{6}}$$ However, I have a feeling this is incorrect.

Answer 1

Choose 1 of 2 colors.
Choose one Ace of the chosen color.
Choose one King of the chosen color.
Choose one Queen of the chosen color.
Choose one Jack of the chosen color.

It sounds like the pair can be either color but both cards in the pair must be the same color.

Lemmon points out an error in my calculation.

If the remaining pair is of the same color as the royals, there are 9 values that this pair could have and if it is of the opposite color there are 13 values.

$\frac {{2\choose 1}{2\choose 1}{2\choose 1}{2\choose 1}{2\choose 1}(9+13)}{{52\choose 6}}$

Answer 2

There are two cases to consider, depending on whether the pair is from among the ranks ace, king, queen, jack or not.

The single pair is from a lower rank: There are two ways to choose the colour of the ace, king, queen, and jack. For each of these four ranks, there are two ways to choose a suit in that colour. There are nine ways to select a lower rank, two ways to select the colour for the pair of that rank, and one way to select both suits of that colour. Hence, there are $$\binom{2}{1}\binom{2}{1}^4\binom{9}{1}\binom{2}{1}\binom{2}{2}$$ such hands.

The single pair is from the ranks ace, king, queen, and jack: Notice that the pair must be of the same colour as the ace, king, queen, and jack of the same colour, for otherwise we would have three of a kind. There are four ways to select the rank of the pair from among the aces, kings, queens, and jacks, two ways to choose the colour of the pair, and one way to choose both suits of that colour for that rank. Choosing the colour of that pair also determines the colour of the other three ranks among the aces, kings, queens, and jacks. For each of those three ranks, there are two ways to select the suit of that colour. The remaining card must be selected from among the $9$ lower ranks. There are four ways to select a suit for the card of that rank. Hence, there are $$\binom{4}{1}\binom{2}{1}\binom{2}{2}\binom{2}{1}^3\binom{9}{1}\binom{4}{1}$$ such hands.

Since there are $\binom{52}{6}$ ways to select six of the $52$ cards in the deck and the two favorable cases are mutually exclusive and exhaustive, the probability of selecting an ace, a king, a queen, and a jack of the same colour and a single pair of the same colour is $$\frac{\dbinom{2}{1}\dbinom{2}{1}^4\dbinom{9}{1}\dbinom{2}{1}\dbinom{2}{2} + \dbinom{4}{1}\dbinom{2}{1}\dbinom{2}{2}\dbinom{2}{1}^3\dbinom{9}{1}\dbinom{4}{1}}{\dbinom{52}{6}}$$