Given a 52 card deck of playing cards, I am trying to calculate the probability of the following in a 4 person game.
I am the dealer, the person to my left (player 1) is to be dealt the first card and this goes around the circle.
We are dealing 6 cards in total to each player, one by one around the group.
What is the probability of the person to my left getting 9(suit), 10(suit), J(suit), Q(suit), K(suit), A(suit), with me then declaring that the suit "of choice" (a rule in the game) is the suit that they were dealt (let's assume 1/4)
I started off with:
$${\binom{6}{52}\binom{5}{48}\binom{4}{44}\binom{3}{40}\binom{2}{36}\binom{1}{32}\binom{1}{4}}$$
Is that all I have to do? Or am I missing some other calculation? Do I need to calculate that the other players aren't dealt one of those cards also as they come before player 1 is dealt their next card?
There are four ways to deal the six highest cards of any suit to the player to your left. The probability of the player getting dealt these cards thus equals:
$$\frac{4}{52 \choose 6} = \frac{4}{20358520} \approx 1.965 \cdot 10^{-7}$$
Assuming the choice for the suit is $\frac{1}{4}$ (independent of the cards the dealer gets), we arrive at:
$$\frac{1.965 \cdot 10^{-7}}{4} \approx 4.912 \cdot 10^{-8}$$
Alternatively, look at it this way. If the dealer always chooses a suit with probability $\frac{1}{4}$, we can choose the suit first. Then, there is only one combination of cards the player to your left should get, or again:
$$\frac{1}{52 \choose 6} \approx 4.912 \cdot 10^{-8}$$