Question

The game of euchre (YOO ker) is played using only the 9s,10s,jack,queen,kings, and aces from a standard desk of cards .Find the probability of being dealt a 5 card hand containing all four suits.

Answer 1

The way I see it, you could solve it like so: Since you need all four suits, you have to choose two cards from one suit, and one from the others. Thus, you have 4 choices for the suit with two cards, two choices of cards from that suit, and six choices from the other suits. Thus, I think the formula should be\[\left(\begin{matrix} 4 \\ 1 \end{matrix} \right)\left(\begin{matrix} 6 \\ 2 \end{matrix} \right)\left(\begin{matrix} 6 \\ 1 \end{matrix} \right)\left(\begin{matrix} 6 \\ 1 \end{matrix} \right)\left(\begin{matrix} 6 \\ 1 \end{matrix} \right) = 4*15*6*6*6=12960\]

Answer 2

can you explain I still don't get

Answer 3

total pick 5?

Answer 4

9s,10s,jack,queen,kings, and aces total 6, how you get 4?

Answer 5

First, you know that only one suit can contain more than one card, and the other three suits must contain only one card. Thus, you have 4 choices for the suit that has 2 cards. (The 4-choose-1 term)

Then, from the suit that has 2 cards, you have 6 possible choices, and you choose two of those 6 possible cards. (The 6-choose-2 term)

Then, for the three remaining suits, you have to choose only 1 card from each from 6 possible cards. From this you get the 3 different 6-choose-1 terms.

The 4's that I use come from the number of suits there are; diamonds, hearts, clubs, spades.

Answer 6

I've chosen 5 cards because one of suits has two cards, so I have 6-choose-2 possibilities for that suit, and then three others have 6-choose-1 possibilities. \(2+1+1+1=5\)

Answer 7

ty

Answer 8

you're welcome.