Question

Still looking for help: A Euchre deck consists of 24 cards; each card has a unique combination of one of 6 ranks (9,10,J,Q,K,A) and one of 4 suits (C,S,H,D). Suppose we draw a hand of five cards from a Euchre deck, assuming that all cards are equally likely:
(a) What is the probability that our hand is all one suit?
(b) What is the probability that our hand has all four suits?
(c) what is the probability that our hand is all of one rank?
(d) what is the probability that our hand contains 5 different ranks?
(e) suppose now that we draw an additional card. What is the probability that the suit of the new card mat.

Answer 1

(e) suppose now that we draw an additional card. What is the probability that the suit of the new card matches a suit already in our hand? (Hint: you may as well assume that the last card was drawn first, and think in terms of the complementary event.)

Answer 2

I started to work on this and will put my work here....any help is appreciated.

Answer 3

For (a) all one suit:
P(1) = 6^4/24
P(2) = 5/23
P(3) = 4/22
P(4) = 3/21
P(5) = 2/20
and I multiplied them together to get 155520/5100480 or .00001293

Answer 4

For (b) all 4 suits:
\[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}6 \\ 2\end{matrix}\right)\left(\begin{matrix}6 \\ 1\end{matrix}\right)\left(\begin{matrix}6 \\ 1\end{matrix}\right)\left(\begin{matrix}6 \\ 1\end{matrix}\right)\rightarrow4 \frac{ 6(5) }{ 2! }6(6)(6)=12960\]
\[\left(\begin{matrix}24 \\ 5\end{matrix}\right)\] in all = 42504
12960/42504=.305

Answer 5

The chance of getting a suit in the first card is 1. Then
5/23, 4/22,3/21,/2/20, and1/19 for the nest 5 cards. Your P(1) should be 1

Answer 6

@NoelGreco good point

Answer 7

(c) all one rank
P(1) = 4/24
P(2) = 3/23
P(3) = 2/22
P(4) = 1/21
P(5) = 20/20
24/255024=1/10626
but there could be 6!/4!2! different orders
15 (1/10626) = .001412

Answer 8

nice job

Answer 9

@noelgreco Thank you - that makes sense for part a

Answer 10

Any thoughts on the other parts that I've worked out @noelgreco?

Answer 11

anyone?

Answer 12

just a thought on this... the probabilities will all be dependent on where you are in the deal... seeing the hands are being dealt to 4 people in the game.

Answer 13

good point @campbell_st however I don't think it's a factor to consider for my homework problem. I think it's more general. What's hanging me up right now is the thought of how many different ways each one of these hands could be drawn and trying to figure out if I need to consider that.