Question

The game of euchre (YOOker)is played using only the 9s ,10s ,jacks ,queens ,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

think this is
\[\frac{\dbinom{6}{1}\times \dbinom{6}{1}\times \dbinom{6}{1}\times \dbinom{6}{1}}{\dbinom{24}{5}}\]

Answer 2

simplifies a great deal get
\[\frac{6^4\times 5\times 4\times 3\times 2}{24\times 23\times 22\times 21\times 20}\]

Answer 3

how you get 24 and 6 ?

Answer 4

well maybe i am wrong. there are 6 times 4 = 24 cards in the deck. the number of possible 5 card hands is therefore
\[\dbinom{24}{5}\]

Answer 5

there are 6 of each suit, and you want one of each, so you have to choose one out of the six spades, one out of the six hearts etc . one card is irrelevant.

Answer 6

6 cards , 4 suite
6x4=24
Ok, I get it TY

Answer 7

if this is right, the answer is
\[\frac{24}{1771}\] so you know the correct answer?

Answer 8

I don't have the key

Answer 9

i think it is right, but i would not bet much money on it

Answer 10

ty

Answer 11

This problem was tricky for me... I kept coming up with arguments, that like a slick lawyer, sounded plausible, but not obviously true. To settle the matter, I wrote a program and methodically searched through all possible hands which contained all four suits. Knowing the answer helped to do the analysis.

The universe of 5 card hands, where order does not matter is 24 choose 5.
Now for the numerator. Order does not matter. There are 6 spades, 6 hearts, 6 clubs, 6 diamonds, or 6^4 possibilities. The 5th card is one of the (24-4)= 20 remaining cards.
Now the tricky part. Consider
S C H D 5th_card
the 5th_card is (obviously) one of the 4 suits. Say it's a spade (S). We can search our list to find exactly the same C H D cards, but the S and the 5th_card swapped. That is, for every 5 card set, there is a duplicate. So the 20 is really 10!
The final answer is
\[\frac{\left(\begin{matrix}6 \\ 1\end{matrix}\right)^{4}10}{\left(\begin{matrix}24 \\ 5\end{matrix}\right)}= \frac{12960}{42504}=30.49\%\]
This answer matches the results of my computer program.

Answer 12

gj

Answer 13

you got my answer so you are probably correct ;)

http://openstudy.com/users/zarkon/updates/4e3f4d690b8b41fed9a92f1d

\[\frac{_4C_1\cdot_6C_2\cdot(_6C_1)^3}{_{24}C_{5}}=\frac{540}{1771}\]

Answer 14

@Zarkon, I like your answer. Once you know the answer, the logic is not too tough...but it sure wasn't obvious the first time round....

Answer 15

yeah it is right. i wrote
\[\frac{\dbinom{6}{1}^4\times 5\times 4\times 3\times 2}{24\times 23\times 22\times 21\times 20}\]

Answer 16

but i was off by a factor of 10

Answer 17

forgot about the choices for fifth card