Combinations of cards in a hand of euchre

[kimmy_koo51]

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces from a standard deck of cards. How many five-card hands have:
(a) all red cards?
(b) at least two red cards?
(c) at most two red cards?

 

[pka]

Re: Combinations

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces from a standard deck of cards. How many five-card hands have:
(a) all red cards? (b) at least two red cards? (c) at most two red cards?

The deck of ‘euchre’ has 24 cards.
a) Half of those are red \(\displaystyle {{12} \choose 5}.\)
b) \(\displaystyle \sum\limits_{k = 2}^5 {{12} \choose k}{{12} \choose {5-k}}\)

 

[galactus]

There are 6 denominations, therefore, 12 red cards. There are 2 red cards per denomination, hearts and diamonds.

So, if the hand is all red, we must select 5 cards from the 12 red. C(12,5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

C(24,5)-(C(12,0)C(12,5)+C(12,1)C(12,4))

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

At most 2 red cards:

C(12,0)C(12,5)+C(12,1)C(12,4)+C(12,2)C(12,3)


Your task is to interpret. See how they work?.

 

[Denis]

A true Canadian does not play euchre, but plays "500" :shock:

I spent many winter nights playing 500 with my 3 brothers:

"Canadian Five Hundred
500 is played, under the name Cinq Cents, in French Canada, especially Montreal. It is played with a 46-card pack, made by throwing out the twos and threes from a 52-card pack and adding two distinguishable jokers, which are the highest trumps. There are four players in fixed partnerships, each of whom receive 10 cards; the high bidder takes the kitty of six cards and discards six."

Of course we developed forms of cheating, like put a card down sideways if
we wanted our partner to come back in same suit :idea:

 

[soroban]

Hello, kimmy_koo51!

The game of euchre uses only 9s, 10s, jacks, queens, kinds, and aces
from a standard deck of cards.
How many five-card hands have:

(a) all red cards?
(b) at least two red cards?
(c) at most two red cards?

There are: \(\displaystyle \:{24\choose5} \:=\:\frac{24!}{5!19!} \:=\:42,504\) possible hands.


(a) pka is absolutely right.
There are 12 red cards and we must select 5 of them.
. . There are: \(\displaystyle \:{12\choose5} \:=\:\frac{12!}{5!7!} \:=\:792\) possible all-red hands.


(b) "At least two red cards" (two or more red cards)
The opposite of "2 or more red cards" is "less than two red cards".
. . This means: "0 red cards" or "1 red card".

"No red" means "all black"; we must choose 5 black cards.
. . \(\displaystyle n(0\text{ red})\:=\:n(\text{5 black}) \:=\:{12\choose5} \:=\:792\) ways.

"1 red (and 4 black)"
. . \(\displaystyle n(\text{1 red, 4 black}) \:=\:{12\choose1}{12\choose4}\:=\:5940\) ways.

Hence, there are: \(\displaystyle \,792\,+\,5940\:=\:6732\) ways to get less than two reds.

Therefore: \(\displaystyle n(\text{2 or more red})\:=\:42,504\,-\,6,732\:=\:35,772\) ways. **


(c) "At most two red cards"
This means: "0 red", "1 red", or "2 red".

From part (b) we have: \(\displaystyle n(\text{0 red}) \:=\:792,\;n(\text{1 red}) \:=\:5,940\)

We need "2 red (and 3 black)".
. . \(\displaystyle n(\text{2 red, 3 black}) \:=\:{12\choose 2}{12\choose 3} \:=\:\frac{12!}{2!10!}\cdot\frac{12!}{3!9!} \:=\:14,520\)


Therefore: \(\displaystyle \,n(\text{at most 2 red}) \:=\:792\,+\,5,940\,+\,14,520 \:=\:21,252\) ways.

**
Edit: corrected a silly subtraction error . . .

 

[pka]

(a) pka is absolutely right.

Well thank you. It is a great relief to me to have your approval.
But please tell me, is the rest of the post showing the student how to do the question or is a complete working out of the question? If it is the latter, then I am sure you will be voted the best helper here. After all, who wants to learn if the job is just handed to me?