Combinations of card hands

[OnTarget]

The game of euchre uses only 9s, 10s, jacks, queens, kings, and aces from a standard deck of cards
Cards are ranked 9->aces.

How many X sized card hands contain:
- (at least one spade and no hearts) or (at least a heart that is >= a jack)

 

[OnTarget]

This is what I have so far, it seems correct but is there a simpler way to get to the result?


x = number of cards per hand (1<=x<= clubs+diamonds)
clubs = 6
diamonds = 6
hearts = 6
spades = 6
validHearts = 4
worseThanJack = 24 - 4 = 20


ms = maxSpadesPerHand = Min(6, x)
mvh = maxValidHeartsPerHand = Min(validHearts, x) = Min(4,x)


Total x card combinations = \(\displaystyle \:{24\choose x} \:=\:\frac{24!}{x!(24-x)!} \)


A = Combinations of spades with no heart cards is the sum of


"1 spade (and x-1 clubs/diamonds)" . . \(\displaystyle n(\text{1 spade, x-1 clubs/diamonds}) \:=\:{spades\choose1}{clubs+diamonds\choose x-1}\)
...
"ms spades (and x-ms clubs/diamonds)" . . \(\displaystyle n(\text{ms spade, x-ms clubs/diamonds}) \:=\:{spades\choose ms}{clubs+diamonds\choose x-ms}\)


B = Combinations of better cards is the sum of:


"1 valid heart (and x-1 worseThanJack)" . . \(\displaystyle n(\text{1 valid heart, x-1 worseThanJack}) \:=\:{validHearts\choose1}{worseThanJack\choose x-1}\)
...
"mvh valid heart (and x-mvh worseThanJack)" . . \(\displaystyle n(\text{mvh valid heart, x-mvh worseThanJack}) \:=\:{validHearts\choose mvh}{worseThanJack\choose x-mvh}\)


Total hands = A+B

 

[pka]

The game of euchre uses only 9s, 10s, jacks, queens, kings, and aces from a standard deck of cards Cards are ranked 9->aces.
How many X sized card hands contain: - (at least one spade and no hearts) or (at least a heart that is >= a jack)

I have read your own reply #2. But I must say I think you have not considered all the cases.
I think that if \(\displaystyle X<6\) and \(\displaystyle X\ge 6\) are two different cases.

Let's say that \(\displaystyle X=8\) then the number of ways to have "at least one spade and no hearts" is
\(\displaystyle \sum\limits_{k = 1}^6 {\left[\binom{6}{k}\binom{12}{X-k}\right]} \)

How would you handle cases where \(\displaystyle X<6\text{ or }X>12~?\)

 

[JeffM]

I for one understand nothing you're saying/doing...
(no finger-pointing meant...it's probably me that's dense!)

Can the same problem be represented using:
6 black cards numbered 1 to 6; similarly:
6 white cards...
6 blue cards...
6 yellow cards...

Or are the "rules" of euchre required to be known?

BUT perhaps Jeff (who I hear is quite apt at this game...cheats
with no one noticing!) will step in and set me straight

If you lived in Pittsburgh, which may be the euchre capital of the world, you would understand that the game is usually played in a bar, and no one knows whether you are cheating or just drunk or both. It has been the secret of my success.

Hands that contain no hearts and no spades must contain no more than 12 cards.

So x < 13.

Number of hands = \(\displaystyle \dbinom{24}{x} = \dfrac{24!}{x! * (24 - x)!}.\)

Number of hands with no hearts = \(\displaystyle \dbinom{18}{x} = \dfrac{18!}{x! * (18 - x)!}.\)

Number of hands with no heart face cards = \(\displaystyle \dbinom{24 - 4}{x} = \dfrac{20!}{x! * (20 - x)!}.\)

Number of hands with no hearts or spades = \(\displaystyle \dbinom{24 - 2 * 6}{x} = \dfrac{12!}{x! * (12 - x)!}.\)

Number of hands that have at least one heart face card = \(\displaystyle \dfrac{24!}{x! * (24 - x)!} - \dfrac{20!}{x! * (x - 20)!}.\)

Number of hands with no hearts and at least one spade = \(\displaystyle \dfrac{18!}{x! * (18 - x)!} - \dfrac{12}{x! * (12 - x)!}.\)

Now the overlap between hands that have no hearts and those that have at least one heart face card is zilch.

Number of hands that contain at least one spade and no hearts or at least one heart face card =

\(\displaystyle \dfrac{18!}{x! * (18 - x)!} - \dfrac{12}{x! * (12 - x)!} + \dfrac{24!}{x! * (24 - x)!} - \dfrac{20!}{x! * (x - 20)!}.\)

No time to check. So please someone look it over.

 

[OnTarget]

My calculation and yours give the same result so its likely that its correct.

Yours however is a lot more elegant than my brute force solution

Thanks for the help.