Combinations, a Deck with Separate Hands and Players The main issue I'm having is trying to solve this scenario below where we have multiple decks between multiple players, with left over cards that remain in no player's hand. With this I've tried to determine the probabilities of the scenarios below, but I am almost certain that my answers are wrong. I'll show how I came to my answers, pointing out any flaws in my logic would be greatly appreciated.
A euchre deck contains the cards {9, 10, jack, queen, king, ace} of each of the four suits (diamonds, spades, hearts, clubs). A hand contains 5 cards. There are four players. One card of the undealt four cards is revealed (which is relevant to the game, but we don't need to know how the game works to answer these questions).
Given that one king is revealed and you have no king, what is the probability that each other player has one king?
$\frac{{18 \choose 3}*3}{{24 \choose 5}}$ = 5.76%
24 cards in deck, 5 in own hand subtracted from available total = 19, -1 from one revealed = 18, 3 kings still left over out of 18 cards with 3 players, out of the available 5-card decks.
The most obvious problem with this answer is the fact that you can't simply multiply by 3 in the numerator like that I believe, as it would be more likely to be $\frac{{18 \choose 3}^3}{{24 \choose 5}}$, which would be 1278323%, so obviously that doesn't work. Also, how would we account for the fact that each of their individual decks have 5 cards? Would the 3 cards that nobody has need to be factored separately instead of just subtracted from the total?
What is the probability that your hand has only red (hearts and diamonds) cards?
$\frac{{12 \choose 24}}{{24 \choose 5}}$ = 6362%
So obviously this approach doesn't work, as ${12 \choose 24}$, which would be the 12 red cards out of the 24 cards in the deck, is much greater than the ${24 \choose 5}$ total 5-card hands that can be configured from the 24 card deck. So how would we incorporate that the 3 other players who each also have their own 5-card deck and the fact that there would be 4 cards left that belong to no player if none are revealed?
Would it perhaps be that you need to account for each card as you go such that
$\frac{{12 \choose 1}{11 \choose 1}{10 \choose 1}{9 \choose 1}{8 \choose 1}}{{24 \choose 5}}$? That would equal 356.47%, so there's still something.
How many ways can every player have only cards of a one suit? (For example, player 1 has only spades, player 2 only diamonds, player 3 only hearts, player 4 only clubs in a single deal.)
$\frac{{12 \choose 4}^4}{{24 \choose 5}^4}$ = 1.84*10^(-6)% or $\frac{{24 \choose 5}{19 \choose 5}{14 \choose 5}{9 \choose 5}}{{24 \choose 5}^4}$ = 0.003820%
The second between these 2 options seems to be much more correct and intuitive, and I think perhaps the way to account for the 4 separate decks my be as I've done here, but I'm not sure if it takes into account the 4 cards claimed by no player, or if that is a factor that we need to be concerned with at all.
What is the probability your hand contains only 9s and 10s?
$\frac{{24 \choose 12}}{{24 \choose 5}^4}$ = 2.25*10^(-11)% or $\frac{{8 \choose 1}{7 \choose 1}{6 \choose 1}{5 \choose 1}{4 \choose 1}}{{24 \choose 5}}$ = 224.07%
Here I tried a both approaches again trying to incorporate the 4 separate hands by setting the total configurable decks to the 4 power, and the percentage probability again seems much too small, and it still doesn't account for the 4 left over cards, this is, if that factor needs to be accounted for at all. The second method was again from the second half of the second question.
So for all these example would it be mostly like the second example for the third question? That one seemed the most intuitive of them all but I'm still not sure if I'm tackling this in the correct manner.
 A: 
Given that one king is revealed and you have no king, what is the probability that each other player has one king?

$$ \frac{\binom{15}{4}\binom{11}{4}\binom{7}{4} \times (3!)}
{\binom{18}{5}\binom{13}{5}\binom{8}{5}}. \tag1 $$
Note that no information regarding your hand is available, other than that it does not contain a King.  You can assume, without loss of generality therefore, that your hand specifically contains $(4)$ 9's and the 10-spades.
This very handy assumption does not alter the computations.  That is, if instead, your hand contained $4$ Jacks and a Queen, that would not alter the computation of the probability that each of the other $3$ players has a King.
In (1) above, the denominator indicates how many different distributions of the $(18)$ remaining cards there are, to the $(3)$ players, distributing the cards $(5)$ at a time.
Note also that for convenience, when computing the denominator, the order of distributions of the cards is deemed relevant.  So, player-1 getting a certain group of $5$ cards is deemed distinct from player-2 getting these same $5$ cards.
Consequently, the numerator must be computed in a consistent manner.  Note that of the $(18)$ cards remaining, $15$ of them are non-Kings, and the $3$ Kings can be assigned in $(3!)$ ways.


What is the probability that your hand has only red (hearts and diamonds) cards?

The probability is
$$\frac{\binom{12}{5}}{\binom{24}{5}}.$$
That is, the numerator indicates how many ways there are of selecting $(5)$ cards out of $(12)$.


How many ways can every player have only cards of a one suit? (For example, player 1 has only spades, player 2 only diamonds, player 3 only hearts, player 4 only clubs in a single deal.)

There is confusion about whether you are asking about the number of different distributions (where presumably) the players are distinguishable from each other, or the probability of the event occurring.
If you are asking about the probability of the event occurring, then the answer will be
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where
$$D = \binom{24}{5} \times \binom{19}{5} \times \binom{14}{5} \times \binom{9}{5}.$$
So, it only remains to compute $N$, which will correspond to the number of different distributions, if this is what you are asking.
First, you have a factor of $(4!)$, which represents that the $(4)$ suits can be distributed among the players in $(4!)$ ways.  Then, since each player will get $5$ of the $6$ cards in that suit, you also have a factor of $~\displaystyle \left[\binom{6}{5}\right]^4.$
Putting this all together,
$$\frac{N}{D} = \frac{(4!) \times 6^4}{\binom{24}{5} \times \binom{19}{5} \times \binom{14}{5} \times \binom{9}{5}}.$$


What is the probability your hand contains only 9s and 10s?

This is straightforward.  You have to select $5$ cards from the 9's and 10's.
$$\frac{\binom{8}{5}}{\binom{24}{5}}.$$
A: 
Given that one king is revealed and you have no king, what is the probability that each other player has one king?

The deck contains 4 kings, and 20 non-kings. Four hands of five cards are dealt, one card is revealed, and 3 cards remain.  One among the hands is "yours", three are "not yours".
So, you seek the probability for obtaining one place among each of three hands of five cards (and none among the three remainders), when given that these three kings may placed among any from these $18$ places in the deck that are not "yours" or the revealed card ($24-5-1$).
$$\dfrac{\dbinom 51^3\dbinom 30}{\dbinom{18}{3}}\qquad\raise{5ex}{\gets 5\times 3+3=18\\\gets1\times 3+0=3\\\text{check sum }\checkmark}$$

Solve the rest of your problems similarly.  Make sure the favoured event being counted in the numerator is a subset of the "total" event being counted in the denominator, and that they are being counted in the same general manner.
In the above, we are selecting 3 things from 18 in the denominator and the numerator, just with the favoured selection being a more specific arrangement of that : 1 from 5 in each of 3 hands, and 0 from 3 in the 1 group of 'unrevealed'.
Including factors that equal 1, such as $\binom 30$, will help make your counting logic clearer, and ensure the "check sum" test works.


What is the probability that your hand has only red (hearts and diamonds) cards?

What is the probability for obtaining 5 from 12 cards in "your" hand (and none from the other 12 cards), when "your" hand may select 5 from all 24 cards?
What is the probability for obtaining 5 among 5 places in "your" hand and 7 among 19 places elsewhere, when selecting 12 among 24 places in the deck for the red cards?
A: For the main question of "do I need to worry about other hands?  do I need to worry about what's buried?"  The answer is generally no.  "Do I need to worry about the order in which cards appear in my hand (or other hands)?"  Again, generally no.
Could you make it such that those things are relevant?  Sure, but you need to ensure that your numerator and your denominator are both encapsulating the same type of information to the same degree of specificity.  If you treat order as relevant in the numerator you must also treat order as relevant in the denominator.  If you worry about the cards in the other players' hands in the numerator, you must also keep track of cards in the other players' hands in the denominator.  If done correctly, regardless what decisions you make with regards to these the answers will be correct and you are free to make decisions as to the relevance however you like to make things easier on yourself so long as the question you are trying to answer does not require certain levels of specificity.

Problem 1:  What is the conditional probability that the three remaining kings are sent one each to the other three players given the turn is not a king and you have no king in your hand.
We can do this more carefully with $\Pr(A\mid B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}$, but I will ignore that approach and do this directly instead.  We have treat this as though we intentionally searched for five non-king cards and one king and gave five non-kings to us and one king placed as the turn and shuffled the remaining $18$ cards and dealt them out.
There are $\binom{18}{5,5,5,3}=\frac{18!}{5!5!5!3!}=\binom{18}{5}\binom{13}{5}\binom{8}{5}\binom{3}{3}$ ways to distribute the remaining cards among the other players and the buried deck where we do care about who gets which card but don't care about the order in which they received said cards.  Note... this is not $\binom{24}{5}^3$ or $\binom{24}{5}^3\binom{24}{3}$ or some other expression like you were trying to use... once a card is in someone's hand, that card is not eligible to appear in anyone else's hand.
There are $3\cdot \binom{15}{4}\cdot 2\cdot \binom{11}{4}\cdot 1\cdot \binom{7}{4}\cdot \binom{3}{3}$ ways to choose a king to go to the second player and then four more cards from those remaining, then 2 ways to choose the king to go to the third player and four more cards from those remaining, the last king goes to the fourth player and choosing the other four cards to them, and then the final three cards to go the buried deck.  This gives us a probability of:
$$\frac{3\cdot \binom{15}{4}\cdot 2\cdot \binom{11}{4}\cdot 1\cdot \binom{7}{4}\cdot \binom{3}{3}}{\binom{18}{5}\binom{13}{5}\binom{8}{5}\binom{3}{3}} = \frac{3!\binom{15}{4,4,4,3}}{\binom{18}{5,5,5,3}}$$
Again, I will reemphasize that we could just as well have done this as order in which the cards appear in people's hands as mattering and gotten a larger expression  which I leave to you to fiddle with if you like... however you can also treat order as relevant and only care about the kings having ignored where any of the rest of the cards are sent.
With order relevant within the hand, there are $\binom{18}{3}$ different collections of positions where the kings could have been sent.  Among those, there are $5^3$ different sets of positions which would have been favorable where all kings go in different hands, giving a probability of $\dfrac{5^3}{\binom{18}{3}}$.  Similarly, we could have done this by distributing the cards to random positions and gone one card at a time having started by distributing the kings.  The first king is sent to a "good" spot with probability $\frac{15}{18}$.  Given the first king is sent to a good spot, there are $10$ good spots remaining and we will continue by sending the second king to a good spot with probability $\frac{10}{17}$ and then the third after this with probability $\frac{5}{16}$ giving the final probability as $\frac{15\cdot 10\cdot 5}{18\cdot 17\cdot 16}$.  Of course, all of these approaches give the same final answer.

Problem 2 is a textbook example of a hypergeometric distribution.  The cards in the other players' hands don't need to be relevant so we can focus on our hand only.
There are $12$ red cards and we want to choose five of them to go to us.  There are $\binom{12}{5}$ ways to do this.  This is compared to the $\binom{24}{5}$ ways we could have given ourselves five cards.  This gives our probability as $\dfrac{\binom{12}{5}}{\binom{24}{5}}$.
If you insist on making the other hands relevant... then again, the correct way to do this is with multinomial coefficients rather than taking the denominator to a higher power because, again, the cards can not be reused multiple times.  If you really insist, the answer will look closer to $\dfrac{\binom{12}{5}\binom{19}{5,5,5,4}}{\binom{24}{5,5,5,5,4}}$ but this is far messier to work with and the extra arithmetic needed is not worth it.

Problem 3: What is the probability every player has cards entirely of a single suit?  (I assume we will ignore the special behavior of Jacks as left-bowers in Euchre)
Your numerator here in your second attempt was actually what the denominator was supposed to be.  The numerator was supposed to be the number of ways of sending five of the same suit to the first player, second player, etc...
For this, first decide who is designated to receive which suit in $4!$ ways.  Then, choose which ranks those cards were for the first player in $\binom{6}{5}$ ways.  Similarly for the other players.  This is in comparison to the $\binom{24}{5,5,5,5,4}$ ways in which the deck can be distributed (order within hands not mattering) giving a probability of $\dfrac{4!6^4}{\binom{24}{5,5,5,5,4}}$

Problem 4: you only get 9's and 10's, again this is a textbook example of hypergeometric probability.  $\dfrac{\binom{8}{5}}{\binom{24}{5}}$
