Calculate your expected winnings (or losses) if you throw away all but the K♣. You might get a pair, 2-pair, 3-of-a-kind, 4-of-a-kind, full house, straight, flush, straight-flush, royal flush, or nothing.

So now we have discarded J♦, 4♦, 5♠, 6♠ and we are keeping the K♣. Let R be the set of cards {2,3,7,8,9,10,Q,A} which are still in the deck, and S be the set of {J, 4, 5, 6} which are still in the deck (not the ones we have discarded). Here is how we can complete the hand:

pair of Kings: this will consist of one of the 3 kings and 3 other cards which do not form a pair or a three of a kind.

Those cards are either all from R, or

two from R and one from S, or

one from R and two from S, or

all three from S

pair of Jacks: your hand will have two Jacks from the three remaining, the K♣, and two other cards which don't form a pair and are not jacks.

The remaining two cards will be both in R, or

one card from R and one card from {4, 5, 6}, or

both cards from {4, 5, 6}

pair of Queen or pair of Ace (these are treated in the same way so just do Queens and multiply by 2). Your hand will have two of the four possible Queens, the K♣, and two other cards which don't form a pair.

The remaining two cards will both be in {2,3,7,8,9,10,A}, or

one card from {2,3,7,8,9,10,A} and one card from S, or

both cards from S.

two pair where one pair is a King: this will consist of K♣, another K, another pair, one other card.

the second pair comes from R, the last card can be one of the 40 remaining (not K, not one of J♦, 4♦, 5♠, 6♠, and not the same as the second pair)

the second pair comes from S, the last card can be one of the 41 remaining

two pair where one pair is not a King: this will consist of K♣ and two other pair

both pair are from R, or

one pair is from R and one from S, or

both pair are from S.

three-of-a-kind Kings: this will consist of K♣ and two other kings, plus two other cards which do not form a pair

the last two cards can both be in R, or

one last card is in R and one in S, or

both cards are in S.

three of a kind from R: this will have a 3-of-a-kind, the K♣, and one card from one of the remaining 40

three of a kind from S: this will have a 3-of-a-kind, the K♣, and one card from the remaining 41

4-of-a-kind of Kings: all four kings and the last card is one of the remaining 44

4-of-a-kind from R: the K♣ and a 4-of-a-kind from the 8 possible in R

full house 3 Kings, pair from R

full house 3 Kings, pair from S

full house 2 Kings, 3-of-kind from R

full house 2 Kings 3-of-kind from S

straight where K is the high card: there is a 9,10,J (not ♦),Q and the K♣, don't include the straight flush

straight where A is the high card: there is a 10,J (not ♦),Q and the K♣,A, don't include the royal flush

flush: all 4 cards must be from the 12 remaining ♣ in the deck but not the straight flush and not the royal flush

straight-flush: there is one possible with the K♣ as the high card

royal flush: there is one possible that includes the K♣

nothing: there are C(47,4) minus all possible other hands that will count as 'nothing.' You should verify that this is the same as:

four different cards from R (not all ♣ because that would be a flush)

three different cards from R one from S (not all ♣ because that is a flush and not 10,J,Q,A because that is a straight)

two different cards from R and two different from S (not all ♣ because that would be a flush)

one card from R and three different from S (not all ♣ because that would be a flush)

all four different cards from S (not all ♣ because that would be a flush)

pair of 4s or pair of 5s or pair of 6s and two different cards from R

pair of 4s or pair of 5s or pair of 6s and one card from R and one card from S

pair of 4s or pair of 5s or pair of 6s and two different cards from S

pair of {2,3,7,8,9,10} and two different cards from R

pair of {2,3,7,8,9,10} and one card from R and one card from S

pair of {2,3,7,8,9,10} and two different cards from S