I have a 54 card deck, with
2 jokers counting as zero
aces counting as 1
numbered cars counting as their value
picture or royalty cards counting as 10
What is the most likely sum of cards in a set, if the set contains 2 or 3 or 4 cards from this deck?
Since the 10s, jacks, queens, and kings make up 16/54 or about 30% of your deck, you will draw a card with a value of ten more frequently than any other card.
Therefore, with two cards and assuming no replacement, the most probable sum would be 20 and would occur with probability: $\frac{16}{54}\cdot\frac{15}{53}=\frac{40}{477}$
Likewise, for three cards the most probable sum is 30 and would occur with probability: $\frac{16}{54}\cdot\frac{15}{53}\cdot\frac{14}{52}=\frac{140}{6201}$
And for four cards the most probable sum is 40, occurring with probability: $\frac{16}{54}\cdot\frac{15}{53}\cdot\frac{14}{52}\cdot\frac{13}{51}=\frac{140}{24327}$
For three cards, a sum of $20$ can come from two $10$s plus a joker, $10$ plus two spots, or three spots. Two $10$s and a joker can be done in $3\cdot 2 \cdot 16 \cdot 15$ ways where the $3$ picks which card is the joker, the $2$ which joker, and the others are which $10$. $10$ plus two spots is $3 \cdot 16 \cdot (32\cdot 4+4 \cdot 3)$ where the $32$ is drawing a non-$5$ and the $4$ is drawing a matching card and the $4 \cdot 3$ is drawing two $5$s. There are four combinations of pair plus one and four combinations of three different ranks among the spot cards. A given pair plus one comes in $3\cdot 4 \cdot 4 \cdot 3$ ways and three different specific ranks come in $6 \cdot 4^3$, so we have $$3\cdot 2 \cdot 16 \cdot 15+3 \cdot 16 \cdot (32\cdot 4+4 \cdot 3)+4\cdot 3\cdot 4 \cdot 4 \cdot 3+4\cdot 6 \cdot 4^3=10272$$ Dividing by $54 \cdot 53 \cdot 52$ give $\frac {428}{6201}$, which is over three times as likely as a sum of $30$
Four cards summing to 30 can come from three tens plus a joker, two tens plus two spots or four large spots. The first two give $\frac {4 \cdot 2 \cdot 16 \cdot 15 \cdot 14+6\cdot 16 \cdot 15 (32 \cdot 4 + 4\cdot 3)}{54\cdot 53\cdot 52 \cdot 51}=\frac {560}{18603}$ The four large spots is a fair mess to compute. There are three $3+1$ combinations $(9993,8886,7779)$, two $2+2$s, two $2+1+1$ and one $1+1+1+1$ These account for $3\cdot 4 \cdot 4 \cdot 3 \cdot 2 \cdot 4+2\cdot 6 \cdot 4 \cdot 3 \cdot 4 \cdot 3+2 \cdot 2 \cdot 4 \cdot 3 \cdot 4 \cdot 4 + 24 \cdot 4^4=9792$ more combinations, giving a total chance of $\frac {584}{18603}$ for a sum of $30