It'll be a substantial amount of work, but it's doable. My intuition says the probability should be pretty high - let's see how that works out.
Final result: Approximately $57.8\%$. Somewhat lower than I would have initially guessed, but still quite large.
And now, the calculation. I'll sort by hand type.
Many of a kind. We count all of the ten-card stacks in which it's possible to make 4 of a kind, full house, or 3 of a kind.
To get four of a kind, we choose the denomination and then choose six other cards, for $13\cdot \binom{48}{6}$ ways. Some of these double up with two denominations each having four; that happens in $\binom{13}{2}\cdot \binom{44}{2}$ ways, which we subtract off for their double-counting.
We won't separately track full house hands; every hand that could be a full house is already three of a kind, so these add nothing to the count.
To get three of a kind, choose a denomination, then choose three cards from that denomination and seven cards of other denominations. That's $13\cdot \binom{4}{3}\cdot\binom{48}{7}$ ways. We can also have three of two kinds in $\binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}$ ways, or three of three kinds in $\binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1}$ ways. Counting by inclusion-exclusion, we subtract the "three of two kinds" options and add back the "three of three kinds" options. It's also possible to get three and four in $13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3}$ ways, or three, three, and four in $13\cdot \binom{12}{2}\cdot\binom{4}{3}^2$ ways. We subtract the former since they're already counted as four of a kind hands, and add the latter back by inclusion-exclusion.
So far, with just counting hands of four of a kind, full house, or three of a kind, we're at
\begin{align*}M &= 13\cdot \binom{48}{6} - \binom{13}{2}\cdot \binom{44}{2} + 13\cdot \binom{4}{3}\cdot\binom{48}{7} - \binom{13}{2}\cdot \binom{4}{3}^2\cdot\binom{44}{4}\\
&\quad + \binom{13}{3}\cdot\binom{4}{3}^3\cdot\binom{40}{1} - 13\cdot 12\cdot\binom{4}{3}\cdot\binom{44}{3} + 13\cdot \binom{12}{2}\cdot\binom{4}{3}^2\end{align*}
ways out of the $N=\binom{52}{10}$ ways to choose ten cards. Computing it, $N\approx 1.582\cdot 10^{10}$ and $M\approx 3.811\cdot 10^9$, for a probability $\frac MN$ of about $24.1\%$.
Straights. Our count will exclude all "many of a kind" hands.
To get at straights, we look at the list of denominations among the cards we've drawn. We can have one, two, three, or four cards per denomination - but we've already counted everything with three or more of a kind, so we'll just look at ones and twos. The options:
- Five denominations, two of each. There are $9$ ways to choose those five denominations to make a straight, and $\binom{4}{2}$ ways to choose the cards in each denomination, for $9\cdot\binom{4}{2}^5$ stacks with a straight.
- Six denominations, $2+2+2+2+1+1$. There are nine ways to pick a straight and eight ways to pick the other denomination, minus eight ways to double up with a length-6 straight. Then there are $\binom{6}{2}$ ways to choose which denominations only get one card, and either $\binom{4}{2}$ or $\binom{4}{1}$ ways to choose cards in each denomination. We get $(9\cdot 8-8)\cdot\binom{6}{2}\cdot\binom{4}{2}^4\cdot\binom{4}{1}^2$ stacks with straights this way.
- Seven denominations, $2+2+2+1+1+1+1$. There are $9\cdot \binom{8}{2}$ ways to pick a 5-straight and two others, $8\cdot 7$ ways to pick a 6-straight and one other, and $7$ ways to pick a 7-straight. This one isn't an inclusion-exclusion count - just subtracting the 6-straights from the 5-straights will already give us the proper weighting for the 7-straights. That's $9\cdot \binom{8}{2}-8\cdot 7$ for the number of ways to pick denominations with a straight. Then we have $\binom{7}{4}$ ways to choose which denominations have only one card, and $\binom{4}{2}$ or $\binom{4}{1}$ choices within each denomination. We get $\left(9\cdot \binom{8}{2}-8\cdot 7\right)\cdot \binom{7}{4}\cdot \binom{4}{2}^3\cdot \binom{4}{1}^4$ stacks with a straight.
- Eight denominations, $2+2+1+1+1+1+1+1$. There are $9\cdot\binom{8}{3}$ ways to pick a 5-straight and three others, and $8\cdot\binom{7}{2}$ ways to pick a 6-straight and two others, for $9\cdot\binom{8}{3}-8\cdot\binom{7}{2}$ ways to choose the denominations. Then there are $\binom{8}{6}$ ways to choose which denominations get only one card, and the usual $\binom{4}{2}$ or $\binom{4}{1}$ choices in each denomination, for $\left(9\cdot\binom{8}{3}-8\cdot\binom{7}{2}\right)\cdot \binom{8}{6}\cdot\binom{4}{2}^2\cdot\binom{4}{1}^6$ stacks with a straight.
- Nine denominations, one with a pair and the rest singles. There are $9\cdot\binom{8}{4}$ ways to pick a 5-straight and four others, and $8\cdot\binom{7}{3}$ ways to pick a 6-straight and three others, for $9\cdot\binom{8}{4}-8\cdot\binom{7}{3}$ ways to choose the denominations. Then there are $\binom{9}{8}$ ways to choose which denominations get only one card, and the usual $\binom{4}{2}$ or $\binom{4}{1}$ choices in each denomination, for $\left(9\cdot\binom{8}{4}-8\cdot\binom{7}{3}\right)\cdot \binom{9}{8}\cdot\binom{4}{2}\cdot\binom{4}{1}^8$ stacks with a straight.
- Ten denominations, one of each. There are $9\cdot\binom{8}{5}$ ways to pick a 5-straight and five others, and $8\cdot\binom{7}{4}$ ways to pick a 6-straight and four others. There are also six ways to choose two separated 5-straights (2-6 and 8-Q, 2-6 and 9-K, 2-6 and 10-A, 3-7 and 9-K, 3-7 and 10-A, 4-8 and 10-A). Accounting for both, we have $9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6$ ways to choose the denominations. Then there are $\binom{4}{1}$ ways to choose a card in each denomination, for $\left(9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6\right)\cdot \binom{4}{1}^{10}$ stacks with a straight.
Adding these up, straights that aren't also 3 or more of a kind account for
\begin{align*}S &= 9\cdot 6^5 + (9\cdot 8-8)\cdot\binom{6}{2}\cdot 6^4\cdot 4^2 + \left(9\cdot \binom{8}{2}-8\cdot 7\right)\cdot \binom{7}{4}\cdot 6^3\cdot 4^4\\
&\quad +\left(9\cdot\binom{8}{3}-8\cdot\binom{7}{2}\right)\cdot \binom{8}{6}\cdot 6^2\cdot 4^6 + \left(9\cdot\binom{8}{4}-8\cdot\binom{7}{3}\right)\cdot \binom{9}{8}\cdot 6\cdot 4^8\\
&\quad +\left(9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6\right)\cdot 4^{10}\end{align*}
Computing that, it's about $3.253\cdot 10^9$, for an added probability $\frac{S}{N}\approx 20.6\%$. We're up to nearly half.
Since straight flushes, including the royal variety, are already counted as both straights and flushes, we won't be treating them separately. Instead, we'll look at stacks that contain both a straight and a flush - even if those aren't the same five-card hand. That can wait, though.
Flushes. We sort by the length of the long suit, then account for the elements outside that suit. The elements off-suit may cause us to double-count with three or more of a kind, so we'll subtract them.
- Ten-card flushes, no off-suit cards. There are four suits and $\binom{13}{10}$ denomination choices, for $4\cdot\binom{13}{10}$ stacks.
- Nine-card flushes, one off-suit card. There are four suits and $\binom{13}{9}$ denomination choices. Three of a kind is still impossible, so we get $4\cdot\binom{13}{9}\cdot 39$ stacks.
- Eight-card flushes, two off-suit cards. There are four suits and $\binom{13}{8}$ denomination choices. We can have three of a kind by choosing our two off-suit cards to be in one of the flush denominations, so that's $\binom{39}{2}-8\cdot\binom{3}{2}$ good off-suit choices. Combine them, and we get $4\cdot\binom{13}{8}\cdot\left(\binom{39}{2}-8\cdot\binom{3}{2}\right)$ stacks.
- Seven card flushes, three off-suit cards. There are four suits and $\binom{13}{7}$ denomination choices. There are $\binom{39}{3}$ ways to choose the off-suit cards. Of these, we can reach three of a kind with all three off-suit cards in the same denomination in $13$ ways, or with a pair in a flush denomination and one other in $7\cdot \binom{3}{2}\cdot 36$ ways. Combined, that's $4\cdot\binom{13}{7}\cdot\left(\binom{39}{3}-13-7\cdot \binom{3}{2}\cdot 36\right)$ stacks.
- Six-card flushes, four off-suit cards. There are four suits and $\binom{13}{6}$ denomination choices. There are $\binom{39}{4}$ ways to choose the off-suit cards. Of these, we can reach three of a kind with a triple and one other in $13\cdot 36$ ways, a pair in a flush denomination and two others in $6\cdot\binom{3}{2}\cdot\binom{36}{2}$ ways, or (overcounting) two pairs in flush denominations in $\binom{6}{2}\cdot\binom{3}{2}^2$ ways. Combined, that's $4\cdot\binom{13}{6}\cdot\left(\binom{39}{4}-13\cdot 36-6\cdot \binom{3}{2}\cdot \binom{36}{2}+\binom{6}{2}\cdot\binom{3}{2}^2\right)$ stacks.
- Five-card flushes, five off-suit cards. There are four suits and $\binom{13}{5}$ denomination choices. There are $\binom{39}{4}$ ways to choose the off-suit cards. Of these, we can reach three of a kind with a triple and two others in $13\cdot\binom{36}{2}$ ways, a pair in a flush denomination and three others in $5\cdot\binom{3}{2}\cdot\binom{36}{3}$ ways, a pair in a flush denomination and a triple in $5\cdot\binom{3}{2}\cdot 12$ ways, or two pairs in flush denominations and one other in $\binom{5}{2}\cdot\binom{3}{2}^2\cdot 33$. Combined, that's $4\cdot\binom{13}{5}\cdot\left(\binom{39}{5}-13\cdot\binom{36}{2}-5\cdot\binom{3}{2}\cdot\binom{36}{3}+5\cdot\binom{3}{2}\cdot 12+\binom{5}{2}\cdot\binom{3}{2}^2\cdot 33\right)$ stacks.
- Two different five-card flushes. There are $\binom{4}{2}$ pairs of suit and $\binom{13}{5}$ denomination choices in each suit. Three of a kind is impossible, so that's $\binom{4}{2}\cdot\binom{13}{5}^2$ stacks. These are of course an overcount that we'll subtract from the others.
Adding these up, flushes that aren't also 3 or more of a kind account for
\begin{align*}F &= 4\binom{13}{9}\cdot 39 + 4\binom{13}{8}\left(\binom{39}{2}-8\binom{3}{2}\right) + 4\binom{13}{7}\left(\binom{39}{3}-13-7\binom{3}{2}36\right)\\
& +4\binom{13}{10}+ 4\binom{13}{6}\left(\binom{39}{4}-13\binom{36}{1}-6 \binom{3}{2} \binom{36}{2}+\binom{6}{2}\binom{3}{2}^2\right) - \binom{4}{2}\binom{13}{5}^2\\
& + 4\binom{13}{5}\left(\binom{39}{5}-13\binom{36}{2}-5\binom{3}{2}\binom{36}{3}+5\binom{3}{2} 12+\binom{5}{2}\binom{3}{2}^2 33\right) \end{align*}
Computing that, it's about $2.922\cdot 10^9$, for an added probability of about $18.5\%$. We're up past $60\%$ overall.
Straights and flushes. Unfortunately, we still have some overcounting to deal with. It's possible to have both a straight and a flush in the same stack, and it's possible with any level of overlap. We'll work with the straight count from earlier to find the doubles. Sorting in the same way:
- Five denominations, two of each. Choosing two suits at each denomination, we can reach a five-card flush in $4\cdot 3^5$ ways, and a double flush in $\binom{4}{2}$. That gives $9\cdot \left(4\cdot 3^5 - \binom{4}{2}\right)$ stacks with both straights and flushes.
- Six denominations. We can reach a six-card flush in $4\cdot 3^4$ ways, and a five-card flush in $4\left(4\cdot \binom{3}{2}\cdot 3^3 + 2\cdot 3\cdot 3^4\right)$ ways, sorting by how many cards are in the excluded denomination. To simplify future expressions, note that we have $3$ non-flush suit choices at a denomination with two cards regardless of whether one of those cards is in the flush. At a denomination with only one card, the suit is determined if it's in the flush and we have three choices otherwise. It's also possible to reach a double flush, in $\binom{4}{2}\cdot\binom{2}{1}$ ways (choose the suits, choose which single-card denomination goes with each). That gives $(9\cdot 8-8)\cdot \binom{6}{2}\cdot 4\left(5\cdot 3^4 + 2\cdot 3^5 - 3\right)$ stacks.
- Seven denominations. We can reach a seven-card flush in $4\cdot 3^3$ ways, a six-card flush in $4\left(3\cdot 3^3 + 4\cdot 3^4\right)$ ways, and a five-card flush in $4\left(\binom{3}{2}\cdot 3^3 + 3\cdot 4\cdot 3^4 + \binom{4}{2}\cdot 3^5\right)$ ways. Double flushes are possible in $\binom{4}{2}\cdot\binom{4}{2}$ ways. That gives $\left(9\cdot 28-8\cdot 7\right)\cdot\binom{7}{4}\cdot 4\left(7\cdot 3^3+16\cdot 3^4+6\cdot 3^5 - 9\right)$ stacks.
- Eight denominations. We can reach an eight-card flush in $4\cdot 3^2$ ways, a seven-card flush in $4\left(2\cdot 3^2 + 6\cdot 3^3\right)$ ways, a six-card flush in $4\left(\binom{2}{2}\cdot 3^2 + 2\cdot 6\cdot 3^3 + \binom{6}{2}\cdot 3^4\right)$ ways, and a five-card flush in $4\left(\binom{2}{2}\cdot 6\cdot 3^3 + 2\cdot \binom{6}{2}\cdot 3^4 + \binom{6}{3}\cdot 3^5\right)$. Double flushes are possible in $\binom{4}{2}\cdot\binom{6}{3}$ ways. That gives $\left(9\cdot 56-8\cdot 28\right)\cdot\binom{8}{6}\cdot 4\left(4\cdot 3^2+24\cdot 3^3+45\cdot 3^4+20\cdot 3^5-30\right)$ stacks.
- Nine denominations. We can reach a nine-card flush in $4\cdot 3$ ways, an eight-card flush in $4\left(3+8\cdot 3^2\right)$ ways, a seven-card flush in $4\left(8\cdot 3^2+\binom{8}{2}\cdot 3^3\right)$ ways, a six-card flush in $4\left(\binom{8}{2}\cdot 3^3+\binom{8}{3}\cdot 3^4\right)$ ways, and a five-card flush in $4\left(\binom{8}{3}\cdot 3^4+\binom{8}{4}\cdot 3^5\right)$ ways. Double flushes are possible in $\binom{4}{2}\cdot\binom{8}{4}$ ways. That gives $\left(9\cdot 70-8\cdot 35\right)\cdot\binom{9}{8}\cdot 4\left(2\cdot 3+16\cdot 3^2+56\cdot 3^3+112\cdot 3^4+70\cdot 3^5 - 105\right)$ stacks.
- Ten denominations. We can reach a ten-card flush in $4$ ways, a nine-card flush in $4\cdot 10\cdot 3$ ways, an eight-card flush in $4\cdot \binom{10}{2}\cdot 3^2$ ways, a seven-card flush in $4\cdot\binom{10}{3}\cdot 3^3$ ways, a six-card flush in $4\cdot\binom{10}{4}\cdot 3^4$ ways, and a five-card flush in $4\cdot\binom{10}{5}\cdot 3^5$ ways. Double flushes are possible in $\binom{6}{2}\cdot\binom{10}{5}$ ways. That gives $\left(9\cdot\binom{8}{5} - 8\cdot\binom{7}{4} - 6\right)\cdot 4\cdot \left(1+10\cdot 3+45\cdot 3^2+120\cdot 3^3+210\cdot 3^4+252\cdot 3^5 - 378\right)$ stacks.
Adding these up, stacks with both straights and flushes but not three or more of a kind come to
\begin{align*}X &= 9\cdot \left(4\cdot 3^5 - \binom{4}{2}\right) + (9\cdot 8-8)\cdot \binom{6}{2}\cdot 4\left(5\cdot 3^4 + 2\cdot 3^5 - 3\right)\\
& + \left(9\cdot 28-8\cdot 7\right)\cdot\binom{7}{4}\cdot 4\left(7\cdot 3^3+16\cdot 3^4+6\cdot 3^5 - 9\right)\\
& + \left(9\cdot 56-8\cdot 28\right)\cdot\binom{8}{6}\cdot 4\left(4\cdot 3^2+24\cdot 3^3+45\cdot 3^4+20\cdot 3^5-30\right)\\
& + \left(9\cdot 70-8\cdot 35\right)\cdot\binom{9}{8}\cdot 4\left(2\cdot 3+16\cdot 3^2+56\cdot 3^3+112\cdot 3^4+70\cdot 3^5 - 105\right)\\
& + \left(9\binom{8}{5} - 8\binom{7}{4} - 6\right) 4 \left(1+10\cdot 3+45\cdot 3^2+120\cdot 3^3+210\cdot 3^4+252\cdot 3^5 - 378\right)\end{align*}
Computing that, it's about $0.848\cdot 10^9$, for a probability of about $5.4\%$.
The overall probability we seek is $\dfrac{M+S+F-X}{N} = \dfrac{9139533260}{15820024220}\approx 57.8\%$.
Best Answer
Try breaking the problem into the case where you get the joker, and the case where you don't (this is then identical to the "joker free" problem that you already understand).
For example, if you've drawn the joker, you have a 100% chance of having at least one pair.