Winning instantly at Phase 10 Assuming in a deck there are $96$ cards, two of each rank from $1$ through $12$ in each of four colors, from which $10$ cards are drawn. What is the probability that there is (at least)

*

*two sets of three $(0.032)$

*a run of eight $(0.010)$

*a set of four and a run of four $(0.004)$
I found this blog post but the described method doesn't agree with my numerical simulations above (which I double-checked). For example the first case is expressed like that:
$$\frac{\binom{12}{2} \binom{8}{3}^2 \binom{90}{4}}{\binom{96}{10}} = 0.047$$
Is this method correct? How would the other cases be calculated? Thanks in advance.
 A: Simulation is a great way to get a handle on the correct answer.  You can quickly see if what you've derived does not agree with the simulation.  For two sets of three, for instance, simulating 10,000,000 hands takes a minute or two in Python, and yields a probability of $p_1 \approx 0.0326 \pm 0.0001$.  This doesn't agree with your $0.047$, so you've done something wrong.  What you're doing wrong specifically is that you're overcounting certain hands.  For instance, hands where you have a set of strictly more than $3$ cards and another set of at least $3$ cards are counted multiple times in your enumeration.
Suppose you have exactly $m \ge 3$ cards in the smallest-rank set, and exactly $n\ge 3$ cards in the largest-rank set.  There are ${12}\choose{2}$ ways to pick the ranks, and ${8}\choose{m}$ ways to pick the cards in the first set, and ${8}\choose{n}$ ways to pick the cards in the second set; and there are then ${80}\choose{10-m-n}$ ways to pick the remaining cards (from other ranks).
This gives a count of
$$
\sum_{m=3}^{7}\sum_{n=3}^{10-m}{{12}\choose{2}}{{8}\choose{m}}{{8}\choose{n}}{{80}\choose{10-m-n}}=372250482912
$$
hands, for a probability of $0.0330$. But this is still (slightly but decidedly) inconsistent with the simulation result, because we're still overcounting hands with three sets of at least three cards.  Specifically, we're counting each such hand three times.  By analogous reasoning to the previous equation, there are
$$
\sum_{l=3}^{4}\sum_{m=3}^{7-l}\sum_{n=3}^{7-l-m}{{12}\choose{3}}{{8}\choose{l}}{{8}\choose{m}}{{8}\choose{n}}{{72}\choose{10-l-m-n}}=2926640640
$$
of these $3$-sets-of-at-least-$3$ hands; we need to subtract twice this number from our count.  So the desired (and now exact) probability is
$$
p_1=\frac{372250482912 - 2 \cdot 2926640640}{{96}\choose{10}}=\frac{693934094}{21363497077}\approx 0.03248223.
$$
