Odds of drawing any 3 identical cards from 27-card deck (9 unique x 3 copies each) in 9 draws I have a 27-card deck with 3 copies of each of 9 distinct cards.  If I'm drawing 9 cards, what are the odds that I draw all 3 copies of any one of the 9 possible options?  [Also known as "What are the odds of drawing all three of the same city in the Pandemic Legacy season 2 prologue game?"]
My initial approach was to count the number of possible matching hands versus the ${27 \choose 9}$ total possible number of hands.
I figured that there are ${24 \choose 6}$ possible ways to pick the rest of the 9-card draw after assuming one full set of 3 cards, so that makes $9 * {24\choose6}$ starting hands.
However, this double-counts hands that have two or more sets of 3 cards so I need to subtract those.  Calculating that follows a similar approach, so we have to subtract $8 * {21 \choose 3}$ cards.  Of course that latter bit double-counts the hands where our six cards are two sets of three.  So we have to subtract the 7 such hands from that.
So my final formulation was:
$\frac{9 * {24\choose6} - (8 * {21\choose3} - 7)}{27\choose9}$ ~= 25.6%
Does that seem right?  If not, what am I doing wrong?
 A: Note:  "probability" and "odds" are not the same thing. $1:1$ odds corresponds to probability $\frac 12$, for example.  It looks like you are computing the probability, so that's what I have done below.
The approach is fine, but the arithmetic is not.  There are $\binom 92=36$ ways to choose two cities and $\binom 93=84$ ways to choose three.  Thus your numerator should be $$9\times \binom {24}6-36\times \binom {21}3+84$$  making your final answer $$\frac {1163568}{4686825}=.2483$$
As another approach:
Let's work backwards.  Let $p$ be the probability that no three of a kind is drawn. 
Then each city must be represented by $0,1,2$ cards.
If there are exactly $n$ cities for which you drew $2$ cards then there must be $9-2n$ cities for which you drew exactly $1$.  
Assuming you drew exactly $n$ cities for which you drew two cards, then we need to choose those $n$ cities ($\binom 9n$), choose the two cards from each city (3), then choose the $9-2n$ cities for which one card will be drawn ($\binom {9-n}{9-2n}$), and choose the one card from each of those cities.  Thus $$\binom 9n \times 3^n\times \binom {9-n}{9-2n}\times 3^{9-2n}$$ 
We need to sum these from $n=0$ to $n=4$ and divide by the total number of hands, namely $\binom {27}9$.  Thus $$p=\sum_{n=0}^4 \binom 9n \times 3^n\times \binom {9-n}{9-2n}\times 3^{9-2n}\Bigg / \binom {27}{9}=\frac {3523257}{4686825}=.7517$$
So, barring arithmetic error, the answer to your question is $$1-p=\boxed {.2483}$$
A: Sure, inclusion-exclusion (as you've done) is one way to do it.  It's nice to check by doing it another way when you can.  In this case, consider the ways to choose $9$ cards without choosing all three of any one "rank".  You must either have $9$ singletons, or $7$ singletons and a pair, or $5$ singletons and two pairs, or $3$ singletons and $3$ pairs, or one singleton and $4$ pairs.  Each singleton or pair can be chosen in three ways from its "rank".  So this amounts to
$$
3^9+\frac{9!}{7!}\cdot3^8+\frac{9!}{5!2!2!}\cdot3^7+\frac{9!}{3!3!3!}\cdot3^6+\frac{9!}{4!4!}\cdot3^5=3523257
$$
ways to not get three of any one "rank".  Dividing this by the ${{27}\choose{9}}=4686825$ draws gives a
$$
\frac{3523257}{4686825}\approx 75.174\%
$$
chance of not getting three of a kind, which doesn't quite agree with your stated chance of getting one.  You've done two small things wrong.  First, when subtracting the double-counted cases (where you've drawn two sets of three and three leftovers), there are ${{9}\choose{2}}=36$ (not just $8$) ways to choose the two ranks.  Second, when adding back in the double-counted cases from that set (where you've drawn three sets of three), there are ${{9}\choose{3}}=84$ possibilities (not just $7$).  So you should get
$$
\frac{9\cdot{{24}\choose{6}} - 36\cdot {{21}\choose{3}} + 84}{{27}\choose{9}}=\frac{1163568}{4686825}\approx 24.826\% ...
$$
this is now in exact agreement with the other approach.
A: Generating Function Approach
For each "suit" there is $\binom30=1$ way to choose $0$ cards, $\binom31=3$ ways to choose $1$ card, $\binom32=3$ ways to choose $2$ cards, and $\binom33=1$ way to choose $3$ cards.
Therefore, I counted the number of ways to draw no more than $2$ of each suit to be
$$
\begin{align}
&\left[x^9\right]\left(1+3x+3x^2\right)^9\\[6pt]
&=\left[x^9\right]\left((1+x)^3-x^3\right)^9\\[3pt]
&=\left[x^9\right]\left(\binom90(1+x)^{27}-\binom91(1+x)^{24}x^3+\binom92(1+x)^{21}x^6-\binom93(1+x)^{18}x^9+\dots\right)\\
&=\binom90\binom{27}{9}-\binom91\binom{24}{6}+\binom92\binom{21}{3}-\binom93\binom{18}{0}\\[6pt]
&=3523257
\end{align}
$$
Whereas, the number of ways to draw $9$ cards is
$$
\begin{align}
\left[x^9\right]\left(1+3x+3x^2+x^3\right)^9
&=\left[x^9\right](1+x)^{27}\\
&=\binom{27}{9}\\
&=4686825
\end{align}
$$
this gives a probability for drawing at least one triple to be
$$
1-\frac{3523257}{4686825}\doteq0.24826359
$$

Inclusion-Exclusion
Number of ways to draw a triple
$$
\overbrace{\overbrace{\ \ \ \binom{9}{1}\ \ \ }^{\substack{\text{choices}\\\text{of suit}}}\overbrace{\ \ \binom{24}{6}\ \ }^{\substack{\text{choices}\\\text{for rest}}}}^{\text{one triple}}-\overbrace{\overbrace{\ \ \ \binom{9}{2}\ \ \ }^{\substack{\text{choices}\\\text{of suits}}}\overbrace{\ \ \binom{21}{3}\ \ }^{\substack{\text{choices}\\\text{for rest}}}}^{\text{two triples}}+\overbrace{\overbrace{\ \ \ \binom{9}{3}\ \ \ }^{\substack{\text{choices}\\\text{of suits}}}\overbrace{\ \ \binom{18}{0}\ \ }^{\substack{\text{choices}\\\text{for rest}}}}^{\text{three triples}}=1163568
$$
Number of hands
$$
\binom{27}{9}=4686825
$$
Giving the probability of getting a triple to be
$$
\frac{1163568}{4686825}=0.24826359
$$
