Odds of drawing any 3 identical cards from 27-card deck (9 unique x 3 copies each) in 9 draws

Question

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?

Answer 1

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.

Answer 2

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 $$1-p=\boxed {.2483}$$

Answer 3

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 $$ 1-\frac{3523257}{4686825}\doteq0.24826359 $$

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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 $$ \frac{1163568}{4686825}=0.24826359 $$