1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.