Why is this probability > 1?

Question

I'm going to blow my brains out.

I have 27 cards of red, blue and green cards. There is 9 of each color. I draw 12 cards. What is the probability that I have AT LEAST 6 blues, AT LEAST 1 red and AT LEAST 1 green.

I saw other problems online that I thought were similar ("...at least one ace in 5 draws") and yet this problem eludes me.

My reasoning: There is a total of (27 choose 12) ways to pick any combination of 12 cards, hence the denominator. There is (9 choose 6) ways to pick from blue, (9 choose 1) ways to pick from red and green. We have used 6+1+1=8 cards leaving 27-8=19 cards remaining in the deck. We still have to draw 4 more cards and since our conditions at this point are satisfied, any 4 of the remaining 19 cards will do, so we append (19 choose 4).

$$ \frac{\binom{9}{6}\binom{9}{1}\binom{9}{1}\binom{19}{4}}{\binom{27}{12}}=1.5\dots $$

I don't care for HOW to solve the problem. I want to know WHY this is wrong. What am I misunderstanding about combinations that is causing this.

Answer 1

Think about a smaller example: 2 red cards and 2 blue cards in the deck: $r_1,r_2,b_1,b_2$. Draw 2 cards. What's the probability you get at least one red card?

All possible 2-card hands are equally likely, so the answer is $\frac{\text{number of different 2-card hands with at least one red card}}{\text{total number of different 2-card hands}}.$

The denominator is $\binom42=|\{r_1r_2,r_1b_1,r_1b_2,r_2b_1,r_2b_2,b_1b_2\}|=6$ and the numerator is $|\{r_1r_2,r_1b_1,r_1b_2,r_2b_1,r_2b_2\}|=5$, so the answer is $\frac56$. (In this case it would have been easier to count the complement.)

The problem you're having is with the numerator. Your approach says: the number of ways to first choose one red card and then choose any one of the remaining 3 cards is $\binom21\binom31=6.$ This is true, but it counts "$r_1$ then $r_2$" and "$r_2$ then $r_1$" as two different ways, whereas you only want to count the hand $r_1r_2$ once (because it was only counted once in the denominator).

Counting ordered pairs instead of unordered pairs is a valid approach, but you have to do it correctly in both the numerator and the denominator. In that case, the denominator is $P(4,2)=12$ and the numerator is (number of ways to first choose a red card and then any other card) + (number of ways to first choose a non-red card and then a red card) = $\binom21\binom31+\binom21\binom21=10$, so the probability is $\frac{10}{12}=\frac56.$

Answer 2

Your approach is incorrect because it double-counts certain distributions.

Here is an example of how your formula double-counts. For each color, number the cards in that color from $1$ to $9$. I will use $B5$ to refer to the fifth blue card, and $G7$ for the seventh green card. Here is one way to follow your instructions:

1. Choose six blue cards. Choose $B1,B3,B4,B7,B8,B9$.

2. Choose one red card. Choose $R2$.

3. Choose one green card. Choose $G4$.

4. Choose four cards of any color. Choose $B5,B6,R6,R9$.

Here is a different way to follow your instructions:

1. Choose six blue cards. Choose $B1,B5,B6,B7,B8,B9$.

2. Choose one red card. Choose $R9$.

3. Choose one green card. Choose $G4$.

4. Choose four cards of any color. Choose $B3,B4,R2,R6$.

Your method of counting would count these as two separate hands, but they produce the exact same hand of cards, so this hand has been double-counted. Actually, this particular hand will be counted more than two times.

Answer 3

Other responses have addressed the issue of where the OP (i.e. original poster) went wrong.

Any approach that I can think of is going to be at least somewhat tedious.

One approach is Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

My initial approach will avoid using Inclusion-Exclusion because the alternative (direct) approach will make the answer very easy to understand.

At the end of my answer, I will provide a 2nd hybrid approach that mixes the direct approach with Inclusion- Exclusion.

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For $~b,~r ~$ each in $~\{0,1,2,\cdots,9\}~$:

• Let $~p(b,r)~$ denote the probability that there were exactly $~b~$ blue cards drawn out of the $~12~$ and exactly $~r~$ red cards drawn out of the $~12.$

Then, for all ordered pairs $~(b,r),~$ where $~3 \leq (b+r) \leq 12~$ :

$$p(b,r) = \frac{\binom{9}{b} \binom{9}{r} \binom{9}{12 - b - r}}{\binom{27}{12}}.$$

Per the constraints of the problem, one way of calculating the answer is then:

$$\sum_{b=6}^9 ~\left[ ~\sum_{r=1}^{11-b} p(b,r) ~\right]. \tag1 $$

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In actual practice, using the formula in (1) above is more tedious than the computation needs to be, because of the use of a double summation.

In my 2nd approach, I will capitalize on the fact that the number of required red and green cards, are $~1~$ each, which is a very small number.

Let $~p(b)~$ denote the probability that exactly $~b~$ blue cards were used, for $~0 \leq b \leq 9.$

Then,

$$p(b) = \frac{\binom{9}{b}\binom{18}{12-b}}{\binom{27}{12}}. \tag2 $$

So, the following computation is $~\color{red}{\text{inaccurate}},~$ because it fails to accommodate the constraint that there must be at least $~1~$ red card, and at least $~1~$ green card.

$$\sum_{b=6}^9 p(b). \tag3 $$

To correct the flaw, I will examine each case of $~b \in \{6,7,8,9\}~$ separately, and for the specific pertinent value of $~b,~$ calculate the probability that there was at least $~1~$ red card and at least $~1~$ green card.

Let $~q(b)~$ denote the probability of having exactly $~b~$ blue cards, where $~b~ \in \{6,7,8,9\} ~$ and having the other $~(12 - b)~$ cards all be red.

Then

$$q(b) = \frac{\binom{9}{b}\binom{9}{12-b}}{\binom{27}{12}}. \tag4 $$

Similarly, by symmetry, $~q(b)~$ is also equal to the probability of having exactly $~b~$ blue cards, where $~b~ \in \{6,7,8,9\} ~$ and having the other $~(12 - b)~$ cards all be green.

Notice that when there are $~b~$ blue cards used, the event that the rest of the cards are all red, and the event that the rest of the cards are all green are disjoint events.

Because of this, I can now compute the final answer as

$$\sum_{p=6}^9 \left[ ~p(b) - 2q(b) ~\right]$$

$$= \sum_{p=6}^9 \left\{ ~\left[ ~\frac{\binom{9}{b}\binom{18}{12-b}}{\binom{27}{12}} ~\right] - \left[ ~2 \times \frac{\binom{9}{b}\binom{9}{12-b}}{\binom{27}{12}} ~\right] ~\right\}$$

$$= \sum_{p=6}^9 ~\left\{ ~\frac{\binom{9}{b}}{\binom{27}{12}} \times \left[\binom{18}{12-b} - 2 \times \binom{9}{12-b} ~\right] ~\right\}.$$