Computing card probabilities

Question

$1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

1. All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.

2. Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.

3. Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.

4. There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).

5. All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.

Are my solutions correct? If not, where did I go wrong?

Answer 1

Your solutions to questions 1, 2, 3, and 5 are correct.

Let's examine problem 4. If we draw at least one card with the number 12 on it but no cards with the number 13 on it, then there are three possibilities:

1. We draw exactly one 12 and two cards with numbers smaller than 12, which can occur in $\binom{4}{1}\binom{44}{2}$ ways.
2. We draw exactly two 12s and one card with a number smaller than 12, which can occur in $\binom{4}{2}\binom{44}{1}$ ways.
3. We draw three 12s, which can occur in $\binom{4}{3}$ ways.

Since these three cases are mutually exclusive and exhaustive, the number of favorable cases is $$\binom{4}{1}\binom{44}{2} + \binom{4}{2}\binom{44}{1} + \binom{4}{3}\binom{44}{0}$$ Alternatively, we must draw three cards numbered less than 13 without drawing three cards less than 12, so the number of favorable cases is $$\binom{48}{3} - \binom{44}{3}$$

By designating one 12 as the 12 in the hand, you count each case with more than 12 multiple times, once for each way of designating one of the 12s in the hand as the 12 in the hand. Therefore, your method counts each hand with two 12s twice and each hand with three 12s three times.

To illustrate, suppose your hand consists of 12 blue, 12 green, and 7 red. You count this hand twice:

\begin{array}{l l} \text{designated}~12 & \text{two additional cards}\\ \hline 12~\text{blue} & 12~\text{green}, 7~\text{red}\\ 12~\text{green} & 12~\text{blue}, 7~\text{red} \end{array}

If your hand consists of 12 blue, 12 green, and 12 red, you count this hand three times:

\begin{array}{l l} \text{designated}~12 & \text{two additional cards}\\ \hline 12~\text{blue} & 12~\text{green}, 12~\text{red}\\ 12~\text{green} & 12~\text{blue}, 12~\text{red}\\ 12~\text{red} & 12~\text{blue}, 12~\text{green} \end{array}

Notice that $$\color{red}{\binom{4}{1}\binom{47}{2}} = \binom{4}{1}\binom{44}{2} + \color{red}{\binom{2}{1}}\binom{4}{2}\binom{44}{1} + \color{red}{\binom{3}{1}}\binom{4}{3}\binom{44}{0}$$

Answer 2

The first three and the last one are correct.

In the fourth one you’ve overcounted. Specifically, you’ve counted each set of $3$ that contains exactly two $12$s twice, once for each $12$, and each set of $3$ that includes exactly three $12$s three times, once for each $12$: any one of the $12$s can be the one that you singled out at the beginning, and you’re counting each of those possibilities separately.

It’s easier here to count the sets of $3$ that contain no $12$s or $13$s and subtract that from number of sets of $3$ that contain no $13$s. There are $\binom{44}3$ ways to choose $3$ cards that are neither $12$s nor $13$s, and there are altogether $\binom{48}3$ ways to choose $3$ cards that are not $13$s, so there are $\binom{48}3-\binom{44}3$ ways to choose $3$ cards that include at least one $12$ but no $13$s.