What are the chances of drawing this unique combination of 3 coloured dice from 6 dice?

Question

I'm having trouble understanding why one (or both) of my methods is wrong. Here is the problem.

Let's say I have 6 dice: 2 yellow, 2 green, 2 red. I'm going to draw 3 dice from the 6 dice, and I want to know the chance that I draw one of each colour. Here is the first method that I think is correct:

I number each dice 1 to 6. I figure out every combination of 3 numbers from the 6 numbers. I translate the numbers back into colours. Since there are pairs of numbers that are the same colour, I then go through the list counting the permutations, so I end up with a list of unique combinations of 3 colours. I get 7 combinations:

RRG x 2

RRY x 2

RGG x 2

RGY x 8

RYY x 2

GGY x 2

GYY x 2

Now, here's the first method I use to calculate the chance of drawing, say, RYG: $P(RYG) = \frac{2}{6}\frac{2}{5}\frac{2}{4}(duplicates) = \frac{1}{15}2=0.0666$

Here's the second method: There are 20 total combinations and 8 of them are RYG, so surely $P(RRG)=2/5=0.4$

Now, I'm certain that my work here is so sloppy that it makes your skin crawl. I admit I'm fumbling about here. Sorry. If someone can point me in the right direction, I'd appreciate it.

Answer 1

You have all of the combinations correct, so you're basically there.

There are $8$ ways to draw one of each color (you wrote "$RGY \times 8$"). (Choose which of two yellows, choose which of two reds, and choose which of two greens.) There are $_6C_3 = 20$ total ways to draw three dice from six. (You instead wrote all of the combinations explicitly, for which the total is $20$.)

Hence your probability for drawing one of each color with your three is $P = \frac{8}{20} = 0.4$.

Another way to look at it is to take $1$ minus the probability that you don't get one of each color. You have that information too; there are $12$ combinations that have a repeated color.

The answer is the same:

$$P = 1 - \frac{12}{20} = 0.4.$$

Answer 2

The probability that the first die you select is a distinct color is $1$. Since $4$ of the $5$ remaining dice are of a different color than the first one you selected, the probability that the second die you select is of a different color than the first die you selected is $4/5$. After those two dice are selected, $2$ of the remaining $4$ dice are of the third color. Therefore, the probability of selecting three different colors is $$1 \cdot \frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$$