Probability Question on a dice roll - Conditional probability

Question

I have a question which I have a hard time wrapping my head around it.

If I roll a dice three times, and I know that I got three different numbers, what is the probability that one of the numbers is $2$?

I thought to first calculate the probability to get different numbers which is:

$$\frac{6\cdot5\cdot4}{6^3} = \frac{5}{9}$$

Then I know you need to use here Bayes' law since we know we got different numbers. So it is:

$P$(getting $2$| getting different numbers) = $P$(getting different numbers $\cap$ getting $2$ | getting different numbers)

Here I got stuck. I think it is the probability of getting $2$ ($1/6$) and then divide it by $5/9$.

Answer 1

If you take into account the order of the rolls,

• there are $6 \times 5 \times 4 =120$ equally likely ways of rolling three dice and getting different values
• Of these, there are different ways of getting a $2$:
  • $1 \times 5 \times 4=20$ if the $2$ comes first
  • $5 \times 1 \times 4=20$ if the $2$ comes second
  • $5 \times 4 \times 1=20$ if the $2$ comes third

So the overall conditional probability is $\frac{20+20+20}{120}=\frac12$ of getting a $2$ if three different values are rolled.

A faster way is to say that there are three different values shown and three not shown. By symmetry, $2$ is equally likely to be in either set so has a conditional probability of $\frac36=\frac12$ of being rolled.

Answer 2

Let's think about the probability that you don't see a 2. On the first roll of the dice, you have 6 options: $1,2,3,4,5,$ and $6$, which means your probability of not rolling a $2$ is $5/6$. Suppose you roll a $6$. Then, since you require all rolls to be distinct, you have 5 options for the next roll: $1,2,3,4,$ and $5$, meaning your probability of not rolling a $2$ is $4/5$. Say you roll a $3$. Then you have 4 options for your next roll: $1,2,4,$ and $5$, which means your probability of not rolling a $2$ is $3/4$. So, out of three distinct rolls, the probability of never rolling a $2$ is $$\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} =\frac{1}{2}. $$ Then, since the probability of never rolling a $2$ is $1/2$, the probability of seeing at least one $2$ is $1-\frac{1}{2}$=$\frac{1}{2}$.