Probability of rolling a "1" on a die conditioned on when all rolls are different. So I have this problem:
I am rolling a six sided die 3 times. 
Conditioned on the rolls all being different, what's the probability at least one die is a "1"
So I worked it out like this:
probability of not getting a 1 on the first roll is 5/6
probability of not getting a 1 on the second roll is 4/6
probability of not getting a 1 on the second roll is 3/6
I then just did (5/6) * (4/6) * (3/6) to get 60/216 possible conditions where you would not roll a 1. 
Doing 216-60 you get that 158/216 possible solutions (or 13/18) possible solutions for rolling a "1" when all numbers are different. Does this make sense? The number seems a bit large and I am not sure how to check it. 
Thank you in advance. 
 A: You're almost there, but you made a small error:
Hint: How many possible outcomes are there on the second roll?
A: You can interpret this as electing $3$ distinct elements of set $\left\{ 1,2,3,4,5,6\right\}$. The probability for a specific number to be elected is $\frac{3}{6}=\frac{1}{2}$. This is also true for number $1$.
Alternatively the number of subsets $\left\{ i,j,k\right\} $ with
$\left|\left\{ i,j,k\right\} \right|=3$ is $\binom{6}{3}$, and the
number of subsets $\left\{ 1,j,k\right\} $ with $\left|\left\{ 1,j,k\right\} \right|=3$
is $\binom{5}{2}$ leading to probability: $$\binom{5}{2}\binom{6}{3}^{-1}=\frac{1}{2}$$
A: This question is probably simplest if we disregard the order in which the dice are rolled. Imagine instead of rolling the same die three times, you have three identical dice in your hand and throw them all at once.
A successful outcome is one in which one of the dice shows a "1" and the other two dice are distinct members of the set $\{2, 3, 4, 5, 6\}$. There are $\binom{5}{2}$ successful outcomes.
The number of outcomes in which all faces are distinct correspond to any selection of three distinct members of the set $\{1, 2, 3, 4, 5, 6\}$. There are $\binom{6}{3}$ total outcomes.
The probability of a success is therefore
$$
\frac{\binom{5}{2}}{\binom{6}{3}} = \frac{1}{2}.
$$
