TITLE: Dice throwing probability of three different faces.
QUESTION [8 upvotes]: A standard six side die is rolled three times. 
Find the chance that three different faces appear?
My thoughts:
Probability p(same face pops up is)= $1- P$. You roll first one and record what you get. The probability of next (2) two rolls give you same face is...? My chances of rolling any given face is $1/6$. It does not matter what you get first role the next two rows have to the same as the first one role. Therefore$= (1-(1/6)^2)= 1-(1/6^2)=1-(1/36)$

REPLY [3 votes]: Record the throws as described in the post. There are $6^3$ equally likely possibilities.
Call a record good if it has $3$ different entries. We count the number of good records. The first entry can have any of $6$ values. For each such value, the second entry can have any of $5$ values. And for every choice of the first $2$, there are $4$ choices for the third, for a total of $(6)(5)(4)$. Thus our probability is
$$\dfrac{(6)(5)(4)}{6^3}.$$