# Dice throwing probability of three different faces.

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$= (1-(1/6)^2)= 1-(1/6^2)=1-(1/36)$

• Welcome to MSE! It really helps readability to format questions using MathJax (see FAQ). I recommend making titles to be representative of the problem in case others are looking for similar problems. Regards – Amzoti Apr 18 '13 at 1:11

You’re right that the first roll can be anything and that the probability of getting something different on the second roll is $1-\frac16=\frac56$. However, the third roll has to differ from both of the first two in order for you to get three different numbers, and that happens with probability $\frac46=\frac23$. Moreover, you need to multiply the probabilities, so that even if the correct probability for the third roll were $1-\frac16$, your answer would be $\left(1-\frac16\right)^2=\left(\frac56\right)^2$, not $1-\left(\frac16\right)^2$.
Thus, the actual probability of getting three different numbers is $$\frac56\cdot\frac23=\frac59\;.$$
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 $$\dfrac{(6)(5)(4)}{6^3}.$$