Four standard dice are rolled.
What are the chances of:
1.) Rolling three sixes and one five?
2.) Two sixes and two fives?
3.) Exactly one six?
4.) Four different numbers?
Here is my thought process for them, please help me develop an intuition for these sort of questions and/or correct me:
1.) Is it safe to assume that on the first dice I roll and get a five. The probability of that is $\frac{1}{6}$, and then the probability of rolling three sixes is $\frac{1}{6^3} $, with the final probability being $\frac{1}{6^4}$.
2.) Again… I'm not sure I'm doing this correctly. Wouldn't it be the same as 1?
3.)Exactly one six would be $\frac{1}{6}*\frac{5^3}{6^3}$
4.) Four different numbers would be $\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6}$
Best Answer
There are $6^4=1296$ different outcomes.
There are $\frac{(3+1)!}{3!\cdot1!}=4$ ways to do it, hence the probability is $\frac{4}{1296}$
There are $\frac{(2+2)!}{2!\cdot2!}=6$ ways to do it, hence the probability is $\frac{6}{1296}$
There are $\binom41\cdot(6-1)^{4-1}=500$ ways to do it, hence the probability is $\frac{500}{1296}$
There are $\binom64\cdot4!=360$ ways to do it, hence the probability is $\frac{360}{1296}$