If a fair dice is rolled $5$ times, what is the probability that some number appears more than once? A fair dice is rolled $5$ times. Find the probability that at least one of the possible outcomes $(1,2,3,4,5,6)$ appears more than once. A hint is the Birthday Problem.
I keep getting $0.9074$, but that is apparently wrong. Can someone help me?
 A: This probability is $1$ minus the probability that all outcomes are different, just like in the birthday problem. 
The first throw can be anything (probability 1), the second must be unequal the first so probability $\frac{5}{6}$, the third unequal to the previous 2, so $\frac{4}{6}$, and so on till the fifth, which has two options left, so $\frac{2}{6}$. Multiply these, and take one minus that.
This comes to (if I computed correctly) to $\frac{49}{54} = 0.9074074074\ldots$, so I think your answer was correct in the first place (maybe an exact answer was asked for instead of a decimal approximation?) It's also the same as the answer by ADG.
A: $${\rm P}=1-\frac{\binom655!}{6^5}$$
A: You have $6^5$ total possible outcomes
But you want outcome of everything but every number appears exactly once.
12345
23456
13456
12456  etc...
There are $_6P_5$ possible ways to have such outcomes (since you roll a die at a time, order matters, so it is permutation)
Therefore, as ADG answered,
\begin{align}
P &= 1-\frac{_6P_5}{6^5} = 1-\frac{\frac{6!}{(6-5)!}}{6^5} \\
&= 1-\frac{5!}{6^4} = 0.907407407
\end{align}
hope it helps.
