TITLE: Compute the probability that the outcome of a die's fifth throw is one of the previous 4
QUESTION [5 upvotes]: A die is randomly rolled five times, the probability that outcome of
the fifth throw is one of the outcomes of the first 4 throws is?

Attempt:
For first 4, there are following possibilities:  (after that we will multilply the possibilities for the fifth one)

All different

2 same 2 different

3 same 1 different

All same


$1.$ All different:
Number of ways: $(6*5*4*3)\times\dbinom 41 = 1440$
$2$. $2$ same, $2$ different:
Number of ways: $\dfrac{4!}{2!}\dbinom 63 \times \dbinom 31 = 720$
$3$. $3$ same, $1$ different:
Number of ways: $\dfrac{4!}{3!}\dbinom 62 \times \dbinom 21 = 120$
$4$. All same:
Number of ways: $\dbinom 61 \times \dbinom 11 = 6$
Thus, $p(E) = \dfrac{@1+@2+@3+@4}{6^5} = \dfrac{381}{6^4}$
But answer given is: $\dfrac{671}{6^4}$
What's my mistake?
Edit:
Forgot to use: (as pointed out by @anryvian in comments)
$5.$ 2 same, 2 same:
Number of ways: $\dfrac{4!}{2!2!} \times \dbinom 6 2 \times 2 = 180$
Which gives the answer as: $\dfrac{411}{6^4}$ (still incorrect :( )

REPLY [6 votes]: An easier approach: the complementary event is "none of the first four throws is the same as the fifth throw."

 $$1-\left (\frac{5}{6}\right )^4$$

I haven't yet found how to fix your attempt (which is a valid approach), but hopefully I (or someone else) catches it soon.

Corrected computations:

$1440$ is correct.
$\frac{4!}{2!} 6 \cdot \binom{5}{2} \cdot 3 = 2160$ (You have to pick which number is the pair and which are the singles)
$\frac{4!}{3!} 6 \cdot  5 \cdot 2 = 240$. (You have to pick which number is the triple and which is the single)
$6$ is correct.
$180$ is correct.