Probability - mixing cards The package contains 8 different cards, two of each color (that means 4 colors). We will mix the package. How likely we get combination in which no two cards of the same color are next to each other? 
There are $8!$ combinations how to mix the package. Now, I have to compute the number of combinations in which no two cards of the same color are next to each other. No idea how to do that.
Maybe we place 4 cards of different color randomly (in $8*7*6*5$ ways) and then - I don't know. Maybe some Inclusion-Exclusion Principle?
 A: Solution by the Principle of Inclusion/Exclusion: There are $$N = \frac{8!}{2^4} = 2520$$ ways to arrange the 8 cards, each of which we assume is equally likely.  Say an arrangement has "Property $i$" if the two cards of color $i$ are adjacent.  Let $S_i$ be the number of arrangements which have $i$ of the properties.  Then
$$\begin{aligned}
S_1 &= \binom{4}{1} \frac{7!}{2^3} \\
S_2 &= \binom{4}{2} \frac{6!}{2^2} \\
S_3 &= \binom{4}{4} \frac{5!}{2^1} \\
S_4 &= \binom{4}{4} \frac{4!}{2^0} 
\end{aligned}$$ and the number of arrangements which have none of the properties, i.e. have no two cards of the same color adjacent, is
$$N_0 = N - S_1 + S_2 - S_3 + S_4 = 864$$
So the probability that no two cards of the same color are adjacent is
$$\frac{N_0}{N} \approx 0.342857$$
A: You have incorrectly counted the number of distinguishable arrangements. 
Suppose the colors are blue, green, red, and yellow.  A particular arrangement is determined by selecting two of the eight positions for the blue cards, two of the remaining six positions for the green cards, two of the remaining four positions for the red cards, and then filling the two remaining positions with yellow cards.  These selections can be made in 
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{8!}{2!2!2!2!}$$
The four factors of $2!$ in the denominator represent the number of ways we can permute cards of the same color among themselves without creating an arrangement that is distinguishable from the given arrangement.  
From these, we must exclude those arrangements in which two cards of the same color are adjacent.  We count those arrangements.
Suppose two cards of the same color are adjacent.  Place them in a box of that color.  We now have seven objects to arrange, the box and the six cards of the remaining three colors.  A similar argument to that given above shows that these objects can be arranged in 
$$\frac{7!}{2!2!2!}$$
distinguishable ways.  Since there are four possible choices for the color of the adjacent cards, the number of arrangements in which two adjacent cards are the same color is 
$$\binom{4}{1}\frac{7!}{2!2!2!}$$ 
Suppose that two pairs of adjacent cards of the same color appear.  Place each pair of cards in a box of the corresponding color.  We are left with six objects to arrange, the two boxes and the four cards of the two remaining colors - two from each of those colors.  They can be arranged in 
$$\frac{6!}{2!2!}$$
distinguishable ways.  Since there are $\binom{4}{2}$ ways to select two of the four colors, there are 
$$\binom{4}{2}\frac{6!}{2!2!}$$
arrangements in which two pairs of adjacent cards of the same color appear.
Now suppose three pairs of adjacent cards of the same color appear.  Place each such pair of cards in a box of that color.  That gives us five objects to arrange, the three boxes and the two remaining cards of the other color.  These objects can be arranged in 
$$\frac{5!}{2!}$$
distinguishable ways.  Since three of the four colors can be selected in $\binom{4}{3}$ ways, the number of arrangements in which three pairs of adjacent cards of the same color appear is 
$$\binom{4}{3}\frac{5!}{2!}$$
Finally, suppose that all four pairs of adjacent cards of the same color appear.  Place each pair in a box of that color.  The boxes can be arranged in 
$$4!$$ 
ways. 
By the Inclusion-Exclusion Principle, the number of arrangements of eight cards, consisting of two cards each of four different colors, in which at least two consecutive cards are of the same color is
$$\binom{4}{1}\frac{7!}{2!2!2!} - \binom{4}{2}\frac{6!}{2!2!} + \binom{4}{3}\frac{5!}{2!} - \binom{4}{4}4!$$ 
so the number of arrangements in which no two cards of the same color are adjacent is
$$\frac{8!}{2!2!2!2!} - \binom{4}{1}\frac{7!}{2!2!2!} + \binom{4}{2}\frac{6!}{2!2!} - \binom{4}{3}\frac{5!}{2!} + \binom{4}{4}4!$$ 
The probability that no two cards of the same color are adjacent is found by dividing this quantity by 
$$\frac{8!}{2!2!2!2!}$$
Notice that the probability is equal to $1$ minus the probability that at least two cards of the same color are adjacent.
A: Comment: Originally, I misread the Question. This is a revised Answer mutatis mutandis, showing a 
simulation in R statistical software of the actual problem, auxiliary to the elegant
combinatorial solution of @awkward.
With a million random arrangements of the contents of a box, the result should be
accurate to two or more places.
[Notes on simulation code: First, arr is a permutation of the "colors" 1. 2, 3 and 4. Second, if two adjacent
numbers are the same, then the vector diff(arr) will contain a 0.
Finally, the logical vector no.adj has TRUE when there are no
adjacencies, otherwise FALSE.] 
m = 10^6;  box=c(1,1,2,2,3,3,4,4)  # colors 1 through 4
no.adj = logical(m) # at end, m-vector of TRUE's and FALSE's
for(i in 1:m) {
   arr = sample(box, 8)
   no.adj[i] = (min(abs(diff(arr))) > 0) }
mean(no.adj)  # mean of logical vector is proportion of its TRUE's
## 0.342915   # aprx P(No Adjacencies)

