probability of throwing three adjacent numbers When throwing one dice 3 times in a row, what is the probability of getting adjacent numbers in right order, for example 2,3,4 or 4,3,2?
 A: Record the results of the throws as an ordered triple $(a,b,c)$, where each of $a,b,c$ is a positive integer between $1$ and $6$.
There are $6^3$ such triples, all equally likely.
Now we count the number of favourables, that is, ordered triples in which the numbers obtained are consecutive, in ascending or descending order.
We can choose the order in $2$ ways, ascending or descending. For ascending, the smallest number can be any of $1,2,3,4$, and then the rest are determined. The same is true of descending. So the number of favourables is $(2)(4)$.
It follows that the required probability is $\dfrac{(2)(4)}{6^3}$.
A: There are exactly $8$ of these series of adjacent numbers. Any series has probability $6^{-3}$ to occur so...
A: Simply put, the posible combinations are:
$$\begin{array}{ccc}
\{1,2,3\}&&
\{3,2,1\}\\
\{2,3,4\}&&
\{4,3,2\}\\
\{3,4,5\}&&
\{5,4,3\}\\
\{4,5,6\}&&
\{6,5,4\}\\
\end{array}$$
So, there's $8$ useful combinations, out of $6^3=216$ total posible combinations. Then
$$P=\frac{8}{216}$$

We can generalize the result, obviously: say we want a set of $k$ adjacent numbers out of $n$ total sides.
Then, we have a total of $n^k$ total possible combinations ($k$ throws of an $n$-sided dice), and 
$$2(n-k+1)$$ 
useful combinations: you only can form $n-k+1$ ordered subsets of adjacent numbers (the $k$ first adjacent numbers, the $k$ second, and so on); and we are allowing adjacent numbers forwardly and backwardly ($2$ times).
Finally, the probability is
$$P_n(k)=\frac{2(n-k+1)}{n^k}$$
