Rolling dice statistics, probability of rolling a straight 

Suppose you roll $n$ die at the same time. Let $E_i$ be the event that we don't roll any $i$, $1\leq i\leq 3$ Then 

\[P(E_i)=1-(\frac{5}{6})^n,\]
\[P(E_i \cap E_j)=1-(\frac{4}{6})^n,      i\neq j,\]
\[P(E_1 \cap E_2 \cap E_3)=1-(\frac{3}{6})^n.\]

The probability of not having a straight 123 is
\[P(E_1\cup E_2 \cup E_3)=P(E_1)+P(E_2)+P(E_3)\]
\[-P(E_1\cap E_2)-P(E_1\cap E_3)-P(E_2\cap E_3)+P(E_1\cap E_2\cap E_3)\]
\[=3(1-(\frac{5}{6})^n)-3(1-(\frac{4}{6})^n)+1-(\frac{3}{6})^n\]
\[=1+3((\frac{4}{6})^n-(\frac{5}{6})^n)-(\frac{3}{6})^n\]
\[=\frac{6^n+3\times4^n-3\times 5^n-3^n}{6^n}.\]
So the probability of rolling a straight 123 is
\[1-\frac{6^n+3\times4^n-3\times 5^n-3^n}{6^n}=\frac{-3\times4^n+3\times 5^n+3^n}{6^n}.\]
Probability of rolling 234, 345, and 456, are the same as probability of rolling 123. 

The probability of failure is
\[(*)   P_{failure}={6 \choose 1}\times(\frac{1}{6})^n+{6 \choose 2}\times(\frac{2}{6})^n+({6 \choose 3}-4)\times(\frac{3}{6})^n+(\frac{4}{6})^n.\]
And the probability of success is simply $1-P_{failure}$.

Indeed the first term in (*) is the probability of failing by getting the same number for all n die. The second terms is the probability of getting only two numbers. The third term is the probability of getting exactly three numbers and only 4 cases (123, 234, 345, 456 ) will be a success. The fourth term is probability of getting exactly 4 numbers and only one case would lead to a failure 1245. If we get 5 or 6 numbers we will succeed.