<s>[INST] Probability that I will get a $4$ consecutive numbers I have five dice to roll. I roll them. What is the probability that I will get a straight with exactly four consecutive numbers and not $5$?
There are three options: $1,2,3,4$ or $2,3,4,5$ or $3,4,5,6$.
I have $1,2,3,4,*$ where $*$ can be either $1/2/3/4/6$. It cannot be five. Now we do $4!*5$.
Next, I take $2,3,4,5,*$ where $*= 2,3,4,5$. We get $4!*4$ and $4!*5$ for $3/4/5/6$.
I get the probability $(4!*14)/6^5$ but there are some complexities involved. Can anyone explain this clearly? I tried $4!*5*5$ since the position of * is variable. Now, I get $(4!*70)/6^5$ which is not correct either.
Can someone explain it systematically?
 [/INST] As you have noticed, there are $14$ possible multisets of $5$ numbers that are "good" in your sense, of containing four consecutive numbers but no five consecutive numbers. These $14$ multisets can be divided into two kinds:


*

*$12$ of them with some number repeated, namely: $\{1, 2, 3, 4, 1\}, \{1, 2, 3, 4, 2\}, \{1, 2, 3, 4, 3\}, \{1, 2, 3, 4, 4\}, \{2, 3, 4, 5, 2\}, \{2, 3, 4, 5, 3\}, \{2, 3, 4, 5, 4\}, \{2, 3, 4, 5, 5\}, \{3, 4, 5, 6, 3\}, \{3, 4, 5, 6, 4\}, \{3, 4, 5, 6, 5\}, \{3, 4, 5, 6, 6\}$

*$2$ of them with all five numbers distinct, namely: $\{1, 2, 3, 4, 6\}$ and $\{3, 4, 5, 6, 1\}$.


For each of the latter two multisets, it can be ordered in $5!$ ways, so the probability that you get exactly that multiset is: $$\frac{5!}{6^5}$$
For each of the former twelve multisets, it can be ordered in $\frac{5!}{2!}$ ways, so the probability that you get exactly that multiset is:
$$\frac{5!}{2! 6^5}$$
So the answer you want is:
$$ 2\frac{5!}{6^5} + 12\frac{5!}{2! 6^5} = \frac{2(5!) + 6(5!)}{6^5} = \frac{960}{6^5} = \frac{960}{7776} = \frac{10}{81}$$
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