<s>[INST] Probability of rolling 2 and 3 numbers in a sequence when rolling 3, 6 sided dice I am wondering what the probability is to roll numbers in a sequence with 3, 6 sided dice.
" If 3 fair six-sided dice are rolled. What is the probability that at least two of the numbers will be in a sequence (ex: 1,2 or 5,6). Also, what is the probability that all 3 of them will be in a sequence. These dice are all rolled at one time and can be put in any order."
I've tried solving it but I couldn't get it. What is the probability to get these outcomes.
 [/INST] An approach to this problem, a bit lengthy but having the advantage to provide a clear picture,
might be the following.
Start from considering the dice marked.
The set of possible, equi-probable, outcomes is represented by $6^3=216$ triples.
Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever
$$
\begin{array}{c|ccc}
   {die} &  &  1 & 2 & 3  \\
   {result} &  &  {1 \le k \le 5} & {k + 1} & \forall   \\
   {prob} &  &  {5/6} & {1/6} & 1  \\
\end{array}
$$
the probability of getting such a scheme is $5/36$.
Now, since in our problem order does not matter, we shall swap (permute) the above configuration.
But we cannot perform that without considering the  value of die 3 (call it $j$) compared with the result of die 1 and 2.
In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways.
Moreover, we shall exclude the permuted triples that  fall within the range of those already considered.
So, the prospect of the possible ordered configurations and number of ways to swap them is the following
$$ \bbox[lightyellow] {  
\begin{array}{*{20}c}
   {config.} & {N.\,ordered} & {N.\,swaps} & {Tot.}  \\
\hline
   {\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60}  \\
   {\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15}  \\
   {\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15}  \\
   {\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0  \\
   {\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36}  \\
\hline
   {{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm  = } & {126}  \\
\end{array}
} $$
We see that  the fourth configuration is cancelled as being already included in the first.
The prospect for the complementary case (no consecutive outcomes) will give
$$ \bbox[lightyellow] {  
\begin{array}{*{20}c}
   {config.} & {N.\,ordered} & {N.\,swaps} & {Tot.}  \\
\hline
   {\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6  \\
   {\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30}  \\
   {\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30}  \\
   {\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18}  \\
   {\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6  \\
\hline
   {{\rm none}\,{\rm consecutive}} & {} & {\rm  = } & {90}  \\
\end{array}
} $$
In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have
$$
\begin{array}{c|ccc}
   {die} &  &  1 & 2 & 3  \\
   {result} &  &  {1 \le k \le 4} & {k + 1} & {k + 2}   \\
   {prob} &  &  {4/6} & {1/6} & {1/6}  \\
\end{array}
$$
and since each possible triple has distinct values, we can permute them to obtain: 
$$ \bbox[lightyellow] {  p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$.
You can verify by direct counting that the values above are correct.
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