# 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.

• Hint for the first question: converse probability. What is the probabilty that none of the outcomes of the three die are in sequence ? – callculus May 5 '17 at 18:05
• I'm not sure if I could figure that out I would know the answer to the probability of rolling the desired outcome – Aaron May 5 '17 at 18:29

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.

• So there would be a 83% chance to roll 2 numbers in a row in a linear sequentiality? – Aaron May 5 '17 at 18:34
• @Aaron: yes $5/6$, when order is not important – G Cab May 5 '17 at 21:32
• The person below said there is a 94.4% chance? – Aaron May 5 '17 at 21:42
• @Aaron: I totally recasted my answer: results have been checked vs. direct computation. – G Cab May 6 '17 at 14:29

Sequences with the numbers $1,3,5$ and $2,4,6$ have no consecutive numbers.

$1,3,5$ can be ordered in $3!=6$ ways. Similar for $2,4,6$. In total we have $12$ outcomes where the numbers are not consecutive. And we have overall $6^3=216$ possible outcomes.

Thus the probacbility that the sequence has no consecutive numbers is $P(\overline C)= \frac{12}{216}$.

Consequently the probabbility that the sequence has at least consecutive two numbers is

$P( C)=1-P(\overline C)=1- \frac{12}{216}=\frac{216}{216}-\frac{12}{216}=\frac{204}{216}$

• Awesome thanks so much for the help! I thought it was very likely to get 2 in a sequence but wasn't sure how likely. Is there any way you can help with getting three in a row? – Aaron May 5 '17 at 19:45
• The consecutive sequences has the numbers $(1,2,3), (2,3,4), (3,4,5),(4,5,6)$ They all can be ordered in 6 ways each. Does it help ? – callculus May 5 '17 at 20:10
• So there would be a total of 24 rolls that would lead to three numbers in sequence? Meaning that there is a 11.11% chance? – Aaron May 5 '17 at 20:15
• @Aaron Yeah, or in fractional notation $\frac19$. – callculus May 5 '17 at 20:28
• @callculus you're forgetting sequences like 1,1,3 and 4,4,6 where one of the numbers is repeated. 24 is the right number for 3-sequences. – stretch May 5 '17 at 22:49

Hint - For sequence on two dice.

You have to consider all different cases. If sequence of 2 numbers on 1st and 2nd dice. Then it should not be 6 on first die. Then number on first die $\frac 56$ and on second $\frac 16$ then on last one it can be not in sequence so it can be any of the remaining 5 numbers so $\frac 56$.

Similarly case of sequence on 2nd and 3rd dice.

The case of 3 sequential numbers is much easier to calculate than the 2 case. There are only 4 possible sequences: 123, 234, 345, 456. Each of them can be rolled 6 different ways: for example, for 345 the dice can be 345, 354, 435, 453, 534, or 543. So there are 24 possible 3-sequences out of 216 possible rolls.

Table of all the possibilities: