If I roll two six-sided dice where the first die is valued simply 1-6, but the second die is valued as 1-2=0, 3-4=1, and 5-6=2, and I total the two dice, will the probability of the numbers 1-8 be evenly distributed? If not what are the probabilities of the numbers 1-8. Assume the dice are fair. Intuitively, I think the chances will be evenly distributed – or close to even – but, I don't know how to do the math to prove it either way.
[Math] Probability of rolling 1-8 using six-sided dice
diceprobability
Related Solutions
For your first question, six-sided die vs. eight-sided, make a $6$ by $8$ table, with values $1, 2, 3, 4, 5, 6$ in one direction and $1, 2, 3, 4, 5, 6, 7, 8$ in the other direction.
The resulting $48$ small squares in the table determine $48$ possible outcomes of the two dice. You can then see for how many squares does six-sided beat eight-sided; how many ties; and how many times eight-sided beats six-sided. Assuming the dice are fair, you can then divide by $48$ to get the desired probabilities.
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.
Best Answer
You have an unbiased six-sided die and an unbiased three-sided die, so are drawing pairs uniformly distributed on $\{1,2,3,4,5,6\}\times\{0,1,2\}$.
$$\mathsf P(X+Y=1) ~=~ \mathsf P((X,Y)\in\{(1,0)\qquad\;\;\,\qquad\;\;\,\})~=~ 1/18 \\\mathsf P(X+Y=2) ~=~ \mathsf P((X,Y)\in\{(2,0),(1,1)\qquad\;\;\,\})~=~ 2/18 \\\mathsf P(X+Y=3) ~=~ \mathsf P((X,Y)\in\{(3,0),(2,1),(1,2)\})~=~ 3/18 \\\mathsf P(X+Y=4) ~=~ \mathsf P((X,Y)\in\{(4,0),(3,1),(2,2)\})~=~ 3/18 \\\mathsf P(X+Y=5) ~=~ \mathsf P((X,Y)\in\{(5,0),(4,1),(3,2)\})~=~ 3/18 \\\mathsf P(X+Y=6) ~=~ \mathsf P((X,Y)\in\{(6,0),(5,1),(4,2)\})~=~ 3/18 \\\mathsf P(X+Y=7) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,(6,1),(5,2)\})~=~ 2/18 \\\mathsf P(X+Y=8) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,\qquad\;\;\,(6,2)\})~=~ 1/18 \\$$