Probability of rolling 1-8 using six-sided dice

Question

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.

Answer 1

You have an unbiased six-sided die and an unbiased three-sided die, so are drawing pairs uniformly distributed on $\{1,2,3,4,5,6\}\times\{0,1,2\}$.

$$\mathsf P(X+Y=1) ~=~ \mathsf P((X,Y)\in\{(1,0)\qquad\;\;\,\qquad\;\;\,\})~=~ 1/18 \\\mathsf P(X+Y=2) ~=~ \mathsf P((X,Y)\in\{(2,0),(1,1)\qquad\;\;\,\})~=~ 2/18 \\\mathsf P(X+Y=3) ~=~ \mathsf P((X,Y)\in\{(3,0),(2,1),(1,2)\})~=~ 3/18 \\\mathsf P(X+Y=4) ~=~ \mathsf P((X,Y)\in\{(4,0),(3,1),(2,2)\})~=~ 3/18 \\\mathsf P(X+Y=5) ~=~ \mathsf P((X,Y)\in\{(5,0),(4,1),(3,2)\})~=~ 3/18 \\\mathsf P(X+Y=6) ~=~ \mathsf P((X,Y)\in\{(6,0),(5,1),(4,2)\})~=~ 3/18 \\\mathsf P(X+Y=7) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,(6,1),(5,2)\})~=~ 2/18 \\\mathsf P(X+Y=8) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,\qquad\;\;\,(6,2)\})~=~ 1/18 \\$$

Answer 2

One (of many) methods to simulate the roll of a single eight-sided die using only six-sided dice:

We roll three six-sided dice. One of the three will be a special color, say red.

If all three dice are greater than or equal to four, we call the outcome an $8$

If all three dice are less than or equal to three, we call the outcome a $7$

Otherwise, we call the outcome whatever the value on the red die is.

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Checking that the probabilities are in fact even: Temporarily assume all three dice are colored, say one is blue, the other is green, and the third is red (still the special one).

There are $6^3=216$ equally likely outcomes in the sample space.

To get a result of $8$ we need for all three dice to each be greater than or equal to $4$, so we count how many outcomes correspond to this: There are $3$ options for the blue die (4,5,6), similarly $3$ options for the green die, and $3$ options for the red die. This gives $3\cdot 3\cdot 3=27$ total possibilities. Out of $216$ total outcomes in the sample space this means our probability is $\frac{27}{216}=\frac{1}{8}$

The same logic and calculations apply for finding the probability of getting an result of $7$.

Now, for an outcome of $6$, we need the red die to be a six and the blue and green dice to have at least one low number. Ignoring needing at least one low number, there are $6\cdot 6\cdot 1=36$ possibilities. Removing the bad possibilities where both numbers were actually high (of which there are $3\cdot 3\cdot 1=9$ of them) we have a total of $36-9=27$ of them. This occurs then with probability $\frac{27}{216}=\frac{1}{8}$

The logic and calculations is similar for all other possible results.

Thus, all results $1-8$ are equally likely with a probability of $\frac{27}{216}$ and there is no need to reroll.

Answer 3

Your second six-sided die is like a fair three-sided die (try constructing that!) with the numbers $0,1,2$.

There are eighteen equally likely outcomes:

$$(1,0), (2,0), (3,0), (4,0), (5,0), (6,0)$$ $$(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)$$ $$(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)$$

Counting up the number of outcomes $r(n)$ is like swiping diagonals down in the grid above from right to left (starting from each element in the top row, then going down the right side):

$$n=1,2,3,4,5,6,7,8$$ $$r=1,2,3,3,3,3,2,1$$

Then the probability of getting a sum of $n$ is just $r(n)/18$. But, as you can see, it's not uniform. Getting a sum of $3$ through $6$ is three times as likely as getting a $1$ or an $8$, and getting a sum of $2$ or $7$ is twice as likely.