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.

  • $\begingroup$ One way is to enumerate all $6 \times 6 = 36$ outcomes and see how many times each sum appears. $\endgroup$ – angryavian Jul 25 '16 at 20:46
  • 2
    $\begingroup$ they are not even. drawing $8$ for example is harder. chances for this is $\frac{1}{6} * \frac{1}{3}$ while drawing $3$ can be done in more than 1 way $\endgroup$ – d_e Jul 25 '16 at 20:47
  • 2
    $\begingroup$ It would be easier to model this instead of rolling a six-sided die and a second six-sided die, instead to model this as using a six-sided die and a three-sided die. Instead of enumerating $36$ outcomes, we instead only enumerate $18$ outcomes. Each of these outcomes are equally likely. As d_e points out, the only way to get a sum of eight is via the outcome $(6,2)$ and so occurs with probability $\frac{1}{18}$ because only one of the equally likely outcomes has this sum. On the other hand, a sum of $3$ occurs as any of $(1,2), (2,1), (3,0)$ and so has probability $\frac{3}{18}=\frac16$ $\endgroup$ – JMoravitz Jul 25 '16 at 20:52
  • 1
    $\begingroup$ If you are asking this in an attempt to model using an eight-sided die in practice having only six-sided dice available, to have the probabilities be even (and avoid the need to reroll), we would require the outcomes to be able to be partitionable into eight equally sized subsets, implying that we would need at least three dice (as thesize of the sample space using only two dice is not divisible by eight, however using three dice it is). $\endgroup$ – JMoravitz Jul 25 '16 at 20:57
  • $\begingroup$ @JMoravitz, yes I was trying to model an eight-side die with a six-sided dice. Throwing more than two dice is ok. I will have to research the 'sample space' because I don't know how these calculations are done. $\endgroup$ – Michael Curtis Jul 25 '16 at 20:59

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

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

  • $\begingroup$ I have a follow up question: how basic is this math? Is this high school level? $\endgroup$ – Michael Curtis Jul 26 '16 at 13:56
  • $\begingroup$ Junior high, mostly. $\endgroup$ – Graham Kemp Jul 26 '16 at 23:01

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.

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.

  • $\begingroup$ Another way of doing generating this with three throws is by treating the dice as coins(e.g. 1-3 as tails, 4-6 as head). Then there are $2^3$ equally likely sequences. But your idea is better since you do not need to remembering the order of the throws. $\endgroup$ – harvey Jul 26 '16 at 23:52

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.

  • $\begingroup$ Clear ! This was the case with a pair of regular dices. Equiprobability of the sum comes back if you under-weight the second dice ( ie 0.1 , 0.2 & 0.3 instead of 1 2 & 3 ). $\endgroup$ – user354674 Jul 26 '16 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.