SOLUTION: Nonstandard dice can produce interesting distributions of outcomes. You have two balanced, six- sided dice. One is a standard die, with faces having one, two, three, four, five, a

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Question 1190863: Nonstandard dice can produce interesting distributions of outcomes. You have two balanced, six-
sided dice. One is a standard die, with faces having one, two, three, four, five, and six spots.
The other die has three faces with one spot and three faces with six spots. Find the probability
distribution for the total number of spots Y on the up-faces when you roll these two dice.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Let's say we have a red die and a blue die
The red die will be the standard die you'd find just about anywhere. It has sides 1 through 6.
The blue die will have faces of either 1 or 6 (three of each)

Here's the 6x6 grid of possible sums when rolling these two dice

+ 1 2 3 4 5 6
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
6 7 8 9 10 11 12
6 7 8 9 10 11 12
6 7 8 9 10 11 12

For example, in the upper left corner we have 1+1 = 2
Or in the bottom right corner we have 6+6 = 12
The rest of the table is filled out in a similar fashion.

The list of unique sums are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
You should notice that nearly each sum shows up exactly 3 times.
This is directly due to the fact that the blue die has repeated faces of this same count (eg: 6 shows up three times).
The only exception to this rule is the sum 7 which shows up 2*3 = 6 times. We still have that multiple of 3 buried in there.

So we have
P(rolling a 2) = 3/36
P(rolling a 3) = 3/36
P(rolling a 4) = 3/36
P(rolling a 5) = 3/36
P(rolling a 6) = 3/36
P(rolling a 7) = 6/36
P(rolling a 8) = 3/36
P(rolling a 9) = 3/36
P(rolling a 10) = 3/36
P(rolling a 11) = 3/36
P(rolling a 12) = 3/36

We can use that data to form this probability distribution table

Y P(Y)
2 3/36
3 3/36
4 3/36
5 3/36
6 3/36
7 6/36
8 3/36
9 3/36
10 3/36
11 3/36
12 3/36

Where Y is the number of dots combined from both dice (i.e. the sum of the dice)

So far I haven't reduced any of the fractions.
In my opinion, it's more descriptive to keep the 36 around in the denominator so we know there are 36 outcomes total.
For later probability distribution problems, it's also helpful to have the same denominator for all the fractions.

If you wanted to reduce the fractions, then this is what the probability distribution looks like

Y P(Y)
2 1/12
3 1/12
4 1/12
5 1/12
6 1/12
7 1/6
8 1/12
9 1/12
10 1/12
11 1/12
12 1/12

Note:
Each P(Y) value is on the interval
Also, the sum of all the P(Y) values yields 1.

Optionally you could convert each of the fractions to the approximate decimal form
1/12 = 0.08333 approximately
1/6 = 0.16667 approximately
I prefer the fraction form because it's exact rather than approximate.