Sample Space for a Pair of Loaded Dice

Question

The following is from page 128 of "Introduction to Mathematical Statistics and Its Applications" (5th ed.) by Larsen and Marx.

Suppose die one has spots $1,2,2,3,3,4$ and die two has spots $1,3,4,5,6,8$. If both dice are rolled, what is the sample space? Let $X =$ total spots showing. Show that the probability mass function (PMF) for $X$ is the same as for normal dice.

My question is about the sample space. Consider the sample space $S=\lbrace (x,y)\rbrace$, where $x,y$ are the values of the first and second dice, respectively. Should I use a multiset and include, for example, $(2,1), (2,1)$ so that the sample space has $36$ outcomes? Since $P(x=1)=1/6$ (and not 1/4), I assume we should also count multiplicities in the denominator as well. On the other hand, I don't remember any axioms allowing me to allow the sample space to be a multiset.

Answer 1

Just as for rolling two ordinary dice, the sample space consists of a $6 \times 6$ of pairs of faces.

Enumeration: For the sum $S$ on the two dice, each of the 36 cells can also be labeled with the total of the two corresponding faces. Then count the cells for each total. (The first two of the six rows are shown below.)

       1    3    4    5    6    8
       --------------------------
1      2    4    5    6    7    9
2      3    5    6    7    8   11
...

Analytic methods: It is easy to show that $$E(S) = E(D_a) + E(D_b) = 15/6 + 27/6 = 42/6 = 3.5,$$ which is the same as for regular dice. A bit more tediously, one can show that $Var(S)$ is the same as for regular dice. 'Probability generating functions' could be used to show that the distribution of $S$ agrees with the (triangular) distribution of the sum of two ordinary dice.

Simulation: The distribution of $S$ can be very closely approximated by simulating the sums on a million rolls of these two special dice and tallying the results. (Simulation in R statistical software gives probabilities accurate to about three places.)

m = 10^6; da = c(1,2,2,3,3,4);  db = c(1, 3:6, 8)
s = replicate(m, sample(da,1)+sample(db,2))
round(table(s)/m, 3)
s
     2     3     4     5     6     7     8     9    10    11    12 
 0.056 0.111 0.167 0.222 0.279 0.334 0.278 0.221 0.167 0.111 0.055 

The plot below shows a histogram of the million simulated totals obtained when rolling a pair of these special dice. The dots show the exact distribution.

hist(s, br=1.5:12.5, prob=T, col="skyblue2", main="Simulated Sums")
x = 2:12;  pdf=c(1:6, 5:1)/36
points(x, pdf, col="red", pch=19)