Joint probability distribution of X and Y using a table, where

Question

Suppose two 4 sided dices are rolled. X is the sum of two resulting numbers after rolling the dice. Y is the absolute difference of two resulting numbers after rolling the dice.

X sample set is {2,3,4,5,6,7,8} and Y sample set is {0,1,2,3}

I have calculated the values of X which are : F(2) = [(1,1)] = 1/16 F(3) = [(1,2), (2,1)] = 2/16 F(4) = [(1,3), (2,2), (3,1)] = 3/16 F(5) = 4/16 F(6) = 3/16 F(7) = 2/16 F(8) = 1/16

For Y following are the values: F(0) = [(1,1),(2,2),(3,3),(4,4)] = 4/16 F(1) = 6/16 F(2) = 4/16 F(3) = 2/16

Please help me in finding joint probability distribution of X and Y using a table. I am confused which values shall I consider?

Answer 1

The table of favorable outcomes: $$\begin{array}{c|c|c|c|c} X/Y&0&1&2&3\\ \hline 2&(1,1)&\emptyset&\emptyset&\emptyset\\ \hline 3&\emptyset&(1,2),(2,1)&\emptyset&\emptyset\\ \hline 4&(2,2)&\emptyset&(1,3),(3,1)&\emptyset\\ \hline 5&\emptyset&(2,3),(3,2)&\emptyset&(1,4),(4,1)\\ \hline 6&(3,3)&\emptyset&(2,4),(4,2)&\emptyset\\ \hline 7&\emptyset&(3,4),(4,3)&\emptyset&\emptyset\\ \hline 8&(4,4)&\emptyset&\emptyset&\emptyset\\ \end{array}$$ The contingency table: $$\begin{array}{c|c|c|c|c} X/Y&0&1&2&3\\ \hline 2&1&0&0&0\\ \hline 3&0&2&0&0\\ \hline 4&1&0&2&0\\ \hline 5&0&2&0&2\\ \hline 6&1&0&2&0\\ \hline 7&0&2&0&0\\ \hline 8&1&0&0&0\\ \end{array}$$ The joint probability table: $$\begin{array}{c|c|c|c|c} X/Y&0&1&2&3\\ \hline 2&1/16&0&0&0\\ \hline 3&0&2/16&0&0\\ \hline 4&1/16&0&2/16&0\\ \hline 5&0&2/16&0&2/16\\ \hline 6&1/16&0&2/16&0\\ \hline 7&0&2/16&0&0\\ \hline 8&1/16&0&0&0\\ \end{array}$$

Answer 2

Comment not answer. I am doing this for six-sided dice. In order to give you an informal view of the joint distribution of $X$ and $Y,$ I did a simulation (in R) of 10,000 performances of the two-die experiment.

In the figure below the probabilities are at the centers of the squares. Points are 'jittered' (randomly displaced) in order to give a clue the relative frequency with which each $(x,y)$ value occurs. The six points (smeared into squares) at the bottom are less likely than the others.

m = 10^4
a = sample(1:6, m, rep=T);  b = sample(1:6, m, rep=T)
x = a+b;  y = abs(a-b)
jx = runif(m, -.4, .4);  jy = runif(m, -.4, .4) # jittering
plot(x+jx, y+jy, pch=".")
  abline(v=2:12, col="green2")
  abline(h=0:5, col="green2")

The probability of each point is a multiple of $1/36.$