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

discrete mathematicsprobabilityprobability distributionsstatistics

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?

Best Answer

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")

enter image description here

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

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