Probability of die thrown repeatedly

[Gregg]

Homework Statement

2. A six-sided die is rolled repeatedly. Let X denote the first score of 4 or
more and let Y denote the first score of 5 or more.

[For the sequence of rolls starting 4, 2, 6, 5, . . . we have X = 4, Y = 6; for the
sequence starting 6, 4, 3, . . . we have X = 6 = Y . ]

(a) Write down in a table the joint probability distribution of (X, Y ) and find the
marginal distributions of X and Y.

The Attempt at a Solution

Can do joint distributions but how to do for this where the number of rolls isn't specified?

 

[Gregg]

I think this makes it more obvious
p(X=4)=1/3
p(X=5)=1/3
p(X=6)=1/3

p(Y=5)=1/2
p(Y=6)=1/2

 

[I like Serena]

I think this makes it more obvious
p(X=4)=1/3
p(X=5)=1/3
p(X=6)=1/3

p(Y=5)=1/2
p(Y=6)=1/2

The easiest way to do this is simply by reasoning it out.

Suppose X=5.
This means that after an unknown number of rolls of less than 4, the number 5 comes up.
What will Y be? Can Y still be 6?

What will Y be when X=6?

And then the tricky one.
Suppose X=4?
What values can Y take and how probable are those?

This should yield a crosstable where you can fill in all the chances...