Possible Values
Since Y is going to a selection from X, we know that it has values within [1,6]. If one of the values is 1, it is clear that Y will take on whatever the other value is, hence Y can be all of 1 - 6.
Distribution of Y
Let's calculate the probability of each event:
For Y = 1, this is easy
#P(Y = 1) = P(X_1 = 1 and X_2 = 1) = 1/9 cdot 1/9 = 1/81#
For Y = 2, it gets messy...
#P(Y = 2) = P({X_1 = 2 and X_2 le 2} or {X_1 < 2 and X_2 = 2}) #
(Note the #le# became a #<# to avoid double counting)
We can make this easier on ourselves. Since #X_1# and #X_2# come from the same distribution, there's a symmetry:
# = 2 * P(X = 2) * P(X le 2) - P(X=2)^2#
This form applies to all of the numbers, so we can actually use it for all of them:
#P(Y= N) = 2 * P(X = N) * P(X le N) - P(X=N)^2#
Applying this,
#P(Y= 1) = 2 * P(X = 1) * P(X le 1) - P(X=1)^2#
# = 2/81 - 1/81 = 1/81#
#P(Y= 2) = 2 * P(X = 2) * P(X le 2) - P(X=2)^2#
#= 2 * 1/6 * (1/6 + 1/9) - 1/36 = 7/108 #
#P(Y= 3) = 2 * P(X = 3) * P(X le 3) - P(X=3)^2#
#= 2 * 1/6 * (2/6 + 1/9) - 1/36 = 13/108 #
#P(Y= 4) = 2 * P(X = 4) * P(X le 4) - P(X=4)^2#
#= 2 * 1/6 * (3/6 + 1/9) - 1/36 = 19/108 #
#P(Y= 5) = 2 * P(X = 5) * P(X le 5) - P(X=5)^2#
#= 2 * 1/6 * (4/6 + 1/9) - 1/36 = 25/108 #
#P(Y= 6) = 2 * P(X = 6) * P(X le 6) - P(X=6)^2#
#= 2 * 2/9 * (1) - 4/81 = 32/81#
Just to check, the sum of these does add up to one.
Probability of Y #ge# 3
#P(Y ge 3) = P(Y=3)+P(Y=4)+P(Y=5)+P(Y=6) #
#= 13/108 + 19/108 + 25/108 + 32/81 = 299/324 #
