Probability of Sum being $4$ with unfair dice You roll two six-sided dice.  Die $1$ is fair.  Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same.  What is the probability of the sum of both dice adding up to $4$?
My answer:
I wrote out the combinations to sum $4$ between both dies:
(F1, U3), (F2, U2), (F3, U1) [F = Fair die, U = Unfair die].
My thinking is we need to find the OR probability of these pairs:
P((F1 and U3) or (F2 and U2) or (F3 and U1)):
$$(\frac16 \times \frac23) + (\frac16 \times \frac12) + (\frac16 \times \frac23) = 0.30555555555.$$
Is this correct?
 A: There is a mistake. For the unfair die, the probability of an odd number is $\frac{2}{3}$ and of an even number is $\frac{1}{3}$. It also says that the probability of each odd number is same and of each even number is same.
So for the unfair die, probability of each odd number is $\frac{2}{9}$ and of each even number is $\frac{1}{9}$.
So the desired probability should be $ = \displaystyle \small 2\cdot \frac{1}{6} \cdot \frac{2}{9} + \frac{1}{6} \cdot \frac{1}{9} = \frac{5}{54}$
A: No.
With the fair die, the chance of $1,2,3$ is $(1/6), (1/6), (1/6)$, respectively.
With the unfair die, the chance of $1,2,3$ is $(2/9), (1/9), (2/9)$, respectively.
Roll of 4 by 1-3 or 2-2 or 3-1.  Chances are 
$(2/54) + (1/54) + (2/54) = (5/54).$
A: The computations of @user2661923 and @mathlover are confirmed (+1 each) by a simulation in R. With a million rolls of the two dice, one can expect two place accuracy.
set.seed(313)
d1 = sample(1:6, 10^6, rep=T)
d2 = sample(1:6, 10^6, rep=T, p = c(2,1,2,1,2,1))
s = d1+d2
mean(s == 4)
[1] 0.092494     # aprx P(S = 4)
5/54
[1] 0.09259259   # exact


cutp = (1:12)+.5
hist(s, prob=T, br=cutp, col="skyblue2",
     main="Two Dice: One Fair")


