Exercise with a dice with 14 faces I'm having trouble solving this problem:

There's a dice with 14 faces, of which 6 are squared, and 8 are triangular. If the probability of getting a squared face is twice the probability of getting a triangular one, then what is the probability of getting two squared sides when throwing the dice twice?

My answer is $\frac49$ (but it's not even in the alternatives), since the probability of getting a squared side is $P(S)=2P(T)$, and I can either get a squared or a triangular one, then the probability of getting a squared side is: $$\frac{2\cdot P(T)}{3\cdot P(T)}=\frac23$$
The solution that is proved to me is $\frac9{25}$, but it doesn't even make sense to me, because that implies that the probability of getting a squared size is $\frac35$. So I want to know if either I'm wrong, or if the solution is correct. Thanks for your help.
 A: Let $s$ denote the probability $P(S)$ of getting a square side, and let $t$ denote the probability of getting a triangular side.  Assuming that all square sides are equally likely and that all triangular sides are equally likely, we have
$$
6s + 8t = 1 \implies\\
6(2t) + 8t = 1 \implies\\
t = \frac 1{20}, \quad s = \frac 1{10}
$$
So, the overall probability of getting a square side on one roll of the die is
$$
6 \cdot s = \frac 3{5}
$$
Perhaps you could take it from there.
A: The critical thing is the interpretation of the question.  You read the question as the total probability of getting one of the square faces is twice the total probability of getting one of the triangular faces, which tells us that the probability of getting some square face is $\frac 23$ and your calculation is correct.  In your reading the number of square and triangular faces is extraneous information.  The intended reading is that the chance of getting any one particular square face is twice the chance of getting any one particular triangular face.  Now the number of each type of face matters and the probability of some square face is $\frac 35$.  I find the problem as written ambiguous on how it should be read. 
A: There are two possibilities to understand the sentence:
The probability of getting a squared face is twice the probability of getting a 
triangular one
The first possibility is the one you understood is that the probability of getting any square is double of getting any triangle:
$P(x\in\{S_1,S_2,S_3,S_4,S_5,S_6\}) = 2\cdot P(x\in\{T_1,T_2,T_3,T_4,T_5,T_6,T_7,T_8\}))$
This leads to your result of $\frac{4}{9}$.
The second possibility is that the probability of getting a specific square is twice the probability of getting a specific triangle:
$P(x=S_1) = ... = P(x=S_6) = 2\cdot P(x=T_1)=...= 2\cdot P(x=T_8)$
With this basis the result is $\frac{9}{25}$
