Throwing dice twice, with unlike probability of occourence? A loaded dice has the property that when the dice is thrown the probability of showing a given number is proportional to the number. For example $2$ is twice as likely to show up compared to $1$ and $3$ is thrice as likely to show up compared to $1$, And so on. What is the probability that when the dice is thrown twice the sum is $4$ or less.
$$
    P =  \frac 36 \cdot \frac 16 +\frac 16 \cdot \frac 36 + \frac 26 \cdot \frac 26 + \frac 16 \cdot \frac 26 + \frac 26 \cdot \frac 16 + \frac 16 \cdot \frac 16  =
    \frac{15}{36} 
$$
Where I am getting it wrong? 

Since $1+2+\cdots+6=21$, the probabilities of $1,2,3,4,5,6$ in $1$ toss are respectively $\frac{1}{21}$, $\frac{2}{21}$, and so on up to $\frac{6}{21}$.   
The probability of a sum of $2$ in $2$ throws is $\frac{1}{21}\cdot\frac{1}{21}$.
The probability of a sum of $3$ is $2\cdot \frac{1}{21}\cdot\frac{2}{21}$.
The probability of a sum of $4$ is $2\cdot \frac{1}{21}\cdot\frac{3}{21}+\frac{2}{21}\cdot\frac{2}{21}$.
Add up. 
Remark: Your numerator, and therefore the basic analysis, was right. The denominators were not. 
