solution verification for biased dice I have the following problem for which I'm not sure my solution is correct:
A dice is constructed in such a way that 1 dot occurs twice more often than the rest of the points. The probabilities for the rest of the dots are mutually equal. The dice is thrown 2 times.
Calculate the probability that the dots on the second dice are more than the dots on the first one.
My solution:
Let x be the probability for 1, and y the probability for anything else.
$$
\left\{ 
\begin{array}{c}
x=2y\\
x+5y=1
\end{array}
\right. 
$$
I get that $x=\frac{2}{7}$ and $y=\frac{1}{7}$. I have four different scenarios for the dots - $(1, 1), (1, i), (i, j), (i, 1)$, where $2 \le i \le 6$ and $2 \le j \le 6$. I have denoted those cases $H_1, H_2, H_3 $ and $H_4$ respectively. For the probability of the desired event I'm using the formula for total probability:
$$P(A)=\sum_{i=1}^4P(H_i)P(A|H_i)=\frac{2}{7}\frac{2}{7}0+\frac{2}{7}\frac{5}{7}1+\frac{5}{7}\frac{5}{7}(\frac{10}{49})+\frac{5}{7}\frac{2}{7}0=\frac{740}{49^2} \approx 0.30$$
Now, is this correct and are there other ways to solve this problem?

The $\frac{10}{49}$ term should be $\frac{10}{25}$: out of the 25 cases where you get two non-1's, 5 cases are where $i=j$, leaving half of the remaining 20 where $i<j$
The fact that you ended up with such a weird fraction for your final answer should have been a clue that something is wrong, as clearly you should end up with something of the form $\frac{x}{49}$
