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?

I don't think the term $\frac57\frac57\frac{10}{49}$ is right. $\frac57\frac57$ is the probability that both are greater than $1$, but if that happens the probability that $i<j$ is quite a bit bigger than $\frac{10}{49}$.
The right way to approach this is that the probability of the second die showing more dots is equal to the probability of the first die showing more dots, so the probability you want is just half the probability that the two dice show different values. So work out the probability they show the same value, subtract from $1$, and halve.
