<s>[INST] expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number? This question is already posted here,but i want to check my approach.
Question

What is the expected sum of the numbers that appear on
two dice, each biased so that a $3$ comes up twice as often
as each other number?


My Approach
Let the probability of getting number other than $3$ is $p$
so
$$\frac{1}{p}+\frac{1}{p}+\frac{2}{p}+\frac{1}{p}+\frac{1}{p}+\frac{1}{p}=1 \Rightarrow p=\frac{1}{7}$$
Proability of getting $3=\frac{2}{7}$
and rest other $=\frac{1}{7}$
let $E(X_1)$ be the expectation of getting sum on rolling $1$ dice.
$E(X_1)=1 \times \frac{1}{7}+2 \times \frac{1}{7}+3 \times \frac{2}{7}+4 \times \frac{1}{7}+5 \times \frac{1}{7}+6 \times \frac{1}{7}=\frac{24}{7}$
Now
expected sum of the numbers that appear on two dice
$$E(X_1 +X_2)=E(X_1)+E(X_2)=\frac{24}{7}+\frac{24}{7}=\frac{48}{7}\approx 6.86$$
Is my approach correct?
 [/INST] This is the same as two seven sided dice with two 3s.
Expected value $E= 2\cdot \frac{1+2+3+3+4+5+6}{7} = 6.8571$
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