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?

Question

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?

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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?

Answer 1

Your approach is correct. To save a bit of the arithmetic and the detour through calculating the probabilities, you could also argue that the problem statement amounts to the result being uniformly randomly drawn from $\{1,2,3,3,4,5,6\}$, so the expected result for one die is the mean of that set,

$$ \frac{1+2+3+3+4+5+6}7=\frac{24}7\;. $$

It's essentially the same solution, just not so many fractions to write :-)

Answer 2

This is the same as two seven sided dice with two 3s.

Expected value $E= 2\cdot \frac{1+2+3+3+4+5+6}{7} = 6.8571$