Htykuut

A fair dice is rolled over and over. What is the probability of getting $2$ or $4$ before getting $3$ or $6$ ?

According to this answer, it is 1/2. As there is a probability of $frac{1}{3} $ of $2$ or $4$ appears before $3$ or $6$ . Out of the total options of them both appearing which should be $$ frac{frac{1}{3}}{frac{1}{3}+frac{1}{3}} = frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2} $$

Is it the correct way to solve?

We can ignore any roll that produces a 1 or a 5 since they have no bearing on which of the numbers 2, 3, 4, 6 appears first. Each of these four numbers is equally likely to appear first among those four. Since two of the four are a 2 or 4, the probability that a 2 or 4 appears before a 3 or 6 is
$$frac{2}{4} = frac{1}{2}$$

Let $W=frac13$ be the probability of a win (3 or 6), similarly for $L$ as a 2 or 4 and again for $D$ a 1 or 5.

Then the probability of a loss is the sum given by $L+DL+D^2L+dots$, which is:

$$frac13+frac13frac13+frac13^2frac13+dots$$
$$=frac13left(dfrac1{1-frac13}right)$$
$$=frac12$$