<s>[INST] Find the probability of each outcome when a biased die is rolled, if rolling a 2 or 4 is three times as likely as rolling each of the other$\dots$ Question:Find the probability of each outcome when a biased die is rolled, if rolling a $2$ or $4$ is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.
My Attempt:
Let $O_1, O_2, \dots, O_6$ be the out come of dies.
Since rolling $O_2$ or $O_4$ is twice as likely, then
$$P(O_2) = P(O_4) = 3P(O_1) = 3P(O_3) = \dots = 3P(O_6)$$
Thus,
$$P(O_1) + P(O_2) + P(O_3) + P(O_4) + P(O_5) + P(O_6) = 1$$
$$P(O_1) + 3P(O_1) + P(O_1) + 3P(O_1) + P(O_1) + P(O_1) = 1$$
$$10P(O_1) = 1$$
$$P(O_1) = \dfrac{1}{10}$$
Therefore,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{10}$$ and,
$$P(O_2) = P(O_4) = \dfrac{3}{10}$$
Problem:
The book gave a different answer such that,
$$P(O_1) = P(O_3) = P(O_5) = P(O_6) = \dfrac{1}{16}$$ and,
$$P(O_2) = P(O_4) = \dfrac{6}{16}$$
Frankly, is $P(O_2) = P(O_4) = 6P(O_1)$ now? I think I'm missing something fundamental here.
 [/INST] Your confusion stems from the obscure description of the question at hand.
The actual meaning of the question is apparently $\displaystyle P(2,4) = 3 \cdot P(1,3,5,6)$.
Therefore, $\displaystyle P(2,4) = \frac{12}{16} = 3 \cdot \frac{4}{16} = 3 \cdot P(1,3,5,6)$ implies the following:


*

*$\displaystyle P(2) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$

*$\displaystyle P(4) = \frac{1}{2} \cdot \frac{12}{16} = \frac{6}{16}$

*$\displaystyle P(1) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(3) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(5) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$

*$\displaystyle P(6) = \frac{1}{4} \cdot \frac{ 4}{16} = \frac{1}{16}$



In other words, you should change this attempt:


*

*$P(2) = P(4) = 3P(1) = 3P(3) = 3P(5) = 3P(6)$


To this attempt:


*

*$P(2) + P(4) = 3P(1) + 3P(3) + 3P(5) + 3P(6)$


Where:


*

*$P(2) = P(4)$

*$P(1) = P(3) = P(5) = P(6)$

</s>