A player selects one number from $1$, $2$, $3$ or $4$ and rolls three standard dice. What is the probability that his chosen number appears on all three dice?
To answer this, I calculated the probability of the player choosing the number $1$, and then the number $1$ appearing on all three dice, and adding this to the probability of the player choosing the number $2$ and then the number $2$ appearing on all three dice, etc.
Probability of player choosing the number $1$, and then the number $1$ appearing on all three dice:
$$P(x=1)=\frac{1}{4}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}.$$
When multiplying this above probability by $4$ to get the answer, I got
$$P(X)=\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6},$$
which is the same probability as the case where the player didn't choose a number from $1$ to $4$ but from $1$ to $6$. Choosing a number from $1$ to $4$ and getting the chosen number three times obviously has a lower probability that choosing a number from $1$ to $6$ and getting the chosen number three times. What is the flaw in my logic?