Probability of rolling a number from 1 to 4 exactly once with 3 identical dice.

In a game, 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 exactly one of the dice?

Even though the player cannot chose from all numbers on the dice $$(1$$ to $$6)$$, I think this should not have any impact on the probability of rolling the chosen number, since whatever number, $$n$$, he choses will theoretically show up $$\frac{1}{6}^{\textrm{th}}$$ of the time. It also seems that the order of the numbers on the dice should not matter for the dice which don't have the chosen number. Therefore, from what I understand, only the following three distinct possible rolls could occur so that the chosen number appears exactly once.

$$n \quad a \quad b$$ $$a \quad n \quad b$$ $$a \quad b \quad n$$

Where $$a, b \in (1, 2, 3, 4, 5, 6)$$. Therefore, the probability should be

$$\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times 3 = \frac{25}{72}$$

But apparently the answer is $$\frac{27}{64}$$, which I can't understand.

Where am I going wrong? Thank you.

• You did the correct computations for a six sided dice (a cube). The result you saw is for a four sided dice, a (tetrahedron). – Ertxiem - reinstate Monica Mar 22 at 18:33
• So you're saying the books wrong? You questioning the book?? =) – Tom Finet Mar 22 at 18:41
• Yup! :) Instead of "standard dice" they should have said "four sided dice" (or something like that) otherwise the book should have said $25/72$ or $75/216$. – Ertxiem - reinstate Monica Mar 22 at 18:43
• Alright cool, thanks for freeing my mind. – Tom Finet Mar 22 at 18:44
• You're welcome. – Ertxiem - reinstate Monica Mar 22 at 18:44