A fair six-sided die is rolled repeatedly I need help with this homework problem.  
A fair six-sided die is rolled repeatedly. Each time the die is rolled, the number showing is written down. Let $X$ be the number of rolls until some number shows twice. The possible values of $X$ are $2,3,4,5,6,$ or $7$. For each of the following parts, explain your reasoning and express your answer as a fraction in reduced form.
Here is the solution I tried to derive. 


*

*$\Pr(3) = \dfrac{30}{216}$


$$6 \times (6-1) \times 1 = 30$$ $$6^3 = 216$$


*

*$\Pr(4) = \dfrac{120}{1296}$


$$6 \times (6-1) \times (6-2) \times 1 = 120$$
$$6^4=1926$$


*

*$\Pr(5) = \dfrac{360}{7776}$


$$6 \times (6-1) \times (6-2) \times (6-3) \times 1 = 360$$
$$6^5=7776$$
Can someone help me out please? Thank

You are on the right track, but the last roll of the die could be any of the numbers already rolled. Then
$$\text{Pr}(2)=\frac66\times\frac16=\frac{6}{36}=\frac{1}{6}$$
$$\text{Pr}(3)=\frac66\times\frac{(6-1)}{6}\times\frac26=\frac{60}{216}=\frac{5}{18}$$
$$\text{Pr}(4)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac36=\frac{360}{1296}=\frac{5}{18}$$
$$\text{Pr}(5)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac46=\frac{1440}{7776}=\frac{5}{27}$$
$$\text{Pr}(6)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac{(6-4)}{6}\times\frac56=\frac{3600}{46656}=\frac{25}{324}$$
$$\text{Pr}(7)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac{(6-4)}{6}\times\frac{(6-5)}{6}\times\frac66=\frac{4320}{279936}=\frac{5}{324}$$
Note that $\text{Pr}(2) + \text{Pr}(3) + \dots + \text{Pr}(7)=1$
Hope it help.
