Rolling a dice $5$ times Rolling a dice $5$ times find the probability that we will get the continuum:



A. $12345$
B. $1234$
C. $434$




My attempt:


A: for each throwing there are $5$ possibilities so 
$P(1)=P(2)=P(3)=P(4)=P(5)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)\cdot P(5)=(\frac{1}{6})^5=1/7776$
B: $P(1)=P(2)=P(3)=P(4)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)=(\frac{1}{6})^4=1/1296$ now we will multiply the result by two because the series could be here $\underline{1}\underline{2}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{1}\underline{2}\underline{3}\underline{4}$ so the final result will be $2/1296$ 
C:$P(4)=P(3)=P(4)=\frac{1}{6}$ so $P(4)\cdot P(3)\cdot P(4)=(\frac{1}{6})^3=1/216$ now we will multiply the result by three because the series could be here $\underline{4}\underline{3}\underline{4}\underline{}\underline{}$ or here 
$\underline{}\underline{4}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{}\underline{4}\underline{3}\underline{4}$  


This is correct?
In part C, you have overcounted because  $43434$ fits both $\underline{434\cdot\cdot}$ and $\underline{\cdot\cdot434}$. You should therefore subtract the probability of getting $\underline{43434}$, which is $1/6^5$, so the answer will be $3 (1/6^3) - 1/6^5$.
Assuming it's a fair die and it has actually 6 faces then


$P(1,2,3,4,5) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \left( \frac{1}{6} \right) ^ 5 = \frac{1}{6^5}$
$P(1,2,3,4) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = 2 \cdot \frac{1}{6^4}$
$P(1,2,3,4) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot 3  - 1 (\text{the event } 43434)= 3 \cdot \frac{1}{6^3} - 1$