A bartender and a customer roll a six-sided, ungalvanized cube. What is his chance of winning? 
A bartender and a customer roll a six-sided, ungalvanized cube. If the
visitor throws a higher number than the bartender, he wins. A guest is
a rascal who cheats. He takes advantage of the waiter's inattention by
secretly looking at his first roll under the dice cup and then, if he
throws a 1-3, he throws the dice again, unnoticed.
What is his chance of winning?

Let $k$ be the number the bartender throws:
For every $k$ there are two ways of winning: Either the visitor has a higher number with his first throw or he throws either a $1$, $2$ or $3$ and than has a higher number.
For $k=6$: $0/6$
For $k=5$: $1/6+3/6\cdot 1/6 $
For $k=4$: $2/6+3/6\cdot2/6$
For $k=3$: $3/6+3/6\cdot3/6$
For $k=2$: $4/6+2/6\cdot4/6$
For $k=1$: $5/6+1/6\cdot5/6$
Adding all these up and dividing them by six, gives $p=\frac{121}{216}$
Is that correct?
 A: To summarize the discussion in the comments:
The stated solution is incorrect because the calculation allows the cheater to (sometimes) consider the bartender's roll, contrary to the specified problem.
To solve the given problem:
We consider probabilities for the various outcomes for the cheater's roll.  For $≤3$ the only way to get that value is to first roll $3$ or less and then roll the specified value.  For $≥4$ the cheater can either get the desired roll initially or get $≤3$ the first time and then roll the desired value.  Thus:
$$ P(X=n) =
\begin{cases}
\frac 1{12}  & \text{if $n≤3$ } \\ \\
\frac 14 & \text{if $n≥4$ }
\end{cases}$$
Where, of course, $X$ is the cheater's roll.
Since the probability of winning if $X=i$ is $\frac {i-1}6$ we see that the probability the cheater wins is $$\frac 1{12}\times \left(\frac 06+\frac 16+\frac 26\right)+\frac 1{4}\times \left(\frac 36+\frac 46+\frac 56\right)=\boxed {\frac {13}{24}}$$
A: No. The key observation here is that the first roll only counts if it is a 4 or above. This does not have any effect on your $k = 3,4,5,6$ cases, but does on the others: you have counted the case where the customer rolls a $2$ or $3$ on the first roll as a win when it exceeds the bartender's roll, but in fact those will be rerolled, and so have only a $\frac{4}{6}$ or $\frac{5}{6}$ chance of being wins, depending on the bartender's roll.
