An unbiased six-faced dice whose faces are marked with numbers 1, 2, 3, 4, 5, and 6 is rolled twice in succession and the number on the top face is recorded each time. The probability that the number appearing in the second roll is an integer multiple of the number appearing in the first roll is ______.
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Fix the first roll and count how many second-roll values are multiples of it, then add up over all six first-roll values.
Step 1: Picture all outcomes as a $6\times6$ grid.
Write the first roll $x$ along the rows and the second roll $y$ along the columns, both running from $1$ to $6$. Each of the $36$ cells is equally likely, since the two rolls are independent and the dice is fair.
Step 2: Go row by row and mark the cells where $y$ is a multiple of $x$.
Row $x=1$: every column from $1$ to $6$ works, since any number is a multiple of $1$. That is $6$ cells. Row $x=2$: columns $2, 4, 6$ work. That is $3$ cells. Row $x=3$: columns $3, 6$ work. That is $2$ cells. Row $x=4$: only column $4$ works. That is $1$ cell. Row $x=5$: only column $5$ works. That is $1$ cell. Row $x=6$: only column $6$ works. That is $1$ cell.
Step 3: Total the marked cells.
Adding the row counts, \[ 6+3+2+1+1+1=14 \] So $14$ of the $36$ equally likely cells satisfy the condition.
Step 4: Turn the count into a probability.
\[ P=\frac{14}{36} \] Divide the top and bottom by $2$: \[ P=\frac{7}{18} \]
Step 5: Sanity check against the answer choices.
A quick check: $\dfrac{1}{6}$ equals $\dfrac{6}{36}$, which only accounts for the row $x=1$ and misses everything else, so it is too small. $\dfrac{5}{6}$ equals $\dfrac{30}{36}$, which is far more than the $14$ cells actually marked, so it is too big. Only $\dfrac{7}{18}$ matches the count of $14$ out of $36$.
