Problem 356

When two standard six-sided dice are rolled, the minimum sum that
can be rolled is $1+1=2$, and so Riah
must move forward at least $2$
squares on each roll.

This means that Riah cannot land on the square numbered $1$, she can land on at most one of the
squares numbered $11$ and $12$, and she can land on at most one of
the squares numbered $16$ and $17$.

To achieve the maximum score possible, the goal is to land on as many
shaded squares as possible, recognizing that a choice must be made
between squares $11$ and $12$, and also between squares $16$ and $17$.

To begin the game, Riah cannot roll a $1$, but she can roll a $3$, thus adding $6$ points to her score. She can then roll
$3$ again, adding $4$ points to her score, followed by a roll
of $2$ to add $5$ more points to her score. Riah is now
on the square numbered $8$, she has
$6+4+5=15$ points, and she has
collected the maximum number of points possible to this point in the
game.

From the square numbered $8$, Riah
can roll a $3$ and collect $7$ points or she can roll a $4$ and collect $8$ points. Since landing on square $11$ or $12$ does not affect her ability to land on
square $16$ or $17$, Riah maximizes her points by rolling
$4$ and collecting $8$ points. She is now on the square
numbered $12$ and has $15+8=23$ points.

From the square numbered $12$, Riah
can roll a $4$ and collect $10$ points or she can roll a $5$ and double her points. Since adding
$10$ points gives Riah $23+10=33$ and doubling her points gives
$23\times2=46$, then rolling a $5$ and landing on the square numbered
$17$ maximizes Riah's points. There
are no more points that can be collected before finishing the
game.

Thus, the greatest number of points that Riah can finish the game with
is $46$.