A strategy is simply given by a set $S\subset\Bbb N$ and amount to "If we have collected $x$ then stop of $x\in S$ and continue if $x\notin S$". We assume there is some optimal strategy $S_0$.
Let $E(x)$ be the expected final payout if we have already collected $x>0$ and make optimal decisions from now on.
Clearly, we have
$$\tag1 E(x)=\begin{cases}x&x\in S_0\\
\frac{E(x+1)+E(x+2)+E(x+3)+E(x+4)+E(x+5)}6&x\notin S_0\end{cases}$$
With strategy $S=\Bbb N$, we will of course stop immediately and have payout $x$. Therefore,
$$\tag2 E(x)\ge x.$$
With $(1)$, this gives us the lower estimate
$$ E(x)\ge \frac{E(x+1)+\cdots+E(x+5)}6\ge \frac{(x+1)+\cdots+(x+5)}6=\frac {5x+15}6,$$
which is certainly $>x$ if $x<15$. In other words, $\{1,\ldots,14\}\cap S_0=\emptyset$.
If $x>k$, let $S=S_0+k=\{\,s+k\mid s\in S_0\,\}$. Then strategy $S$ shows $$E(x)\le E(x-k)+k$$ (namely, we make our decision as if there were $k$ less, hence gain $k$ more than "expected" if we win something, or end up with the same $0$ if we lose)
Combining with $(1)$, we find for $x\in S_0$ that
$$ E(x)\le\frac{5E(x)+15}6$$
and hence
$$ E(x)\le 15.$$
As also $E(x)\ge x$, we conclude that $x\in S_0$ if $x>15$. So
$$ E(x)=x\quad\text{if }x\ge 15.$$
This makes
$$ E(14)=\frac{E(15)+\cdots+E(19)}6=14\tfrac16$$
$$ E(13)=\frac{E(14)+\cdots+E(18)}6=13\tfrac{13}{36}$$
$$ E(12)=\frac{E(13)+\cdots+E(18)}6=12\tfrac{127}{216}$$
and so on until
$$ E(0)=\frac{E(1)+\cdots+E(5)}6=13\tfrac{13}{36}=6\tfrac{ 72285265495}{470184984576}\approx 6.1537.$$
Hence the fair price for the game is $\approx 6.15$ and the best strategy is to keep playing until one reaches $15$ or more.
Alternative approach:
Find $k$ such that you stop after $k$ dollars are on the table.
Your possible profit in rolling one more time, and then stopping is $~\displaystyle \frac{5}{6}.$
Your potential loss is $\frac{1}{6} \times k$. Therefore $k = 5$.
As your analysis indicated, if you roll one more time, when there are exactly $5$ dollars on the table, then your last roll is a break even situation.
Best Answer
The possible outcomes of rolling two die are equal to the sample space of the first die $(6)$ multiplied by the sample space of the second die $(6)$. This creates 36 possible outcomes.
\begin{array}{|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1&(1,1) &(1,2) &(1,3) &(1,4) &(1,5) &(1,6)\\ \hline 2&(2,1) &(2,2) &(2,3) &(2,4) &(2,5) &(2,6)\\ \hline 3&(3,1) &(3,2) &(3,3) &(3,4) &(3,5) &(3,6)\\ \hline 4&(4,1) &(4,2) &(4,3) &(4,4) &(4,5) &(4,6)\\ \hline 5&(5,1) &(5,2) &(5,3) &(5,4) &(5,5) &(5,6)\\ \hline 6&(6,1) &(6,2) &(6,3) &(6,4) &(6,5) &(6,6)\\ \hline \end{array}
Inside those $36$ outcomes, the total is $7$ if the two dice are
$$(1,6), (2,5), (3,4),(4,3),(5,2),(6,1)$$
which is $\frac{6}{36}$ possibilities. The total is $11$ if the two dice are
$$(5,6),(6,5)$$
which is $\frac{2}{36}$ possibilities. Rolling any other total is
$$\frac{36}{36}-\frac{6}{36}-\frac{2}{36}=\frac{28}{36}$$
possibilities. Therefore your calculation is correct since
$$\frac{6}{36}\left($20\right)+\frac{2}{36}\left($50\right)-\frac{28}{36}\left($10\right)\approx-$1.66$$