Decision to play on in perfect square die game Question:You have a six-sided dice, and you will   receive money that equals to sum of all the numbers you roll. After each roll, if the sum is a perfect square, the game ends and you lose all the money. If not, you can decide to keep rolling or stop the game. If your sum is $35$ now, should you keep play? 
Attempt: Should we be incorporating the $35$ already accumulated, in which case we have:
\begin{equation*}
E(\text{Winnings after 1 throw}) = \frac{1}{6}(-35)+\frac{5}{6}(35+4+3.5).
\end{equation*}
The $4$ comes from the fact that if we don't throw a one in the next toss, the average is $20/5 = 4$. And if we do not throw a one on the next toss, we are guaranteed not to hit a perfect square on the second toss after that, so we expect to make an extra $3.5.$
My approach: you lose your 35 dollars with 16.67% chance, or with 5/6 chance you keep your 35, plus gain the average of the next roll (2+3+4+5+6)/5, and also after the first roll, you are guaranteed to be too far away from the next perfect square (49) so you have an added 3.5 to look forward to with certainty
 A: If you count rolling a $1$ as $-35$, then your reference point is your current $35$, and you shouldn't include $35$ in the chance of further winnings.
On the other hand, you can certainly keep rolling while your score is below $43$.
So I would take the value of the next roll to be at least $(1/6)(-35)+(5/6)8$ which is positive.
A: It's up to you whether you incorporate the $35$ already accumulated. You can:  


*

*incorporate it, in which case you compare the expected total winnings $E_{total}$ after throwing again with the $35$ that you win by not throwing again;  

*not incorporate it, in which case you compare the expected extra winnings $E_{extra}$ (positive or negative) against the zero that you gain by not throwing again.


But $E_{total} = E_{extra} + 35$, so the comparisons are really the same.
But your calculation does both! To clear up the confusion, let's use method 1: If we don't throw, we win $35$. If we throw, our expected total winnings are at least $\frac56(35+4+3.5) \approx 35.4$. They may be more than this $-$ it depends on what the optimal strategy is after two more throws. But the strategy of always stopping after two more throws, although not optimal, is still good enough to beat the strategy of stopping at $35$. So you should definitely throw again at $35$.
Just for completeness: method 2 compares the expected extra winnings $\frac{1}{6}(-35)+\frac{5}{6}(4+3.5) \approx 0.4$ against $0$. Which is the same as comparing $35.4$ against $35$.
