Optimal strategy to maximize the expected gains of a probability based game

Question

The game: you start with nothing, and you choose to roll a fair dice as many times as you want. Each of the 6 outcomes is attached with a certain $ value, except for one, where you will lose all that you gained and the game ends.

My question: Mathmaitcally speaking, is there an optimal stopping rule that maximizes the expected gains of a single game?

There are two possible kinds of strategies:

1- Stop after n rounds

2- Stop as soon as your balance exceed $k

The latter makes more sense to me, as the former feels like I'm indulging in some kind of gambler's fallacy.

Answer 1

Let $N_1,\dots,N_5$ be earnings on outcomes $\{1,\dots,5\}$ such that $N_1+\dots+N_5 = N$. Let $X_i$ be the earnings after i'th round.

$\mathbb E [X_i] = \frac{5}{6} N - \frac{1}{6} X_{i-1}$. A fair strategy is to play when you expect to win from playing the game. Thus you play till $\frac{5}{6} N - \frac{1}{6} X_{i-1} \geq 0$ In other words, stop when $\frac{5}{6} N < \frac{1}{6} X_{i-1}$ or $5 N < X_{i-1}$.

Thus your strategy (2) is right. Stop when your balance exceeds $5N = 5(N_1+\dots+N_5)$

Answer 2

Alternative approach
First, I will assume that if you roll a $2,3,4,5,$ or $6$, that you win $2,3,4,5,6$, respectively. I will then illustrate my approach. Then, I will show the adjustment for different fixed winnings attached to the rolls of $2,3,4,5,6$.

If you roll, and don't hit $(1)$, your average gain will be $(4)$, which is the average of $\{2,3,4,5,6\}$. Suppose that your accumulated total so far is $T$. Then your expectation on the next roll is

$$\frac{1}{6}(-T) + \frac{5}{6}(4) = \frac{1}{6}[(4 \times 5) - T].$$

Therefore, if $T < 20$, you should roll, if $T > 20$ you should stop, and if $T = 20$, you are ambivalent.

Note
If $T > 20$, it is impossible to have a positive expectation on any subsequent roll. This is because, if you don't hit $(1)$ on this roll, your accumulated winnings will have necessarily increased from $T$, to $T^+$ (i.e. $T + $ some positive value).

Therefore, if it assumed that $T > 20$, then $T^+$ must also be $> 20.$

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Suppose instead that the winnings for rolls of $2,3,4,5,6$ are $n_2,n_3,n_4,n_5,n_6$ respectively, where $N = (n_2 + \cdots + n_6)$ Then, if you roll and don't hit $(1)$, your average winnings are $(N/5).$

Therefore, the above analysis is paralleled by :

$$\frac{1}{6}(-T) + \frac{5}{6}(N/5) = \frac{1}{6}[N - T].$$

Therefore, if $T < N$, you should roll, if $T > N$ you should stop, and if $T = N$, you are ambivalent.