Probability – When to Stop Rolling a Die in a Game Where 6 Loses Everything

dicegamblingprobability

You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.

When should one stop playing this game? Obviously, one wants to maximize total score.

As I was asked to show my preliminary results on this one, here they are:

If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following:

$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$

which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!

Best Answer

Before deciding whether to stop or roll, suppose you have a non-negative integer number of points $n$.

How many more rolls should you make to maximise the expected gain over stopping (zero)?

Suppose that further number of rolls is another non-negative integer $k$. Now consider the $6^k$ possible sequences of $k$ rolls:

  • In $5^k$ of those sequences there is no six and you win some points. The sum $D_k$ over all such sequences of the sum of dice rolls within each sequence satisfies the recurrence relation $$D_0=0\quad D_{n+1}=5D_n+15\cdot5^n$$ It turns out that this has a closed form: $$D_k=15k\cdot5^{k-1}=3k\cdot5^k$$
  • In the remaining $6^k-5^k$ sequences there is at least one six and you lose the $n$ points you had beforehand.

So the expected gain when you have $n$ points and try to roll $k$ more times before stopping is $$G(n,k)=\frac{D_k-n(6^k-5^k)}{6^k}=\frac{3k\cdot5^k-n(6^k-5^k)}{6^k}$$ For a fixed $n$, the $k$ that maximises $G(n,k)$ is $m(n)=\max(5-\lfloor n/3\rfloor,0)$; if $3\mid n$ then $k=m(n)+1$ also forms a maximum.

Suppose we fix the maximum number of rolls before starting the game. At $n=0$, $k=5$ and $k=6$ maximise $G(n,k)$ and the expected score with this strategy is $$G(0,5)=\frac{15625}{2592}=6.028163\dots$$ But what if we roll once and then fix the maximum rolls afterwards? If we roll 1 or 2, we roll at most 5 more times; if 3, 4 or 5, 4 more times. The expected score here is higher: $$\frac16(1+G(1,5)+2+G(2,5)+3+G(3,4)+4+G(4,4)+5+G(5,4))=6.068351\dots$$ We will get an even higher expected score if we roll twice and then set the roll limit. This implies that the greedy strategy, outlined below, is optimal:

Before the start of each new roll, calculate $m(n)$. Roll if this is positive and stop if this is zero.

When $n\ge15$, $m(n)=0$. A naïve calculation that formed the previous version of this answer says that rolling once has zero expected gain when $n=15$ and negative expected gain when $n>15$. Together, these suggest that we should stop if and when we have 15 or more points.

Finding a way to calculate the expected score under this "stop-at-15" strategy took quite a while for me to conceptualise and then program, but I managed it in the end; the program is here. The expected score works out to be $$\frac{2893395172951}{470184984576}=6.1537379284\dots$$ So this is the maximum expected score you can achieve.

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