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Here's a dynamic dice game. We can roll dice as many times as we want unless a 6 appears. After each roll, if 1 appears, we win \$1, if 2 appears we win \$2 etc. If 6 appears, we lose all money and game ends. We can decide if we want to continue the game after witnessing anything between 1 and 5. How much would you be willing to pay to play the game?

My approach: I first computed the expected payoff of the game. Sa $x$ is the expected payoff. Then we have $$x = \frac{1}{6}(1+x)+\frac{1}{6}(2+x)+\frac{1}{6}(3+x)+\frac{1}{6}(4+x)+\frac{1}{6}(5+x) $$ This gives me $x=15$.

What next? How to find the optimal price to play the game before it begins? Thanks.

ForumWhiner
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  • What are you trying to optimise? The payout doesn't depend on how much you pay, so paying as little as allowed seems to be the answer. – Henrik supports the community Dec 09 '20 at 15:18
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    The LHS of the equation is your payoff if you stop now, and RHS is the payoff if you continue for one more roll. So by comparing these two quantities, you form a stopping rule by maximizing the expected payoff - continue if current payoff is $$14$ or less, and stop otherwise. So what you want to calculate is the expected payoff of this game, which is the fair price of this game. – BGM Dec 09 '20 at 16:03

1 Answers1

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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.

Hagen von Eitzen
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