6
$\begingroup$

I am working on the following problem from a book:

A casino has a dice game. You can roll as many times as you want. For each roll you get paid $M$ dollars where $M$ is the number of dots on the roll as long as you do not roll a 6. The payment for each roll is additive. However, if you roll a 6, the game terminates and you lose your accumulated profit thus far. How much are you willing to spend on this game?

I am looking at the solution provided by the book, and I am confused. The solution is posted below. The part I am confused about is examining the threshhold for $n$ at which

$$ 5/6 \cdot n + 2.5 > n $$

Equality in the above expression holds when $n = 15$. I understand how they determined this solution, but it is not clear to me why this is the most optimal threshhold because the equation $5/6 * n + 2.5$ is derived from assuming you can only roll 1 more time. So if we have $n = \$16$, the solution is telling us that we shouldn't re-roll because the expectation of the profit of an additional roll is less than the current profit. But this assumes that we can only roll 1 additional time. Shouldn't be consider the cases of rolling more than 1 time if we have $n = \$16$ already?

enter image description here

$\endgroup$
7
  • 1
    $\begingroup$ The solution given does not assume that you \textbf{only} have one more roll; it is, on the other hand, trying to figure out whether you should throw it one more time assuming that you made $n$ amount of dollars. You can then apply this logic recursively. $\endgroup$
    – kiyomi
    Sep 6, 2020 at 1:32
  • $\begingroup$ @Denis28, But they do $\frac{1}{6}((n+1) + \ldots + (n+5))$ to determine whether it exceeds $n$, which is used to determine the threshhold at which they stop the recursion. They're saying that if you have more than 14 dollars, then you should not roll again, but that was determined based on checking if an additional roll is beneficial. Why didn't they check if 2+ additional rolls is beneficial when 1 additional roll isn't? $\endgroup$ Sep 6, 2020 at 2:06
  • 1
    $\begingroup$ @user5965026 yes, one should check that 2+ additional rolls is not beneficial when 1 additional roll isn't. To do this, you can show that the repeated iteration of the function $n \mapsto \frac{5}{6}n+2.5$ will just keep decreasing; or maybe there is a simpler argument that the solution is implicitly assuming. $\endgroup$ Sep 21, 2020 at 20:36
  • 1
    $\begingroup$ What is the book? $\endgroup$
    – RobPratt
    Sep 22, 2020 at 0:54
  • $\begingroup$ @RobPratt A practical Guide to Quantitative Finance Interviews $\endgroup$ Sep 22, 2020 at 0:56

2 Answers 2

4
+50
$\begingroup$

Let $V(n)$ represent the expected number of dollars won if you have accumulated $n$ dollars. If you decide to stop, you win $n$ dollars. If you decide to roll, you will be in one of six states, each with probability $1/6$, and by conditioning on the value $r$ of the next roll, we obtain expected value $$\frac{1}{6} \sum_{r=1}^5 V(n+r) + \frac{1}{6}\cdot 0.$$ Hence $$V(n) = \max\left(n,\frac{1}{6} \sum_{r=1}^5 V(n+r)\right) \quad \text{for all $n \ge 0$} \tag1$$ If you can establish a boundary condition $V(n)=n$ for all $n\ge m$, then you can solve the recurrence for $n<m$ to find $V(0)$.

To find such a threshold $m$, note that the recurrence $(1)$ implies that $$m \ge \frac{1}{6} \sum_{r=1}^5 V(m+r) = \frac{1}{6} \sum_{r=1}^5 (m+r) = \frac{5m+15}{6},$$ so $m \ge 15$, but the book solution does not show that $m \le 15$. Indeed, if you set a larger threshold, say $m=16$, you still obtain $V(15)=15$. Implicit in the argument is that $V(n)=n$ implies $V(n+1)=n+1$. In words, if it is optimal to stop rolling in state $n$, then it is optimal to stop rolling in state $n+1$.

$\endgroup$
7
  • $\begingroup$ What do you mean by "if you can establish a BC V(n) = n?" How do you know whether or not this BC is establishable? $\endgroup$ Sep 22, 2020 at 0:33
  • $\begingroup$ Without some boundary condition, you cannot compute $V(n)$ recursively. At the moment, I don't see any argument for why there should be any threshold. $\endgroup$
    – RobPratt
    Sep 22, 2020 at 0:48
  • $\begingroup$ Yeah I don't think I agree with the process they used to obtain their solution. It seems to make the assumption that you roll one more time and compare that with the current roll, but there is no proof of optimality. $\endgroup$ Sep 22, 2020 at 1:00
  • $\begingroup$ Yes, the book solution should have said "expected payoff will become at least" ... The footnote hints that there should be a threshold, but it is conceivable that every roll increases your expected payoff. The book correctly shows that you should roll again if $n \le 14$, but that does not imply that you should stop when $n=15$. $\endgroup$
    – RobPratt
    Sep 22, 2020 at 1:19
  • 1
    $\begingroup$ Yes, the book implies that stopping when $n \ge 15$ is optimal because rolling when $n \le 14$ is optimal, but that is faulty logic. $\endgroup$
    – RobPratt
    Sep 22, 2020 at 1:26
0
$\begingroup$

yes it makes sense to stop rolling after you have accumulated a certain no of dollars lets assume we have n dollars at a step then at the next step we have $\frac{5n}{6}\,+\,2.5$ dollars so this way if we continue at next $m^{th}$ step we have $(\frac{5^{m}}{6^{m}})\,(n-15)\,+\,15$ our objective would be to maximize this quantity we can clearly see that if we tend m to infinity that is we keep on playing the game then we have an expected outcome of 15 dollars no matter what we have at a particular step, so it makes no sense to start the game with more than 15 dollars at hand, at any step if we have >= 15 dollars then we check whether the expected outcome at the next step is greater than the current amount we have or not , if yes then we give another roll, if no then we stop.

$\endgroup$
1
  • $\begingroup$ I guess when at the beginning you write on a quantity which we have at some step your mean its expectation. $\endgroup$ Nov 7, 2020 at 14:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .