3
$\begingroup$

We are playing a game with a die where we roll and sum our rolls. We can choose to stop at any time and take the sum as our score, but if we roll the same face twice in a row (consecutively), we lose all our points. Calculate the range in which the maximal expected score lies?

My approach : Suppose the current sum so far is $S$, the last roll was $k$. Then expected score in next roll is $0 \times \frac{1}{6} + \left(S + \frac{1}{5}(1+2+\ldots+6-k) \right) \frac{5}{6} = \frac{5S}{6} + \frac{1}{6} (21-k)$, this needs to be greater than or equal to $S$ to make the next roll or equivalently $S \leq 21-k$ or $S+k \leq 21$

Thus, condition for making next roll is : current sum + last roll should be less than or equal to 21

This is my stopping strategy, now how do I compute expected score ?

$\endgroup$
7
  • 1
    $\begingroup$ What do you mean by "range in which maximal expected score lies"? if it's expected score, then it's a unique value so asking for the maximal doesn't quite make sense. If it's maximal, then it's again a single value, so asking for the range doesn't quite make sense. $\endgroup$
    – Calvin Lin
    Commented Feb 10 at 17:51
  • $\begingroup$ @CalvinLin I believe "maximal expected score" means the expected score for the optimal stopping strategy. But I also don't understand asking for a range. $\endgroup$
    – user1266745
    Commented Feb 10 at 17:58
  • 1
    $\begingroup$ Does the game stop after we lose all our points? That wasn't stated in the question, and so the optimal strategy is to just keep on rolling forever as with probability 1 you will eventually get higher points. $\endgroup$
    – Calvin Lin
    Commented Feb 10 at 18:05
  • 1
    $\begingroup$ The final strategy is probably : if $S<S_0$, continue, if $S>S_1$, stop, if $S$ between $S_0$ and $S_1$, continue or stop, depending of last roll $k$ ; So the final score is always between $S_0$ and $S_1+6$ , or $0$; this is the reason why we have a 'range'. $\endgroup$
    – Lourrran
    Commented Feb 10 at 18:48
  • 1
    $\begingroup$ I'm not sure your strategy gives the best expected score. Comparing to the expected score after one more roll is not necessarily the same as the expected value of the scores you can get by continuing for one or more rolls. $\endgroup$
    – aschepler
    Commented Feb 10 at 20:43

1 Answer 1

3
$\begingroup$

Assume the game terminates with payoff $0$ if the value of a roll is equal to the value of the previous roll, otherwise, from "live" state $(s,k)$, the game terminates with payoff $s$ if $s+k > 21$.

Let $e(s,k)$ be the expected value of the game from live state $(s,k)$.

Then we have the recursion $$ e(s,k)=s \qquad\qquad\qquad\; $$ if $s+k > 21$, otherwise, if $s+k\le 21$, we have $$ e(s,k) = \sum_{j\in J_k} {\small{\frac{1}{6}}} e(s+j,j) $$ where $J_k=\{1,...,6\}{\setminus}\{k\}$.

Implementing the recursion in Maple we get $$ \qquad\;\;\;\;\; e(0,0) = \frac{678747596227}{78364164096} \approx 8.66 $$

$\endgroup$
1
  • $\begingroup$ really nice dp solution $\endgroup$
    – Harsh
    Commented Feb 11 at 7:12

You must log in to answer this question.

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