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You play the following game: you roll a fair dice then either you stop rolling and take the sum of the rolled numbers so far or you continue rolling the dice. Any time if a 1 is rolled you lose all your money, and has no option to continue the game. The strategy that you follow is that you wait until the cumulated price reaches a given level and then you stop. What should be this level in order to maximize the expected value of the prize?

My answer is 20...is it correct?

Kapes Mate
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    If you have a balance of $20$ then your expected gain on the next throw is $\frac 16\times (2+3+4+5+6-20)=0$ so with that balance you are indifferent to throwing again (assuming your utility function is just "maximize expected gain"). With a balance of $21$ your expectation is negative so I think the decision is clearer. – lulu Nov 19 '19 at 13:26
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    My standard rant: there is no such thing as "a dice". You mean "a fair die". "Dice" is the plural of the singular "die". – user247327 Nov 19 '19 at 13:43
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    Closely related game but where you lose all money having rolled a 6. The maths involved in finding an answer are otherwise the same. – JMoravitz Nov 19 '19 at 13:59
  • Yes, it is a similar game but it is not the only difference between the two exercise ... in that game you have mentioned they ask the optimal number of the rolled dice...here the task is to determine the optimal level of your money when you should stop the game... – Kapes Mate Nov 19 '19 at 14:14

3 Answers3

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Let's say that $k$ points is the place where rolling again and stopping have the same expected value. That means that $$k=\frac16(0)+\frac16(k+2)+\frac16(k+3)+\frac16(k+4)+\frac16(k+5)+\frac16(k+6)\\k=\frac56k+\frac{20}6\\\frac16k=\frac{20}6\\k=20$$

So you are correct.

1

If we stop the game after at least $k$ is reached, then it finishes after finitely many steps with outcomes $0, k,\dots, k+5$. Let $p_0,p_k,\dots,p_{k+5}$ be the probability that the outcome is the respective value. Then the expected outcome for $k$-strategy is $$ E_k = \sum_{i=0}^5 p_{k+i}(k+i). $$ Let me now consider the game for the $k+1$ strategy. The difference to the $k$ strategy is that we have to continue if the outcome is exactly $k$. The probabilities for the $k+1$ strategy are then $$ \tilde p_{k+1} = p_{k+1}, $$ $$ \tilde p_{k+i} = p_{k+i} + \frac 16 p_k \quad i=2\dots 5. $$ $$ \tilde p_{k+6} = \frac 16 p_k. $$ The expected value for the $k+1$ strategy is $$\begin{split} E_{k+1}&=\sum_{i=1}^6 \tilde p_{k+i}(k+i)\\ &= \sum_{i=1}^5 p_{k+i}(k+i) + \sum_{i=2}^6 \frac16p_k (k+i)\\ &=\sum_{i=0}^5 p_{k+i}(k+i) + p_k(-k + \frac 56p_k +\frac{20}6)\\ &= E_k + \frac16(20-k). \end{split}$$ Hence for $k=20$ or $k=21$ the expected outcome is maximal.

daw
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Every time you don't lose you will win an average of 4. Total prize would be $$ W_n = 4 \cdot n \rightarrow_n \infty $$ The probability of winning is 5/6 and after n turns, the probability of still winning something is: $$ P_n = ( \frac{5}{6} )^n \rightarrow_n 0 $$

Total prize expectancy is $$ W_n \cdot P_n = 4 \cdot n \cdot( \frac{5}{6} )^n $$

Which has two maxima at n equal to 5 and 6. $$ W_5 \cdot P_5 = W_6 \cdot P_6 = \frac{4 \cdot 5^6 }{ 6^5 } \approx 8.04 $$

So my strategy would be to play for 5 or 6 turns what will maximize my earnings. Every time I win, I will earn an average of 20 or 24.

Nice game to play with friends. Game fee 10$ ;)

Alex Moreno
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  • You should not set a number of rolls in advance. You should set a total score to stop at in advance. Having two rolls of $2$ is the same as having one of $4$-all that matters is the current total. – Ross Millikan Nov 19 '19 at 14:54