A little game with rolling a fair dice

Question

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?

Answer 1

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.

Answer 2

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$ ;)

Answer 3

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.