A little game with rolling a fair dice 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?
 A: 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.
A: 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.
A: 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$ ;)
