How much money should we take? I'm a new user so if my question is inappropriate, please comment (or edit maybe).
We want to define a dice game. We will be the casino and a customer will roll dice. I will assume customer is man. He can stop whenever he wants and when he stopped he can take the money as much as sum of his dices so far. But, if he dices 1 he must stop and he can't take any money. For the sake of casino, how much money should we take at the beginning of the game, minimum?
For example: If he dices 2-5-1, he can't get any money but if he dices 4 and stop he get 4 dollars.
I don't have any good work for this problem but I guess it is 10. Also, maybe this helps:
If game would be free and we can play it once in a year, when should we stop? Obviously, if we get 1000 dollars we should stop and if we gain 2 dollars we should play.
Answer for this question is 20. Because, in the game, if we get so far $t$, after next move we will have $\frac26 + \frac36 + \frac46 + \frac56 + \frac66 -\frac t6 = \frac{20-t} 6 $.
Please don't hesitate to edit linguistic mistakes in my question. Thanks for any help.
 A: Let $f(n)$ be the expected final win for a player already having a balance of $n$ and employing the optimal strategy. Trivially, $f(n)\ge n$ as he might decide to stop right now.
However, if the optimal strategy tells him to play at least once more, we find that $f(n)=\frac16\cdot 0+\sum_{k=2}^6 \frac16f(n+k)$. Thus
$$\tag1 f(n)=\max\left\{n,\frac16\sum_{k=2}^6f(n+k)\right\}.$$
If the player always plays as if his current balance were one more than it actually is, the expected win with starting balance $n$ will be $f(n+1)-1+p$ where $p$ is his probability of ending with rolling a "1". Therefore $f(n)\ge f(n+1)-1$ and by induction $f(n+k)\le f(n)+k$. Then from $(1)$ we get
$$ f(n)\le \max\left\{n,\frac56f(n)+\frac{10}{3}\right\}$$ 
Especially, $f(n)>n$ implies $f(n)\le \frac56f(n)+\frac{10}3$, i.e. $f(n)\le 20$. Thus $f(n)=n$ if $n\ge 20$ and we can calculate $f(n)$ for $n<20$ backwards from $(1)$. We obtain step by step
$$\begin{align}f(19)&=\frac{115}6\\
f(18)&=\frac{55}3\\
&\vdots\\
f(0)&=\frac{492303203}{60466176}\approx 8.1418\end{align}$$
as the expected value of the game when starting with zero balance, and this is also the fair price.
We have at the same time found the optimal strategy: As $f(n)>n$ iff $n\le 19$, the optimal strategy is to continue until you have collected at least $20$ and then stop (this matches your remarks in your last paragraph).
