Estimate expected payoff of rolling a dice, with choice of rolling up to $50$ times. This is an extended question of the classical rolling dice and give face value question. 

You roll a dice, and you'll be paid by face value. If you're not satisfied, you can roll again. You are allowed $k$ rolls. 

In the old question, if you are allowed two rolls, then the expected payoff is $E[\text{payoff}] = 4.25$. 
If  you are allowed $3$ rolls, the expected payoff is $E[\text{payoff}] = 4.67$. 
If you can roll up to $50$ times, you can calculate the payoff using the formula and get $E = 5.999762$, notice that after $5^\text{th}$ roll, your expected payoff will be greater than $5$, so you'll only stop once you roll $6$. 

So my question here is, without exact calculation(using geometric process), how would you estimate how many $9$s are there in the answer? Or another way to ask will be, is the expected payoff bigger than $5.9$? bigger than $5.99$? etc.

Let $E_k$ be the expected payoff, if you're allowed to roll $k$ times, with the rules as you've described them. We can compute $E_k$ recursively. 

With just $1$ roll, you must take what you get, since there are no more rolls. The expected value is therefore $$E_1 = \frac{1+2+3+4+5+6}{6} = 3.5$$
With $2$ rolls, if your first roll is $4$, $5$, or $6$, you will keep it, otherwise you will reroll and get $E_1$ from your next (and last) roll. Therefore, \begin{align*}E_2 &= \frac{4+5+6}{6}+\frac{1}{2}E_1 \\
&= 2.5+\frac{1}{2}(3.5) = 4.25\end{align*}
With $3$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_2$ from your next two rolls. Therefore, \begin{align*}
E_3 &= \frac{5+6}{6}+\frac{2}{3}E_2\\
&= \frac{11}{6}+\frac{2}{3}(4.25) = 4.\overline{6}
\end{align*}
With $4$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_3$ from your next three rolls. Therefore, \begin{align*}
E_4 &= \frac{5+6}{6}+\frac{2}{3}E_3\\
&= \frac{11}{6}+\frac{2}{3}(4.\overline{6}) = 4.9\overline{4}
\end{align*}
With $5$ rolls, if your first roll is $5$ or $6$, then you will keep it, otherwise you will reroll and get $E_4$ from your next three rolls. Therefore, \begin{align*}
E_5 &= \frac{5+6}{6}+\frac{2}{3}E_4\\
&= \frac{11}{6}+\frac{2}{3}(4.9\overline{4}) = 5.1\overline{296} = \frac{277}{54}
\end{align*}
Now, we have reached the point at which the recursion relation is stable. With more than $5$ rolls, you will always only keep the first roll if it is a $6$. 

With $k$ rolls, $k>5$ if your first roll is $6$, you will keep it, otherwise you will reroll and get $E_{k-1}$ from the next $k-1$ rolls. Therefore,\begin{align*}
E_k &= \frac{6}{6}+\frac{5}{6}E_{k-1}\\
E_k &= 1+\frac{5}{6}E_{k-1}\tag{1}\\\
\end{align*}
Notice that $$E_5 = \frac{277}{54} = 6 - \frac{47}{54}$$
The solution to the recurrence relation in $(1)$, with initial value $E_5 = 6- 47/54$,  is:
$$E_k = 6 - \left(\frac{47 \cdot 144}{5^5}\left(\frac{5}{6}\right)^k\right)$$

Therefore, in general, the maximum expected payoff that you can achieve, when allowed $k$ rolls of a six-sided die, for any $k$, is $$\boxed{\,\,E_k \,=\,\begin{cases}7/2 \qquad &\text{if}\,\,\,k=1\phantom{l^{l^{l^{\overline{l}}}}}\\ 17/4 \qquad &\text{if}\,\,\,k=2\phantom{l^{l^{l^{\overline{l}}}}}\\ 14/3 \qquad &\text{if}\,\,\,k=3\phantom{l^{l^{l^{\overline{l}}}}}\\ 89/18 \qquad &\text{if}\,\,\,k=4\phantom{l^{l^{l^{\overline{l}}}}}\\\\6-\displaystyle\frac{6768}{3125}\left(\displaystyle\frac{5}{6}\right)^k \qquad &\text{if}\,\,\,k\geq 5\phantom{l_{l_{l_{l_l}}}}\\  \end{cases}\,\,\,}$$
