Probability of rolling at least one 6 while rerolling 1's Scenario: You roll a number of 6-sided dice
Success: Roll at least one 6
Conditions: You can re-roll any 1's you get on the first roll
What are the odds of success for n dice?
Example: 7 dice
Roll:            (1, 1, 1, 4, 4, 4, 5)
Re-roll:      (1, 2, 3)
Final dice: (1, 2, 3, 4, 4, 4, 5)
Result:        Failure
My take on this was: For any number of rolled 1's, "replace" those rolls with rolling n+x dice (where x is the number of 1's rolled) and thus reducing the problem to simple combinatorics, but I didn't get very far.
I suppose there is a simple "trick", so I'm looking for other angles into this problem.
However, if it turns out not to be so simple, please try to be as verbose and layman-friendly as you can.
 A: The chance of ending up with a $1$ on throwing one die is $\frac1{36}$ since you throw $1$ the first time and re-roll and get $1$ again.
The remaining numbers are (obviously) equally likely - probability therefore $\frac 15 \cdot \frac {35}{36}=\frac 7{36}$ for each of $2,3,4,5,6$.
Probability of not throwing any $6$s with $n$ dice is therefore the probability that they are all something other than a six, which is $\left(\frac {36-7}{36}\right)^n=\left(\frac {29}{36}\right)^n$
The probability of at least one $6$ is then $1-\left(\frac {29}{36}\right)^n$
A: The best strategy is to reroll any rolled 1, of course.
So using that strategy, what is the chance that you do not end up at a 6 with a single dice?
The probability for an instant 6 is $1/6$, and for a 1 with a rerolled 6 is $(1/6)^2 = 1/36$.
So in total for a single dice, the chance not to get a 6 is
$$1 - (1/6 + 1/36) = 29/36.$$
So for $n$ dice, the probability of getting at least one 6 is
$$
p = 1 - \left(\frac{29}{36}\right)^{\!n}.
$$
For small $n$, we get these approximated probabilities:
$$
\begin{array}{cc}
n & p \\
\hline
1 & 19.44\% \\
2 & 35.10\% \\
3 & 47.73\% \\
4 & 57.89\% \\
5 & 66.08\% \\
6 & 72.67\% \\
7 & 77.99\% \\
8 & 82.27\% \\
9 & 85.72\% \\
10 & 88.49\%
\end{array}
$$
