When five dice are rolled, what is the chance to get five 6's if you can roll the dice that do not show a 6 on the first roll once more? We have $5$ normal dice. What is the chance to get five $6$'s if you can roll the dice that do not show a 6 one more time (if you do get a die with a $6$, you can leave it and roll the others one more time. Example: first roll $6$ $5$ $1$ $2$ $3$, we will roll $4$ dice and hope for four $6$s or if we get $6$ $6$ $2$ $3$ $3$ we will roll three dice one more time). I tried to calculate if you get $1$, $2$, $3$, $4$ dice with $6$ but I don't know how to "sum" the cases.
 A: Just FYI, dice is the plural form of die, e.g. "roll one die" or "roll two dice."
Consider doing the game with a single die. The probability of rolling a six on the first time is $1/6$, and the probability of failing the first time but succeeding the second time is $(5/6) \cdot (1/6)$. So the probability of getting a six is $\frac{1}{6} + \frac{5}{36} = \frac{11}{36}$.
Now, consider the five dice.
Each die is independent, so the probability of getting all sixes is simply $(11/36)^5$.
Effectively, the problem is now the same as "What is the probability of flipping five coins and getting five heads, if the probability of each coin being heads is $11/36$?" If you think of the problem in this way, you can answer more complicated questions like "what is the probability of at least four dice showing a six?" etc.
A: In a very long but straightforward way, we can calculate the probability by breaking it down into scenarios by how many $6$'s appear on the first roll and calculate this probability.
Case 1: Five $6$'s appear on the first roll: Event $A$
$$P(A)=\left(\dfrac{1}{6}\right)^5$$
Case 2: Four $6$'s appear on the first roll and we roll a $6$ for the remaining one die: Event $B$
$$P(B)=\binom{5}{1}\left(\dfrac{1}{6}\right)^4\left(\dfrac{5}{6}\right)\cdot\left(\dfrac{1}{6}\right)$$
Case 3: Three $6$'s appear on the first roll and we roll two $6$'s for the remaining two dice: Event $C$
$$P(C)=\binom{5}{2}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^2\cdot\left(\dfrac{1}{6}\right)^2$$
and we have the other three cases (two sixes on the first roll and three sixes on the second, a six on the first roll and four sixes on the second, and none on the first and five on the second). But by this point we can see a pattern in the probabilities of the cases.
Indeed, the probability of the event $X$ which you are asking is the sum of the probabilities of these six cases.
In other words, $$P(X)=P(A)+P(B)+\cdot\cdot\cdot+P(F)=\sum_{n=0}^5 \binom{5}{n}\left(\dfrac{1}{6}\right)^5\left(\dfrac{5}{6}\right)^n=\dfrac{161051}{60466176}$$
