Expected number of dice that are all 6 Suppose that a group of n fair dice is rolled $4*6^{n-1}$ times.
a) Find the expected number of times that “all sixes” is achieved (i.e., how often among the $4*6^{n-1}$ rolls it happens that all n dice land 6 simultaneously). 
The answer is $\frac{4}{6}$. (Binomial distribution)
b) Same thing but after one normal roll of the n dice, going from one roll to the next, with probability 6/7 keep the dice in the
same configuration and re-roll with probability 1/7. For example, if n = 3 and the 7th roll is (3, 1, 4), then 6/7 of the time the 8th roll remains (3, 1, 4) and 1/7 of the time the 8th roll is a new random outcome. Does the expected number of times that “all sixes” is achieved stay the same, increase, or decrease (compared with (a))? Give a short but clear explanation.
I'd appreciate some hints on how to approach this. My initial thought was to try to come up with the probability for a throw to be all sixes, using conditional probability, but the equation is messy. Is there a better intuition for this ?  

For b)
The first roll can be all sixes if all the dice roll sixes. The second roll can be all sixes if the first roll is all sixes and the second roll does not change or the second roll is a reroll and you roll all sixes (whether or not the first roll is all sixes).
So, for the second roll, the probability that you get all sixes is: $$\dfrac{6}{7}\left(\dfrac{1}{6^n}\right)+\dfrac{1}{7}\left(\dfrac{1}{6^n}\right) = \dfrac{1}{6^n}$$
By induction, you can show that the probability that any specific roll is all sixes is $\dfrac{1}{6^n}$. Thus, the expected number of times you roll all sixes remains unchanged.
