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 ?  

As a different approach, use Linearity of Expectation.
We remark that, whether you add the rules twist or not, the probability of success on a given trial is $p=\frac 1{6^n}$.  The rules don't change that at all.  Thus the expected number is just $p$ times the number of trials in both cases.
Worth remarking: this is a feature of Expected Value.  If, say, the problem asked for Variance, the rules would make a big difference.  The dependence between trials would increase the variance since extreme outcomes would get extended.  But Expected Value is linear, even if the trials are dependent.
