Expected number of dice that are all 6

Question

Suppose that a group of n fair dice is rolled 4ドル*6^{n-1}$ times.

a) Find the expected number of times that "all sixes" is achieved (i.e., how often among the 4ドル*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 ?

• lulu: Ask yourself a simpler question: I have a fair coin. With probability $\frac 12$ I will toss it twice and count the heads. With probability $\frac 12$ I will toss it once and then double the number of heads I see. What's the expected number of heads I get?
• brainfood: Hmm, I am not able to link this to the initial problem.
• lulu: It's the same problem, just simplified. Your problem is hard to write down, even for small $n,ドル because there are so many possible cases. My version strips it down to something you can enumerate. The goal is to persuade yourself that those weird extra rules have no impact at all on the expectation. You can just ignore them.
• lulu: I mean, just use Linearity of expectation. The expected number of successes is the sum of the expected number for each trial regardless of dependence. And the weird rules do not change the expectation of a single trial at all.

Answer 1

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.

• brainfood: Ok this is what I come up with as well, but I picked a random point in time (k) and wrote down the conditional probability given (k-1), If I would have simplified it, I would have noticed that it's $\frac{1}{6^n}$. Thank you for taking your time to answer.

Answer 2

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.

• brainfood: This does make sense - thanks for explaining this in detail.