Rolling a die until equal or higher value

Question

Roll a die once, and denote U the number your rolled. Then continue rolling the die until you either match or exceed U. What is the expected number of additional rolls?

The way I solved this was I treated X : the number of extra rolls ~ Geo (p), where p is the Pr(N $\ge$ U) and N is the number you roll. p = $\sum_{k=1}^{6} P(N \geq u | U = u) P(U = u)$ which I get 7/12 However, I am not too clear on whether that is the right way to interpret it.

Answer 1

The first roll determines $U$ and that stay fixed for the following additional rolls, which are those that we will effectively consider.
Given $U$, the probability of matching or exceeding it at any roll is $p_u =(6-U+1)/6$.

This probability is constant for each roll, and independent from previous outcomes.
In fact, practically, after $U$ has been determined we can paint all the faces of value $U$ or greater in red, and thus the rolls become Bernoulli trials.

The probability that $n \in [1, \infty)$ rolls be required to hit a "red" is thus a geometric distribution with mean value $$ \bar n(U) = {1 \over {p_{\,u} }} = {6 \over {7 - U}} $$

Since $U$ is uniformly distributed over $[1,6]$, then the expected number of rolls will become $$ \bar n = \sum\limits_{U = 1}^6 {{1 \over 6}\bar n(U)} = \sum\limits_{U = 1}^6 {{1 \over {7 - U}}} = \sum\limits_{k = 1}^6 {{1 \over k}} = H_{\,6} = {{49} \over {20}} $$

Answer 2

The probability that a particular roll is equal to or greater than your first roll is indeed $7/12.$

But the number of extra rolls is not a geometric distribution.

It would be a geometric distribution if the probability of matching or exceeding the first roll on any given subsequent roll were independent of the probability on any other roll. But the probabilities are not independent.

For example, let $A$ be the event that the second roll matches or exceeds the first, and let $B$ be the event that the third roll matches or exceeds the first. Then $P(A) = P(B) = 7/12.$ But $$P(A\cap B \mid U = u) = \frac{(7 - u)^2}{36}, $$

and so $$P(A\cap B) = \sum_{x=1}^6 P(A\cap B \mid U = u)P(U=u) = \frac{91}{216}, $$

but $$ P(A) P(B) = \left(\frac7{12}\right)^2 = \frac{49}{144} \neq P(A\cap B). $$

It should not be surprising that the events are not independent. After all, if all you know about the first two rolls is that the second roll was equal to or greater than the first, the first roll is more likely to be a low number than a high number (given that information), and therefore the third roll has a better than $7/12$ chance to match or exceed the first.

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What you can do is, rather than treating $A$ and $B$ as independent events, don't even consider the total probability of success on any given roll. Instead, consider each possible first roll separately and only bring the results together at the end.

So $E(X \mid U=u)$ is equal to the expectation of a geometric variable with $p = (7 - u)/6,$ because given a known first roll the chance of matching or exceeding it on any subsequent roll is independent from the chance on any other roll. And $$ E(X) = \sum_{x=1}^6 E(X \mid U = u)P(U=u). $$

I leave you to work out the result, but it is quite a bit larger than $12/7.$