Is This Expected?

In general, suppose we keep rolling until we get the number $N$ . Then the expected value of each roll up until we roll $N$ is $A_{N} = \dfrac{21 - N}{5}$ . As the die is fair, the probability that we roll $N$ on any given roll is $\dfrac{1}{6}$ , and the probability we don't is $\dfrac{5}{6}$ . If we take $m \ge 1$ rolls to get $N$ then the final sum will be $N + (m - 1)A_{N}$ . The expected sum $E_{N}$ is then

$\dfrac{1}{6}N + \dfrac{1}{6}*\dfrac{5}{6}(N + A_{N}) + \dfrac{1}{6}\left(\dfrac{5}{6}\right)^{2}(N + 2A_{N}) + ... + \dfrac{1}{6}\left(\dfrac{5}{6}\right)^{m-1}(N + (m - 1)A_{N}) + .... =$

$\displaystyle\dfrac{N}{6}\sum_{k=0}^{\infty} \left(\dfrac{5}{6}\right)^{k} + \dfrac{A_{N}}{6}\sum_{k=1}^{\infty} k\left(\dfrac{5}{6}\right)^{k} = \dfrac{N}{6} \times 6 + \dfrac{A_N}{6} \times \dfrac{\dfrac{5}{6}}{\left(1 - \dfrac{5}{6}\right)^{2}} = N + 5A_{N} = N + (21 - N) = 21$ .

So regardless of which number we roll until, the expected (inclusive) final sum is invariant, and thus the answer is $\boxed{\text{A and B are equal}}$ .

(Note: A shortcut could have been to note that the expected number of rolls to get any given number is $6$ , so the expected sum would be $N + 5A_{N} = 21$ as found above, but I wasn't certain that this was airtight so I went with the long version to be sure.)