I'm aware that the $E[X_n]$ where $X$ is the average number rolls needed to roll $n$ consecutive sixes solves to $E[X_2]=42$ when $n=2$… but why does this differ from rolling other numbers (which are distinct from eachother) such as $5$ and then $6$? For example, if I wanted $X_n$ to be the average number of rolls for getting two consecutive numbers in general, which need not be the same number.
If I begin at $0$ rolls and roll a die $2$ times, the probability of getting $2$ sixes on the first two rolls is $\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$ and the probability of rolling a $5$ and then a $6$ is the same probability.
Intuitively, this led me to believe I could extrapolate the result mentioned above regarding $n$ consecutive sixes and apply it to rolling $n$ consecutive numbers (which need not be the same). But the problem I solved did not have the answer as $42$, the answer for the expected value to roll a $5$ followed immediately by a $6$ was $36$.
Best Answer
When you roll a six and then a non-six, there is no way you can "complete" your two sixes with one more roll.
But if you roll a five, and then a non-six, occasionally, that non-six is a five, in which case, you can complete your pair in the next roll.
This makes it easier to get a $56$ earlier than an $66$.
For example, the ways to get your first $66$ in exactly three rolls is the five possible sequences $[1-5]66$, while the ways to get your first $56$ in exactly three rolls is the six sequences $[1-6]56$.
One reason this is counter-intuitive: We play the following game with two players. A die is tossed repeatedly, until we get a six immediately followed by a 5 or 6. Player 1 wins if the last two rolls are 66, player 2 if they were 65.
This game is "even" - each player has a probability of $\frac{1}{2}$ of winning.
So this even-ness makes us intuit that the expectation of the first occurrence of each is also the same.
But that is a fallacy. When the first is 66 at roll $n$, it is possible for the first $65$ to be at roll $n+1$. On the other hand, if the first is $65$ at roll $n$, it takes minimum of $n+2$ rolls to get $66$. So when $66$ wins, the first $65$ is expected to follow a little faster than the reverse case.