If instead of $n=100$ possibilities we have a large $n$, the expectation converges to $n/6$. In fact, if we select numbers without replacement rather than with (which makes a negligible difference when $n$ is large), the expectation is exactly $\frac16(n+1)$. Here's a way to see that:
First, change the game to be:
We have $n+1$ chairs arranged in a circle. Select one chair at random, remove it from the circle. That leaves a chain of $n$ chairs. Select 5 of those chairs at random, and count the gap between the second and third selected chair.
It should be clear that this gives the same answer -- the business with making a circle and cutting it into a line is completely irrelevant. However, now we can modify the rules further to
We have $n+1$ chairs arranged in a circle. Select six chairs at random. Now pick (still at random) one of the six chairs, remove it, unfold the rest to a line, and count the gap between the second and third selected chair.
Doing the random choices in a different order shouldn't change anything, so we still get the sought-for answer. One more reformulation:
We have $n+1$ chairs arranged in a circle. Select six chairs at random. Start from a random chair among the selected six chairs and walk clockwise. Count how many chairs appear between the second and third time you walk past a selected chair.
We have $n+1$ chairs arranged in a circle. Select six chairs at random. Select one of the six gaps between selected chairs at random, and count how long it is.
In the original formulation the expected size of the different gaps were not obviously the same -- but now all of the apparent asymmetry in that description has been removed. By symmetry the expected length of the random gap can only be one sixth of the total length of the circle (because that is the sum of the 6 possibilities we choose uniformly between in the second choice).
It gets somewhat trickier with replacement because the chair where we break the circle cannot be selected again in the next choices of the original game, so reformulating the game doesn't give us perfect symmetry. That must be why Patrick Stevens' exact answers don't show a simple 6 in the denominator.
By this reasoning we also get $n/6$ exactly if you uniformly choose 5 real numbers in $[0,n)$ (in which case replacement doesn't matter at all).