Suppose the time intervals for a give line of bus arriving a fixed bus stop are $ \{x_i \}_{i = 1}^n $, which means that after $x_1$ minutes another bus will arrive and the next one will arrive after $x_2$ minutes, and so on.
So let's do some math calculations: if we assume that we arrive at the bus stop with the same probability at a random time, then ideally we will wait for $ \bar E $ minutes for the next bus: $$ \bar E = \sum_ {i = 1} ^ n x_i / 2 \cdot x_i / L = \sum_{i = 1} ^ nx_i ^ 2 /(2L),\quad L = \sum_ {i = 1} ^n x_i. $$ This is because the probability that we arrive in the $x_i$ interval is given by $x_i/L$, and the expected minutes that we will wait for in the interval (assuming the bus is also arrived at random time with the same probability) is $$ \int_0 ^ {x_i} 1 / x_i dx = x_i / 2. $$
Now, assuming we just missed a bus, then we must be at one of the beginning of the $n$ time intervals with a probability of $1/n$; we will wait for $x_i$ minutes for the next bus, so we the expected waiting time for missing the bus is $$ E = \sum_ {i = 1} ^ n x_i \cdot 1 / n = \sum_ {i = 1} ^ nx_i / n = L / n. $$
Please note that in general, we do not have $$ \bar E \geq E, $$ Since by Cauchy’s inequality $$ n\sum_ {i = 1} ^ nx_i ^ 2 \geq(\sum_ {i = 1} ^ nx_i)^ 2 \implies 2n ^ 2E \bar E \geq n ^ 2E ^ 2 \implies \bar E \geq E / 2. $$ When each $x_i$ equals to each other, the equality holds in the above inequality. We can imagine that when each $x_i$ is "very different" from each other, we can expect that $ \bar E \geq E $, in that case we say we are lucky, although we have just missed one bus!.
We also know that the difference in $ x_i $ s can be defined by their variance as: $$ D ^ 2 = \sum_ {i = 1} ^ n(x_i / 2- \bar E)^ 2 \cdot x_i / L. $$ It is also equal to, by a well-known formula, $$ D ^ 2 = \sum_ {i = 1} ^ n(x_i / 2)^ 2 \cdot x_i / L - (\sum_ {i = 1} ^ nx_i / 2 \cdot x_i / L)^ 2 = \sum_ {i = 1} ^ nx_i ^ 3 /(4L) - (\sum_ {i = 1} ^ nx_i ^ 2)^ 2 /(4L ^ 2). $$
My QUESTION is that can we give some sufficient conditions expressed by $D$, $E$ and $L$, so that $$ \bar E \geq E? $$
To make the the formula for $E$ work, I have to suppose that there are actually $n + 1$ buses, that $x_1$ is the time in minutes between the first and second buses, and that "just missing" the last bus (the one at time $L$ after the first one) is impossible, since the waiting period in that case is undefined.
I will use the notation $\mathbb E(X)$ for the expected value of a random variable $X,$ since writing everything in $\sum$ notation all the time makes the problem more difficult to work.
Let $U$ be a random variable uniformly distributed on the first $n$ positive integers, and let $X$ be a random variable defined so that $X = x_i$ if $U = i.$ I define $X$ in terms of an auxiliary variable $U$ in order to avoid complications that can arise when some values occur more than once in the list $(x_i).$
In other words, $X$ is the waiting time, given that we have just missed a bus. Then $$\mathbb E(X) = \frac1n\sum x_i \qquad \text{and} \qquad \mathbb E\left(X^2\right) = \frac1n\sum x_i^2. $$ It follows that $E = \mathbb E(X),$ $L = n\mathbb E(X),$ and \begin{align} \bar E &= \frac{1}{2L} \sum_{i=1}^n x_i^2 \\ &= \frac{n}{2L} \mathbb E\left(X^2\right) \\ &= \frac{\mathbb E\left(X^2\right)}{2 \mathbb E(X)} \\ &= \frac{\mathbb E\left(X^2\right) - (\mathbb E(X))^2}{2 \mathbb E(X)} + \frac{\mathbb E(X)}{2} \\ &= \frac12\left(\frac{\mathop{Var}(X)}{\mathbb E(X)} + \mathbb E(X)\right) \\ &= \frac12\left(\frac{\mathop{Var}(X)}{E} + E\right). \end{align}
Then $\bar E \geq E$ if and only if $$ \frac12\left(\frac{\mathop{Var}(X)}{E} + E\right) \geq E,$$ that is, $\bar E \geq E$ if and only if $ \mathop{Var}(X) \geq E^2. $
I think that's a nice neat formula, although it does not use the exact parameters you requested.
With regard to a condition expressed by $D,$ $E,$ and $L,$ consider the sequence of waiting times $(10,10,10,30,100,100,120,260).$ In this case $L = 640,$ $E = 80,$ $D \approx 42.8159214,$ and $\bar E \approx 80.625 > E.$
But if we have instead of this the sequence of waiting times $(10,19.93724,20,50.06276,60,90,140,250),$ then $L = 640,$ $E = 80,$ $D \approx 42.8159214,$ and $\bar E \approx 75.94 < E.$
These two examples do not have the exact same value of $D,$ but that's only because I did not compute the exact amount to adjust the waiting times in order to make the values match exactly. I found that I was able to make the value of $D$ in the second example either slightly larger or slightly smaller than in the first example (where "slightly" is something on the order of $10^{-8}$) without making $\bar E$ much closer to $E,$ so I conclude there is a set of values that make $D,$ $E,$ and $L$ all match the first example exactly, yet $\bar E < E.$
Given two sequences of waiting times with the same values of $D,$ $E,$ and $L$ arising from each sequence, any condition relying only on $D,$ $E,$ and $L$ will give the same answer for both sequences. Yet we have two such sequences, one of which has $\bar E \geq E$ and one of which has $\bar E < E.$ Hence there is no formula of $D,$ $E,$ and $L$ that is a sufficient and necessary condition that $\bar E \geq E.$
If all you want is a sufficient (but not necessary) condition that $\bar E \geq E,$ observe that $$ \left(\sum_{i=1}^n \frac{x_i}{2} \cdot \frac{x_i}{L}\right)^2 \geq 0, $$ since the quantity on the left is squared, and therefore by your calculations, $$ D^2 \leq \sum_{i=1}^n \left(\frac{x_i}{2}\right)^2 \cdot \frac{x_i}{L}. $$ But since $\frac{x_i}{L} \leq 1$ for each $i,$ $$ D^2 \leq \sum_{i=1}^n \left(\frac{x_i}{2}\right)^2 = \frac n4 \mathbb E(X^2) = \frac{L}{4E} \mathbb E(X^2). $$ Therefore $$ \mathop{Var}(X) = \mathbb E(X^2) - (\mathbb E(X))^2 \geq \frac{4E}{L} D^2 - E^2, $$ so if we just satisfy the condition that $$ \frac{2 D^2}{L} \geq E, $$ then $$ \mathop{Var}(X) \geq \frac{4E}{L} D^2 - E^2 \geq E^2, $$ which implies that $\bar E \geq E.$
This condition is achievable; for example, if the waiting times are $x_i = 1$ for $1 \leq k \leq 21$ and $x_{22} = 99,$ then $L = 120,$ $E = 5.4545\ldots,$ $D \approx 18.6184,$ and $2D^2/L \approx 5.7774 > E.$ In this case, $\bar E = 40.925,$ which is certainly greater than $E.$