An intuitive, discrete inter-arrival distribution where average time to next arrival is greater than average inter-arrival time.

Question

In the book, "Introduction-to-probability by Joseph K Blitzstein and Jessica Hwang", exercise 42 (b) reads as follows:

Fred then visits Blunderville, where the times between buses are also 10 minutes on average, and independent. Yet to his dismay, he finds that on average he has to wait more than 1 hour for the next bus when he arrives at the bus stop! How is it possible that the average Fred-to-bus time is greater than the average bus-to-bus time even though Fred arrives at some time between two bus arrivals? Explain this intuitively, and construct a specific discrete distribution for the times between buses showing that this is possible

I have some intuition around inter-arrival distributions with decreasing hazard rates creating conditions where the time until next arrival is more than the average inter-arrival time. I've even verified this with simulations (Python code below).

However, the exercise references some kind of "intuitive discrete inter-arrival distribution" which I haven't been able to find. Can someone suggest some simple discrete distribution that will lead to the same effect?

import numpy as np
from scipy.stats import weibull_min


def wait_time(c=.295):
    time = 0
    arrival = 11543.674
    mu = weibull_min.mean(c)
    while time < arrival:
        time += 10/mu*weibull_min.rvs(c)
    return time - arrival


def avg_arrival(c=.295, n=1000):
    print("Mean: " + str(weibull_min.stats(c, moments='m')))
    sum_t = 0
    for _ in range(n):
        sum_t += wait_time(c)
    return sum_t/n


if __name__ == "__main__":
    avg_arrival()