2^n is half of 2^(n+1), so you'd expect that getting 50% coverage of (n+1)-bit sequences would mean that you have 100% coverage of n-bit sequences. In other words, the answer to your question is at most 1 bit lower than the answer given.
But it is slightly more complicated, because this is an example of the Coupon Collector's Problem [1]. It takes about N ln N random samples to cover a set of size N. If there have been 2^74 samples chosen, then we haven't quite gotten enough to cover N = 2^68.
But it is slightly more complicated, because this is an example of the Coupon Collector's Problem [1]. It takes about N ln N random samples to cover a set of size N. If there have been 2^74 samples chosen, then we haven't quite gotten enough to cover N = 2^68.
[1] https://en.wikipedia.org/wiki/Coupon_collector%27s_problem