which is less likely: obtaining no heads when you flip a fair coin n times,
or obtaining fewer than n heads when you flip the coin 4n times?

The probability of getting no heads is 1 case out of all cases for n
flips, so:

P[no heads] = 1/(2^n)

Getting fewer than n heads in 4n times has [(1/4)n-1] (minus 1 for the
less than comment, take the minus one out if it is equal to or less
than) cases out of all cases of 4n flips, so:

P[LT n heads] = [(1/4)n-1]/[2^(4n)]

You could compare the two probabilities by diviing one over the other,
and seeing if the numerator is larger (in that case the resulting
number would be < 1), or just graph them.  I don't have time to do
that right now, but maybe in a bit...  There may be a range where P[no
heads] is greater, or vice versa.


Ryan