Four eights in SHA-256 plus four eights in (say) MD5 is just as likely as eight eights in SHA-256, at least if each hash function is truly a good approximation of a pseudo-random function.
(In fact, since the definition didn't specify where the eights are, and they can be anywhere in the output, I think four eights in each of two different hash functions is more likely, because there's more total output digits in which to look for them.)
Edit: If you don't specify the other hash function in advance, it's even more likely, because you can first find your SHA-256 amulet and then keep searching for a hash function in which it is also an amulet, where you might have a dozen or two potential choices that amulet-viewers might find interesting or impressive, each of which gives a new roughly independent space of possible amuletness.
4 comments
[ 3.9 ms ] story [ 22.9 ms ] thread(In fact, since the definition didn't specify where the eights are, and they can be anywhere in the output, I think four eights in each of two different hash functions is more likely, because there's more total output digits in which to look for them.)
Edit: If you don't specify the other hash function in advance, it's even more likely, because you can first find your SHA-256 amulet and then keep searching for a hash function in which it is also an amulet, where you might have a dozen or two potential choices that amulet-viewers might find interesting or impressive, each of which gives a new roughly independent space of possible amuletness.