So 10 8's is mythic. That's only 2^40 (probably less since the 8's dont have to be in a specific position in the hash). Based on https://gist.github.com/epixoip/a83d38f412b4737e99bbef804a27... you should be able to bruteforce a mythic amulet in 48 seconds with a high end gpu from 2016.
I wrote a CUDA kernel to look for some. My 3080 managed 3 billion/second including population counting the 8s. After 48 seconds it had spat out three 10-eights and one 11-eights amulet:
Here are some excellent hex digits: 251d5b059cefc6f3
Can you please share how you got the 2^40 number? I've been trying to think of how to figure out the odds of these.
Odds of 4 hex 8's in a row given a 4 digit string is (1/16)^4.
Odds of 4 hex 8's in a row given a 5 digit string is
number of ways to arrange 8s in the first 4 digits (1) times 16 possible 5th digits plus number of ways to arrange 8s in the last 4 digits (1) times 16 possible 1st digits, all divided by the number of possible arrangements (16^5)
I did 10 8's (mythic!) at the beggining of the hash (2^4)^10 = 2^40.
I didnt account they could start at any position, so the actual number is probably
(64-10)/2^40 ≈ 1 / 2^35
We should also subtract strings longer than 10 that were double counted. However i think the probability of such things is negligible relative to 2^35.
If instead we were doing 4 8's i think it would be:
(64-4)/((2^4)^4) = 60/2^16 ≈ 2^10
I've always been bad at calc probabilities so i may have messed this up.
> Odds of 4 hex 8's in a row given a 5 digit string is number of ways to arrange 8s in the first 4 digits (1) times 16 possible 5th digits plus number of ways to arrange 8s in the last 4 digits (1) times 16 possible 1st digits, all divided by the number of possible arrangements (16^5)
You're double counting "88888". But then again so am i.
Doesn't sound that mythic to me.