["python","hash","set","cpython","python-internals"]
Why does tuple(set([1,"a","b","c","z","f"])) == tuple(set(["a","b","c","z","f",1])) 85% of the time with hash randomization enabled?

Given Zero Piraeus' answer to another question, we have that x = tuple(set([1, "a", "b", "c", "z", "f"])) y = tuple(set(["a", "b", "c", "z", "f", 1])) print(x == y) Prints True about 85% of the time with hash randomization enabled. Why 85%?
I'm going to assume any readers of this question to have read both: Zero Piraeus' answer and My explanation of CPython's sets. The first thing to note is that hash randomization is decided on interpreter start-up. The hash of each letter will be the same for both sets, so the only thing that can matter is if there is a collision (where order will be affected). By the deductions of that second link we know the backing array for these sets starts at length 8: _ _ _ _ _ _ _ _ In the first case, we insert 1: _ 1 _ _ _ _ _ _ and then insert the rest: α 1 ? ? ? ? ? ? Then it is rehashed to size 32: 1 can't collide with α as α is an even hash ↓ so 1 is inserted at slot 1 first ? 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? In the second case, we insert the rest: ? β ? ? ? ? ? ? And then try to insert 1: Try to insert 1 here, but will ↓ be rehashed if β exists ? β ? ? ? ? ? ? And then it will be rehashed: Try to insert 1 here, but will be rehashed if β exists and has ↓ not rehashed somewhere else ? β ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? So whether the iteration orders are different depends solely on whether β exists. The chance of a β is the chance that any of the 5 letters will hash to 1 modulo 8 and hash to 1 modulo 32. Since anything that hashes to 1 modulo 32 also hashes to 1 modulo 8, we want to find the chance that of the 32 slots, one of the five is in slot 1: 5 (number of letters) / 32 (number of slots) 5/32 is 0.15625, so there is a 15.625% chance¹ of the orders being different between the two set constructions. Not very strangely at all, this is exactly what Zero Piraeus measured. ¹Technically even this isn't obvious. We can pretend every one of the 5 hashes uniquely because of rehashing, but because of linear probing it's actually more likely for "bunched" structures to occur... but because we're only looking at whether a single slot is occupied, this doesn't actually affect us.