Say someone was giving you clues to their private key. For each character they gave you a couple of options eg, for the fifth position is either 1, 3, f or a. How many options would they have to give you per position for it to be crackable by brute force?
Assuming known legacy address, wif private key contains 85 characters, first is always "5", second one of "H", "J", "K", leaving us with 83 characters. Let's assume the second character is know. When we have 4 possibilities for each character, there are 2
166 possible combinations. Direct brute forcing has been solved for private keys with 2
63 possibilities up to now.
We could accelerate the calculations a bit, by dropping the checksum characters (last 5), and doing checksum on the rest. This drops the complexity to 2
156.
If the public key of the address is known (i.e. spent from, or signed with), then it's "easier" to do Pollard Rho on it - complexity is only 2
128.
With 2 possibilities per character, the complexity is 2
78, so few billions of dollars, many years, and giga-watt-hours later it could be cracked.