Is entropy being lost when the hash of a secret has a specific format? Bitcoin's mini private key format works by brute forcing random data to produce a secret so that when appending ? and then hashing with sha256 it will produce a hash that starts with 0x00. This way the user knows the secret is well formed and can proceed to hash it without appending ? to produce their private key.

This technique could be extended to include useful data in the hash "header", up to 2 bytes could be brute forced in a second or two.

However it seems that this scheme is reducing entropy of the secret i.e. if the attacker knows that a key is not valid by using a cheap sha256 hash, she will skip the expensive private to public key calculations. When the hash header is 1 byte around 99.6% of the random candidate keys were rejected, for 2 bytes it was 99.9983282%.

So does this scheme reduces the entropy of initial secret? If so how much?
From the linked page, a minikey is a 30-character string over the base58 alphabet with the first byte fixed to 'S', so effectively 29 characters. This gives a space of $log_2(58^{29}) \approx 169.88$ bits. Assuming that SHA is a random function, the probability of the hash starting with an 0-byte after appending a ? is 1/256, so this check loses 8 bits of entropy giving just over 161 bits of security.