Cops and Robbers: Redacted Primality (Cops' Thread) This challenge was originally sandboxed by Magic Octopus Urn; I adopted and posted it with his permission.

This is the cops' thread. The robbers' thread is here.

The challenge


Step One: Write a piece of code (function or full program) which checks for primality.
Step Two: Remove pieces of your code by replacing characters with the symbol █.
Step Three: Post the redacted code on the cops thread.
Step Four: Wait for your code to be cracked and try to crack other's code.


For example, the Groovy code {it.isPrime()} could become {██.is█████()}.
 (This one would be stupidly easy to crack; also, I know, .isPrime() is not a Groovy method.)



Scoring

You must include your program's score in its submission. The score is defined as the ratio of redacted characters to characters. So if your program had 20 characters and 5 were redacted, your score would be 0.25. The Groovy code above would have a score of 0.5.



Rules


Your program only needs to handle positive integers. It should output a truthy value if the number is prime and a falsy value otherwise. Please specify in your answer what it outputs.
Your code may not contain any comments or unnecessary whitespace.
No hashing or cryptographic obfuscation.
Your code may be no more than 50% redacted (at least 1/2 the characters must be shown). This means that the highest possible score is 0.5.
If your answer is not cracked within a week you may mark it safe and edit in the intended crack.




Winning

The winner will be the lowest-scoring uncracked answer within two weeks of posting. In the case of a tie, whichever has the most votes will win. This thread is always open to more submissions, but the winner chosen after two weeks will be permanent.
8086 DOS COM, 87 bytes, score 19/87 ~= 0.2183

Cracked by NieDzejkob

1└╣██1█╛ü ¼&lt;█t&lt;&lt; u≈¼&lt;█t█,0|/&lt;██+ô≈ßô☺├δδâ√█|█╞█S█Y╣██9┘t█ë╪1╥≈±○╥t█Aδ∩╞█S█N┤█║S█═!├A
$


This is a COM program; expects number as a command line argument, outputs Y or N. Limit: 65535 because 16 bit processor (sizeof(int) would be 2). Newline is 0x0D 0x0A on this platform. Yes you count 20 █ instead of 19 █. One of them's a real █ and has not been substituted. Muhahaha.

The space in position 10 is actually a NUL byte. The symbol for NUL is the same as space in the old VGA font.