What are the chances of hitting VALID mnemonic phrase out of bip39, if someone picks 12, 18, or 24 words from the word list randomly by himself?

Question

So, here is the statement: "Not every combination of words from bip39 wordlist is valid."

What are the chances of hitting a VALID mnemonic phrase out of bip39, if someone picks 12, 18 and 24 words from the word list randomly by himself?

Answer 1

The checksum is 1 bit for every 3 words so 1/16 for 12 words, 1/64 for 18 words and 1/256 for 24 words

Answer 2

If you mean "valid" being the valid checksum word at the end of the phrase, A valid pneumonic is defined as:

The checksum is then the first n / 32 bits of the SHA256 hash of the entropy. This is just concatenated to the end of the entropy. The mnemonic is then encoded by dividing the entropy into groups of 11 bits and using the resulting 11 bit number as an index into a list of 2048 words

So I would think getting a valid checksum for any arbitrary input would be 1/2048

Answer 3

Any private key in existence, I am guessing a few million created by now, in 2²⁵⁶, are your chances of guessing someones mnemonic/private key. 2²⁵⁶ is a very, very large number.

Answer 4

from https://github.com/bitcoin/bips/blob/master/bip-0039.mediawiki:

The following table describes the relation between the initial entropy length (ENT), the checksum length (CS), and the length of the generated mnemonic sentence (MS) in words.

 CS = ENT / 32 MS = (ENT + CS) / 11
 
|  ENT  | CS | ENT+CS |  MS  |
+-------+----+--------+------+ 
|  128  |  4 |   132  |  12  |
|  160  |  5 |   165  |  15  | 
|  192  |  6 |   198  |  18  | 
|  224  |  7 |   231  |  21  | 
|  256  |  8 |   264  |  24  |

so for 12, 18, 24 words, the entropy of the checksum is 4, 6, 8

or 1 in 16, 64, 256