11-year-old selling cryptographically-secure passwords online

How is this better than me randomly selecting 6 words out of a dictionary?
 
How is this better than me randomly selecting 6 words out of a dictionary?

A dictionary brute force will crack your password in seconds.

With that said, decrypting her method will be a bit more difficult but still crackable.
 
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A dictionary brute force will crack your password in seconds.

With that said, decrypting her method will be a bit more difficult but still crackable.
Fine so throw in a random made up word too.

My point is I don't see any real value being added here to be honest - as harsh as that sounds given that its a 11 year old's entrepreneurial venture.
 
A dictionary brute force will crack your password in seconds.

With that said, decrypting her method will be a bit more difficult but still crackable.

You do know her words are from the dictionary as well ? :)

I build long, strong, memorable passwords using strings of words from the dictionary that I select using dice

So her's will be easily cracked.
 
A dictionary brute force will crack your password in seconds.

With that said, decrypting her method will be a bit more difficult but still crackable.

OK, how is 6 random words she chose more secure than 6 random words I would choose?
 
A dictionary brute force will crack your password in seconds.

With that said, decrypting her method will be a bit more difficult but still crackable.

Not really no. A typical oxford dictionary contains 171,476 words. 717476 ^ 6 = 25 422 605 913 456 636 373 066 491 006 976 possible combinations of 6 words. The average english word length is 5.1 characters.

Assuming your wordlist is encoded in ASCII (7/8 bits) and you don't have any new line characters or spaces your wordlist will be 5.1*6*25422605913456636373066491006976*7/8/1024/1024/1024/1024 = 61 908 419 714 454 910 000 0 TB in size.

Breakdown:
5.1 chars per word
6 words
25422605913456636373066491006976 possible combinations of 6 words
7 bits to store every char
/8 bits -> bytes
/1024 bytes -> KB
/1024 KB -> MB
/1024 MB -> GB
/1024 GB -> TB
 
Not really no. A typical oxford dictionary contains 171,476 words. 717476 ^ 6 = 25 422 605 913 456 636 373 066 491 006 976 possible combinations of 6 words. The average english word length is 5.1 characters.

Assuming your wordlist is encoded in ASCII (7/8 bits) and you don't have any new line characters or spaces your wordlist will be 5.1*6*25422605913456636373066491006976*7/8/1024/1024/1024/1024 = 61 908 419 714 454 910 000 0 TB in size.

Breakdown:
5.1 chars per word
6 words
25422605913456636373066491006976 possible combinations of 6 words
7 bits to store every char
/8 bits -> bytes
/1024 bytes -> KB
/1024 KB -> MB
/1024 MB -> GB
/1024 GB -> TB

Um. I have a word list that has 982,963,904 words and is only 13gb.
 
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