Password Combination Probabilities: Solve 10-15 Char Puzzles

[Omnis]

How do I work out password combination probabilities?

For example, I'm looking for a password between 10-15 characters in length, the password can contain any alpha-numeric character (lower/upper alphabet character and 0-9 digits). How would I find out how many possible combinations there are?

 

[DaveC426913]

What would you make for an answer as your first try? What logic did you use?

Hint:
How many possibilities are there for the first number in the sequence?
How many possibilities are there for the second number in the sequence?
Hw many times do you have to do this?

What is a really compact way representing that as a generalized formula?

 

[Omnis]

How many possibilities are there for the first number in the sequence?

62

How many possibilities are there for the second number in the sequence?

62

How many times do you have to do this?

10-15 times depending on the password length

What is a really compact way representing that as a generalized formula?

I have no idea, that's why I came here, I'm wondering if its possible to recover a zip password on one of my files but I need to know how many possiblilities there are and I also like to know how to work it out for myself :)

 

[DaveC426913]

OK, well there's
62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62x62x62

or

62^10 + 62^11 + 62^12 + 62^13 + 62^14 + 62^15

or

a little more than 781 septillion

combinations.

At one try per second, it will take you a mere 24 quadrillion years, a mere 1.8 billion times longer than the universe has existed to-date.

Don't try to do them all in one sitting.

 

[CRGreathouse]

If you try ten billion passwords in a second with a given computer, and you use a million computers, it would take 1238 years on average.

 

[JasonRox]

OK, well there's
62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62x62
+
62x62x62x62x62x62x62x62x62x62x62x62x62x62x62

or

62^10 + 62^11 + 62^12 + 62^13 + 62^14 + 62^15

or

a little more than 781 septillion

combinations.

At one try per second, it will take you a mere 24 quadrillion years, a mere 1.8 billion times longer than the universe has existed to-date.

Don't try to do them all in one sitting.

After the 10th character the user has the option not choose at all, so that is 63 choices.

 

[ben328i]

isn't there a formula for that from allg or something.
what is the equation not worked out like in a, b c, etc. form.

 

[HallsofIvy]

The "formula" DaveC426913 was using earlier is often referred to as simply "The Fundamental Counting Principle": If you can do A in "n" ways and you can do B in "m" ways, then you can do A and B together in "mn" ways.

If you have 62 symbols you can use and there are no restrictions on how they can be used, then you have 62 ways you can write each of the, say, 10 symbols and so have
62¹⁰ possible 10 letter "passwords".

Similarly, there are 62¹¹ possible 11 letter "passwords", 62¹² 12 letter "passwords", 62¹³ l13 letter "passwords", 62¹⁴ 14 letter passwords,, and, finally, 62¹⁵ 15 letter "passwords".

Altogether, then, there are a total of 62¹⁰+ 62¹¹+62¹²+62¹³+62¹⁴+_62¹⁵ "passwords". That's the figure DaveC426913 gave.

JasonRox]After the 10th character the user has the option not choose at all, so that is 63 choices.

But if the user chose no character for the 11th choice, he must make the same choice for the 12th, 13th 14th, and 15th. I don't think that simplifies anything.

 

[ben328i]

ha good luck if this is really a problem not just a question.

 

[DaveC426913]

After the 10th character the user has the option not choose at all, so that is 63 choices.

What? No.

If the pw is 10 characters long, there are 62^10 possible combinations.
If the pw is 11 characters long, there are 62^11 possible combinations.
etc.

Add them together and you get all possible combinations within the parameters set.

So, your scenario where the user decides to not choose at all after the tenth character (or more accurataly: hits return), is simply the scenario where the pw is only 10 characters long.

 

[Omnis]

Thanks to everyone for the help

ha good luck if this is really a problem not just a question.

It is a problem, I'm trying to recover a password for a zip file, but the luck is much appreciated

 

[JasonRox]

What? No.

If the pw is 10 characters long, there are 62^10 possible combinations.
If the pw is 11 characters long, there are 62^11 possible combinations.
etc.

Add them together and you get all possible combinations within the parameters set.

So, your scenario where the user decides to not choose at all after the tenth character (or more accurataly: hits return), is simply the scenario where the pw is only 10 characters long.

You're right!

I never thought about that.

 

[DaveC426913]

Thanks to everyone for the help



It is a problem, I'm trying to recover a password for a zip file, but the luck is much appreciated

If it's your password, that should eliminate about 99.999999% of the possibilities, shrinking it to ones you would have come up with.

If it's not your password... well, that's raises an intriguing question...

 

[JasonRox]

Yeah, you can eliminate the characters you know for sure you didn't use!

 

[wysard]

Assuming order is important, ie:that "BAT" is not the same as "TAB"

then the general answer is n!/r! or:

62!/ (pwd length)! Where "!" means the factorial function.

Just like picking a lottery number.

PS. It's easier to break the password encryption scheme than guess the actual password.

 

[Dantie Neit]

Figured I'd use this one. I'm trying to crack my password for something I got years ago, but I don't know how worth it it is. I need to know the number of possibilities for a 4 character password which has 6 characters to use. Basically 1111 to 6666. I'm good at maths, but that's only at a low level still. If someone can just work out what it is, that would be great.

Cheers!

 

[slider142]

Use the Multiplication Principle of Counting: You have 6 choices for the first character, 6 choices for the second character, 6 for the third, and 6 for the fourth, so altogether you have 6*6*6*6 choices for a string of 4 characters.
The principle comes from the following thought experiment. If you have a string of two characters with m choices for the first character and n choices for the second character, then first counting all possible strings starting with characters in the first space. For each character in the first space, there will be n possibilities for the second space. Ie., if the first character can be any letter in the alphabet, you would have sets like {A_, A_, ...}, {B_, B_, ...}, ... ,{Z_, Z_, ...} where the underscores are replaced by characters from the second set (of size n). Thus, you should have m sets of n strings, or m*n total strings.

 

[HallsofIvy]

Assuming order is important, ie:that "BAT" is not the same as "TAB"

then the general answer is n!/r! or:

62!/ (pwd length)! Where "!" means the factorial function.

Just like picking a lottery number.

PS. It's easier to break the password encryption scheme than guess the actual password.

No. That would be true if the same letter or numeral could not be used twice in the same password and that is not true.

 

[soandos]

If there are any files in the zip archive that you have unencrypted, or it is encrypted using any early version (as you seem to suggest), then there are flaws in the scheme that have been exploited, and can recover the password near instantaneously. if there is an unencrypted files that is also present in the archive, then there is something that is called a plain text attack that can be used. basically, is uses patterns between the encrypted and unencrypted copies to get the password. gets an alphanumeric 10 character password in about an hour on an average desktop.