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Thread: possibilities.

  1. #1 possibilities. 
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    Suppose I wanted to hack someone's email, and all I knew about their password was that there were x characters in the password and there were y characters among the 101 on the keyboard that could be in it (for example, if the password consisted exclusively of alphanumeric characters, y = 36). If I had unlimited time to sit down and try to hack it using just those facts and trial and error, what equation would I use to figure the greatest number of attempts I will possibly have to make to finally get it right? Would it be x^y, y^x, x^(xy)? Help a brother out.


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  3. #2  
    Forum Professor river_rat's Avatar
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    Well you have y choices for the first entry and y choices for the second entry etc

    So the total number of passwords would be y^x

    eg. Suppose you have an 8 character password and a pool of 36 letters to choose from and could check one password every second it would take you 2821109907456 seconds or just short of 90 000 years

    That is why brute forcing is a bitch lol


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3  
    Forum Ph.D. william's Avatar
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    Kind of reminds me of the lottery... what, to win the jackpot our chances are something like 1 in 100<sup>5</sup>....

    I've never bought a lottery ticket by the way.
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  5. #4  
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    okay, so if the password were 8 alphanumeric characters, I would compute 8^36?
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  6. #5  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by davidstebbins
    okay, so if the password were 8 alphanumeric characters, I would compute 8^36?
    36^8
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  7. #6  
    Forum Ph.D. william's Avatar
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    ...of course... you can't neglect capital letters....
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  8. #7  
    Forum Professor river_rat's Avatar
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    No, you have 36 choices for the first entry and 36 choices for the second entry...

    This means there are 36^8 possible passwords - cue my example here.

    William - It depends on the lotto really, it would not surprise me if some powerball lottos had those odds but your average run of the mill lotto (6 numbers drawn out of a pool of 49) usually gives you 1 in 14 million odds
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by river_rat
    William - It depends on the lotto really, it would not surprise me if some powerball lottos had those odds but your average run of the mill lotto (6 numbers drawn out of a pool of 49) usually gives you 1 in 14 million odds
    Oh... I thought there were only 5 balls with the possible numbers between 1 and 100. And I made the calculation "easier" by assuming there were 5 of each number (independent).

    I guess had I ever bought a ticket I might have known better.

    Cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
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  10. #9  
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    Quote Originally Posted by william
    Kind of reminds me of the lottery... what, to win the jackpot our chances are something like 1 in 100<sup>5</sup>....

    I've never bought a lottery ticket by the way.
    Those are ridiculous odds; im glad in the UK the odds are only 1 in 14 million.

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  11. #10  
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    Let's throw a few monkey wrenches into this.

    1. Suppose there was a third trait involved. For example, if I were to roll two dice, then there would be 6^2 possibilities, or 36. With three dice, it would be 36^3, and so on. But imagine if the dice were different colors. If I had x dice, one red, one blue, and so on, and it mattered what each color's number was, how would I compute the possibilities of y dice with z sides on them?

    2. Imagine if each character could only be used up to y times in the sequence. For example, suppose I was playing Mastermind, with x pegs in the code and z different colors to chosse from, but each color can only be used y times, where x > y. What would the equation be then?

    3. Imagine if I didn't know how many characters/dice/pegs/whatever were in the code I was trying to crack, I only knew that it was greater than it was greater than w and less than x, with y being the number of different possible characters, and z being the maximum number of times each character can be used. What would I do then?
    "The only two constants in the universe are the speed of light and human stupidity." - Albert Einstein.
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