# Probability of guessing a password

## Homework Statement

Consider a simple password scheme using only two lowercase letters. A hacker is given 5 chances to guess the pw before being detected. Computer probability hacker is successful.

p = 1/(26*26)

## The Attempt at a Solution

I'm assuming the hacker isn't guessing randomly, but without replacement.

I feel that I may be multiplying the wrong probabilities, but what the heck:

P(attacker is successful) = $$\frac{1}{26^{2}} + \frac{1}{26^{2}*(26^{2}-1)} + \frac{1}{26^{2}*(26^{2}-1)*(26^{2}-2)} + \frac{1}{26^{2}*(26^{2}-1)*(26^{2}-2)*(26^{2}-3)} + \frac{1}{26^{2}*(26^{2}-1)*(26^{2}-2)*(26^{2}-3)*(26^{2}-4)}$$

This was an intuitive guess, since each guess is equally likely, I just subtract one from the sample space for each of the five guesses.

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Dick
Homework Helper
There are 26*26 possible passwords. Only one of them is correct. The hacker gets 5 chances. So, yes, the hacker will pick 5 different possibilities of that group of 26*26 passwords. You are way overthinking this problem.

Wait, what? I don't think 5/(26*26) would be the right answer, since that implies there is a 5 in 26^2 chance that they get the answer right on the first try...

My teacher has since told me to assume the attempts are independent (dumb hacker, I guess) so I'm modeling this as a geometric distribution with p = 1/(26*26). The total probability will be the probability hacker is successful in one try, two tries, three, four, or five tries all summed up. Does that sound right?

Dick