A/B Testing Password Strength Annual Formula
Learn how the calculator estimates password search space, annual brute-force likelihood, and average cracking time for two password designs.
This calculator compares two password designs under the same assumed attacker guessing rate. It estimates theoretical brute-force resistance using password length and character set size, which helps show how much additional search space one design provides over another.
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Annual brute-force compromise likelihood
Where:
First, estimate how many guesses an attacker can make in one 365-day year. Then divide that annual guess budget by the number of possible passwords. The result is the approximate percentage chance of finding a uniformly random password by exhaustive brute force within one year.
Variables Explained
| Variable | What It Means | Unit |
|---|---|---|
| guessesPerSecond - Attacker guesses per second | The effective number of distinct password guesses an attacker can make each second after practical protections are considered. | guesses per second |
| characterSetSize - Character set size | The number of characters that could be selected for each password position. | number |
| passwordLength - Password length | The number of characters in the password design. | number |
| 31536000 - Seconds per year | The calculator uses 365 days, or 31,536,000 seconds, for one year. | seconds |
| annualGuessBudget - Annual guess budget | The estimated number of guesses made continuously during one year. | guesses |
| searchSpace - Password search space | The theoretical number of possible passwords for a design. | number |
Step-by-Step Calculation
Calculate the annual guess budget
Multiply the effective guessing rate by the number of seconds in 365 days.
annualGuessBudget = guessesPerSecond * 31536000
Calculate Password A search space
Raise Password A's character set size to its length.
passwordASearchSpace = pow(passwordACharset, passwordALength)
Calculate Password B search space
Raise Password B's character set size to its length.
passwordBSearchSpace = pow(passwordBCharset, passwordBLength)
Estimate annual likelihood for each design
Divide the annual guess budget by the relevant search space and convert the result to a percentage.
annualProbability = min(100, (annualGuessBudget / searchSpace) * 100)
Compare search spaces
A result above 1 means Password B has a larger theoretical search space.
passwordBSearchSpaceRatio = passwordBSearchSpace / passwordASearchSpace
Estimate average exhaustive-search time
The average-case estimate assumes the correct password is found halfway through the search space.
expectedCrackYears = searchSpace / (2 * annualGuessBudget)
12-character versus 16-character random password
Annual guess budget
1,000 × 31,536,000
31,536,000,000 guesses
Password A search space
94^12
approximately 4.76 × 10^23 combinations
Password B search space
94^16
approximately 3.72 × 10^31 combinations
Password B relative search space
94^16 ÷ 94^12 = 94^4
78,074,896× larger
Password A annual likelihood
(31,536,000,000 ÷ 4.76 × 10^23) × 100
approximately 6.63 × 10^-12%
Password B annual likelihood
(31,536,000,000 ÷ 3.72 × 10^31) × 100
approximately 8.49 × 10^-20%
Final Result
With equally random characters and the same 94-character set, the 16-character Password B design has about 78,074,896 times the search space of the 12-character Password A design.
Assumptions
- ✓Each password is randomly generated, with every allowed character equally likely at every position.
- ✓The attacker makes distinct guesses continuously for 365 days.
- ✓The entered guess rate already reflects relevant hashing, rate limiting, lockouts, and other defenses.
- ✓The model considers exhaustive brute-force guessing rather than password reuse, phishing, malware, or leaked credentials.
- ✓Average crack time assumes the correct password is found halfway through the theoretical search space.
Limitations
- !Human-created passwords may be predictable even when their length and stated character set appear strong.
- !Actual guessing rates vary greatly between online and offline attacks and can change after a breach or security-control change.
- !A brute-force estimate does not measure risks from phishing, credential stuffing, keylogging, or account recovery weaknesses.
- !Multi-factor authentication can affect account compromise risk but is not directly represented by the formula.
- !The annual probability is a simplified ratio-based estimate and is capped at 100%.
Common Mistakes to Avoid
Using the full character set size when the password does not actually use all characters randomly.
Treating a memorable phrase, pattern, or substituted word as if it were uniformly random.
Entering a raw hardware guess rate without accounting for password hashing, login throttling, or account lockouts.
Comparing passwords with different effective attack environments using the same guessing rate without justification.
Entering an actual password instead of only its length and assumed character set size.
Related Formulas
Frequently Asked Questions
How is password search space calculated?
Search space is calculated as character set size raised to password length. A password with 94 possible characters at each of 12 positions has 94^12 theoretical combinations.
Why does adding a few characters make such a large difference?
Each additional random character multiplies the search space by the character set size. With a 94-character set, four extra random characters multiply the search space by 94^4, or 78,074,896.
What does annual compromise likelihood mean in this calculator?
It is the modeled chance that exhaustive brute-force guessing reaches a uniformly random password within one 365-day year at the entered effective guess rate.
Why is expected crack time divided by two?
An exhaustive attacker does not usually need to test every combination. If the password is uniformly random, the correct combination is expected to appear halfway through the search space on average.
Can this formula measure the strength of a password made from words?
Not reliably. Word-based and patterned passwords can be targeted with specialized guesses, so their practical resistance may be much lower than a simple character-count model suggests.
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