Now that we know the basics of the Birthday Problem, we can use this knowledge to understand the security of password hashing.

In the early days, passwords were stored in the server “as-is”. This means that if your username was juan and your password is Password123! then that information is stored in the server like this:

juan,Password123!

Since the theft of a password file is harder to prevent, the passwords are not anymore stored “as-is” (also known as clear-text). The server will apply an algorithm to the original password which outputs a text called a hash. The algorithm is called a hash function. The hash is what’s put in the password file. A thief in possession of the password file will not be able to know the original password just by looking at it.

For example, the information above will now look like this:

juan,2c103f2c4ed1e59c0b4e2e01821770fa

where “2c103f2c4ed1e59c0b4e2e01821770fa” is the has value of the password “Password123!“.

The hash function I’m using is called the MD5 hash function. Given a password, it will produce a hash value. The set of all hash values is not infinite. In fact, the number of possible hash values is $2^{128}$ for md5. Due to this restriction, the birthday paradox will apply.

The birthday paradox tells us that given a hash function $f(x)$, the probability that at least two passwords hash to the same value is given by:

$\displaystyle 1-\frac{N\times N-1\times N-2\times \ldots \times N-k+1}{N^k}$

Since md5 hash function has $N=2^{128}$ possible values, the probability that two passwords hash to the same value is

$\displaystyle 1-\frac{2^{128}\times 2^{128}-1\times 2^{128}-2\times \ldots \times 2^{128}-k+1}{(2^{128})^k}$

We want to compute for k so that this probability is at least 50%.

$\displaystyle 1-\frac{2^{128}\times 2^{128}-1\times 2^{128}-2\times \ldots \times 2^{128}-k+1}{(2^{128})^k} \ge 0.5$

which is equivalent to

$\displaystyle \frac{2^{128}\times 2^{128}-1\times 2^{128}-2\times \ldots \times 2^{128}-k+1}{(2^{128})^k} < 0.5$

Computing for $k$ when N is large is hard so we need to approximate this. To that end, we need some tools to help us.

We can write the probability in the following way:

$\displaystyle 1-\frac{N}{N}\times\frac{N-1}{N}\times\frac{N-2}{N}\times\frac{N-3}{N}\times\ldots\times\frac{N-k+1}{N}$
$= \displaystyle 1-\frac{N}{N}\times (1-\frac{1}{N})\times (1-\frac{2}{N})\times (1-\frac{3}{N}) \times\ldots\times (1-\frac{k-1}{N})$

Since N is large, the quantities

$\displaystyle \frac{1}{N}, \frac{2}{N}, \frac{3}{N}, \frac{k-1}{N}$

are very small. Because of this, we can use the approximation

$e^{-x} \approx 1-x$

The above approximation comes from the Taylor expansion of $e^{-x}$:

$\displaystyle e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} \ldots$

If $x$ is small, the higher order terms like $x^2, x^3, x^4, \ldots$ vanish. Using this approximation, we can write the probability as:

$\displaystyle \frac{N}{N}\times (1-\frac{1}{N})\times (1-\frac{2}{N})\times (1-\frac{3}{N}) \times\ldots\times (1-\frac{k-1}{N})$

$\displaystyle = e^{-\frac{1}{N}}\cdot e^{-\frac{2}{N}}\cdot e^{-\frac{3}{N}}\cdot \ldots\cdot e^{-\frac{k-1}{N}}$

$\displaystyle = e^{-\frac{1+2+3+4+\ldots + k-1}{N}}$

Since

$\displaystyle \sum_1^n j = 1+2+3+4+ \ldots + n = \frac{n(n+1)}{2}$

we have

$e^{-\frac{1+2+3+4+\ldots + k-1}{N}} = e^{-k\cdot (k-1)/2N }$

Computing for k

Let’s compute k so that

$\displaystyle e^{-k\cdot (k-1)/2N} < 0.5$

Taking the natural logarithms of both sides

$\displaystyle \ln e^{-k\cdot (k-1)/2N} < \ln 0.5$

$\displaystyle \frac{-k\cdot (k-1)}{2N} < \ln 0.5$

$\displaystyle k^2 - k + 2N\ln 0.5 > 0$

Using the quadratic equation, we can solve for k:

$\displaystyle k > \frac{-(-1) \pm \sqrt{(-1)^2 -4(1)(2N\ln 0.5}}{2}$
$\displaystyle k > \frac{1 \pm \sqrt{1-8N\ln 0.5}}{2}$

When $N=2^{128}$, we have

$\displaystyle k > \frac{1 \pm 4.343876e+19}{2} \approx 10^{19}$

This is about 10 quintillion. What this means is that when $k > 10^{19}$, there is already a 50% chance that 2 passwords hash to the same value. In fact, the md5 was already cracked in 2004.

There are only 365 days in a year (excluding leap year). Given that there are about 7.4 billion people on earth, this means that there are approximately 20 million people with the same birthday on any given day. You just divide 7,400,000,000 by 365 and you get 20 million. Happy Birthday to all 20 million people celebrating their birthday today!

Suppose you’re in a crowd, on a bus, in a restaurant, or stadium. There is a big chance you might be standing next to a person with the same birthday as you.

In fact, you only need about 23 people to have a 50/50 chance of two people having the same birthday! This may sound unbelievable since there are 365 days in a year but you only need 23 people to have a 50% chance of 2 people with the same birthday. How come?

This is called the Birthday Paradox and is very important in digital security, especially the password security.

Basic Counting

Probability is all about counting the possibilities. Let’s make it simple by using a dice as an example. We all know what a dice looks like.

When a balanced dice is thrown, it can land showing any one of its six sides. We refer to the result of throwing a dice as an outcome and we say that a dice has 6 possible outcomes. If a dice is balanced, every side is equally likely to show up. We define the probability of a face showing up as the number of times that face occurs in the possible outcomes divided by the total number of possible outcomes. For example, out of the 6 possible outcomes, the number “1” occurs only once. Since there are 6 possible outcomes, the probability of getting a 1 is, therefore:

$\displaystyle \text{Probability of getting a "1"} = 1/6$

Let’s add a second dice. To identify our two dice, let’s call one of them Dice A and the other Dice B. Let’s throw the dice together. When they land, dice A and dice B will show numbers. For this scenario, an outcome is now defined as the numbers that Dice A and Dice B show when they land. A possible outcome is Dice A shows a 1 and Dice B shows a 2. We can give this outcome a name and call it 1,2. We should remind ourselves that the first number is the result of Dice A and the second number is the result of Dice B. We can also refer to each outcome as a combination.

Here are the possible outcomes that the two dice will show:

If you count the number of combinations above, you’ll get 36. The reason it’s 36 is because dice A has 6 different outcomes and dice B has 6 different outcomes. Multiplying them together gives $6 \times 6=6^2 = 36$.

If you add a third dice, say dice C, the total number of combinations becomes:

$\displaystyle 6^3 = 216$.

In general, for N dice, the total number of combinations is

$\displaystyle 6^N$

How many combinations have at least 2 same numbers?

Since there are only 2 numbers for each combination, this question is also the same as “How many combinations show the same numbers?”. If you look at the diagonal, these are the combinations that have the same number for Dice A and Dice B.

If you count them, you’ll get 6. Therefore, the probability of getting at least two equal numbers (in our 2-Dice system) is

6/36

How many combinations show different numbers?

If you count all combinations outside the diagonal, you’ll get 30. Therefore, the probability of getting two different numbers is

30/36

Notice that the probability of getting at least 2 same numbers PLUS the probability of getting different numbers is equal to 1:

6/36 + 30/36 = 36/36 = 1

Knowing One gives you the other

If we know the probability of getting different numbers (30/36), then we can compute the probability of getting at least 2 same numbers simply by subtracting it from 1:

$\displaystyle \text{probability of getting at least 2 numbers same} = 1-30/36 = 1/6 = 0.167$

Avoid counting manually

When we counted the number of combinations which show different numbers, we counted it with our fingers. There is another way to count which is by doing it mentally. Since we are counting the number of ways that the 2-Dice system will show different numbers, we start by getting Dice A and asking how many different ways Dice A can land so that the number it shows is not equal to the number shown by Dice B. Since we have not yet thrown Dice B, then Dice A is allowed to show any number when it lands. This means there are 6 possible ways for Dice A to do this.

Number of ways Dice A can land = 6

Whatever number results in throwing Dice A, we cannot allow Dice B to have that number. This means that Dice B can only choose from 5 other numbers different from the one chosen by Dice A.

Number of ways Dice B can land = 5

If we multiply them, we get the number of combinations that Dice A and Dice B can land with different numbers:

6*5 = 30

This agrees with our manual counting.

At this point, pause and take note that the probability of getting at least 2 numbers the same for a 2-Dice system is 0.167. If we add more dice, this probability will increase. The question then is

How many dice do we need to throw so that the probability of getting 2 dice showing the same number is at least 50%?

Our 2-Dice example above shows that the probability of at least 2 dices showing the same number is 0.167, which is less than 50%. Let’s add a third dice and compute the probability.

How to compute the probability?

Let’s follow the pattern for the 2-Dice system. Since there are now 3 dice, the number of ways to get all numbers different is:

6*5*4

The total number of combinations of a 3-Dice system is

$\displaystyle 6^3$

Therefore, the probability of getting at least 2 dice with the same number is

$\displaystyle 1- \frac{6\times 5\times 4}{6^3} = 0.444$

This is still less than 50%.

Let’s now add a 4th dice and compute the probability using the same pattern:

$\displaystyle 1- \frac{6\times 5\times 4\times 3}{6^4} = 0.722$

This is greater than 50%! So the answer is we need 4 dice thrown so that the probability of getting at least 2 dice with the same number is at least 50%.

The general formula for the probability for a k-Dice system is:

$\displaystyle 1- \frac{ 6\times 5\times \ldots \times (6-k+1)}{6^k}$

How does this relate to the Birthday Problem?

Now that we have the foundations, it’s easy to translate Dice to people and numbers to birthdays. In our dice example, there are 6 different numbers (faces) per dice. Translating this to birthdays, each person can have 365 possible birthdays since there are 365 days in a year (not including leap year).

This is the analogy:

Dice -> 6 possible faces
Person -> 365 possible birthdays

We want to compute how many random persons we need so that the probability of at least two persons having the same birthday is at least 50%. Let k be the number of random persons. Following the same pattern as the Dice example, the formula to compute the probability, given k persons, is:

$\displaystyle \text{Probability of at least 2 persons with the same birthday} = 1-\frac{365 \times 364 \times 363 \times \ldots (365-k+1)}{365^k}$

If we compute starting from k=1 to k=30, we can construct the following table:

   probability
1  0.000000000
2  0.002739726
3  0.008204166
4  0.016355912
5  0.027135574
6  0.040462484
7  0.056235703
8  0.074335292
9  0.094623834
10 0.116948178
11 0.141141378
12 0.167024789
13 0.194410275
14 0.223102512
15 0.252901320
16 0.283604005
17 0.315007665
18 0.346911418
19 0.379118526
20 0.411438384
21 0.443688335
22 0.475695308
23 0.507297234
24 0.538344258
25 0.568699704
26 0.598240820
27 0.626859282
28 0.654461472
29 0.680968537
30 0.706316243


Below is the graph of the same data where we indicate at what number of persons the graph is greater than or equal to 50%. When the number of persons becomes 23, there is already a 50% chance that at least 2 of them have the same birthday!

## When Average Is Not Enough: Thoughts on Designing for Capacity

Designing a system from scratch to handle a workload you don’t know is a challenge. If you put to much hardware, you might be wasting money. You put little, then your users will complain of how slow the system is.

If you’re given only a rate, like 6000 hits/hour, you don’t know how these are distributed in a minute by minute or per second interval. We can make a guess and say that there are about 100 hits per minute or 1.67 hits/sec. If hits come uniformly at that rate, then we can design a system that can handle 2 hits/sec and all users will be happy since all requests will be served quickly and no queueing of requests. But we know it’s not going to happen. There will be some interval where the number of hits is less than 3 and some more than 3.

Theoretically, requests to our server come randomly. Let’s imagine 60 bins represented by seconds in one minute. We also imagine that requests are like balls we throw into the bins. Each bin is equally likely to be landed by a ball. It’s possible that all balls land on only one bin!

After throwing the balls into bins, let’s see what we have.

As you can see, some bins have more than 2 balls (which is the average number of balls in a bin). Therefore if we design our system based on the average, 50% of our users will have a great experience while the other 50% will have a bad experience. Therefore we need to find how many requests per second our server needs to handle so that our users will have a good experience (without overspending).

To determine how many requests per second we need to support, we need to get the probability of getting 4, 5, 6 or more request per second. We will compute the probability starting from 3 requests per second and increment by one until we can get a low enough probability. If we design the system for a rate that has a low probability, we are going to spend money for something that rarely occurs.

Computing the Probability Distribution

We can view the distribution of balls into bins in another way. Imagine labeling each ball with a number from 1 to 60. Each number has an equal chance to be picked. The meaning of this labeling is this: the number that was assigned to the ball is the bin (time bucket) it belongs to. After labeling all balls, what you have is a distribution of balls into bins.

Since each ball can be labeled in 60 different ways and there are 100 balls, the number of ways we can label 100 different balls is therefore

$\displaystyle 60^{100}$

Pick a number from 1-60. Say number 1. Assume 2 balls out of 100 are labeled with number 1. In how many ways can you do this ? Choose the first ball to label. There are 100 ways to choose the ball. Choose the second ball. Now there are 99 ways to choose the second ball. We therefore have 990 ways to select 2 balls and label them 1. Since we don’t really care in what order we picked the ball, we divide 990 with the number of possible arrangements of ball 1 and ball 2, which is 2! (where the exclamation mark stands for “factorial”). So far, the number of ways to label 2 balls with the same number is

$\displaystyle \frac{100 \times 99}{2!}$

Since these are the only balls with label 1, the third ball can be labeled anything except number 1. In that case, there are 59 ways to label ball 3. In the same way, there are 59 ways to label ball 4. Continuing this reasoning until ball 100, the total ways we can label 2 balls with number 1 and the rest with anything else is therefore:

$\displaystyle \frac{100 \times 99}{2!} \times 59^{98}$

Notice that the exponent of 59 is 98 since there are 98 balls starting from ball 3 to ball 100.

Therefore, the probability of having two balls in the same bin is

$\displaystyle \frac{100 \times 99}{2!} \times \frac{59^{98}}{60^{100}} = 0.2648$

We can also write this as

$\displaystyle \frac{100!}{2! \times 98!} \times \frac{(60-1)^{98}}{60^{100}} = \binom{100}{2} \frac{(60-1)^{98}}{60^{100}}$

In general, if m is the number of balls, n the number of bins and k the number of balls with the same label, then the probability of having k balls within the same bin is given by

$\displaystyle \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}}$

,

where

$\displaystyle \binom{m}{k} = \frac{m!}{k!(m-k)!}$

is the binomial coefficient.

It turns out that this is a probability distribution since the sum of all probabilities from k=0 to k=m is equal to 1. that is

$\displaystyle \sum_{k=0}^{n} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} = 1$

To see this, recall from the Binomial Theorem that

$\displaystyle \big( x + y \big)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$

If we let x=n-1 and y=1, we can write the above equation as

$\displaystyle \begin{array}{ll} \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} &= \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}\cdot 1^k}{n^{m}}\\ &= \displaystyle\frac{(n-1+1)^m}{n^{m}}\\ &= \displaystyle\frac{n^m}{n^m}\\ &= \displaystyle 1 \end{array}$

Here is a graph of this probability distribution.

Here’s the plot data:

 probability 1 0.315663315854 2 0.264836171776 3 0.146632456689 4 0.060268424995 5 0.019612775592 6 0.005263315484 7 0.001197945897 8 0.000236035950 9 0.000040895118 10 0.000006307552 

We can see that for k=9, the probability of it occurring is .004%. Anything beyond that we can call rare and no need to spend money with.

Just For Fun

What’s the probability that a given bin is empty, that is, there are no balls in it?

Other Probability Distributions

Our computation above was based on a uniform probability distribution. However, there are other distributions that are more suitable for arrival of requests. One of the most widely used is called the Poisson Distribution where you can read from here.

R Code

The R code to generate the simulation:

par(mfrow=c(4,4))
f=function(){
x=c()
for(i in 1:100){
x=c(x,sample(seq(1,60),1,replace=T))
}
plot(tabulate(x),type="h", ylab="tx", xlab="secs")
}

for(i in 1:16){
f()
}


The R code to generate the probability distribution:

p=function(m,n,s){
prod(seq(m,m-s+1))/factorial(s)*(n-1)^(m-s)
}

tt=c()
for(i in 1:10){
tt=c(tt,p(100,60,i)/60^100)
}
plot(tt,type="h",xlab="Number of Balls",ylab="Probability")


Dedication

This post is dedicated to my friend Ernesto Adorio, a mathematician. He loves combinatorial mathematics.

Rest in peace my friend! I miss bouncing ideas with you.