Probability That You Can Use Everyday for Dummies
This post will answer the following questions...
 How many possible 4digit passcodes can you have on your iPhone 6? (Enumerations & Permumations & Combinatorics)
 How many centuries will it take to break a password with an entropy level of 77? (Security)
 If you got heads on the first flip... What is the probability you will get heads again? (Independent vs Dependent variables)
These are some of the many questions that probability helps us answer. <!more>
So... Let's get started!
So let's start off with the iPhone 6 passcode question. How many possible ways of rearranging the 4 digit passcodes are there?
To answer that, we need to understand something important about probability. If we have some random sequence of some length (think of the 4 digit passcode) and we have a set of outcomes (think 0,1,2,3,4,5,6,7,8,9) we can use this information alone to get our answer. So let's start off with an easy example: If we were allowed to make a passcode with length of 1 (1 digit passcode instead of a 4 digit passcode), our passcode would only have 10 possible ways of rearranging it
Total enumerations with OneDigit Passcode
0 1 2 3 4 5 6 7 8 9
But now if we were allowed to make our passcode with length 2 (2 digit passcode) we would now have 100 possible ways of arranging the 2 digit passcodes! This means that if someone was given 100 tries and they tried every single possiblity, they would be able to break your passcode! Here's a cute little python list comprehension to generate all the possible enumerations:
Total enumerations with TwoDigit Passcode
00 01 02 03 04 05 06 07 08 09
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
30 31 32 33 34 35 36 37 38 39
If we were to continue this pattern at the end it would look like:
90 91 92 93 94 95 96 97 98 99
If we were to count all of these out, you would find that there are a 100 different enumerations of a passcode with a sequence of 2 numbers and an outcome of 10 numbers (09 possible numbers)! Pretty cool huh?! But, what if we wanted to generalize it in a way where we could find all the enumerations of any sequence with any number of outcomes? Here is a pattern to keep in mind: outcomes
will be multipled by outcomes
the same number of times as the length of the sequence
.
In short: outcomes
^sequence length
will give you all the possible enumerations
Outcomes (09)  Sequence Length (14)  Calculation  Total Enumerations 

10  1  10^1  10 
10  2  10^2  100 
10  3  10^3  1,000 
10  4  10^4  10,000 
Now 10,000 possible enumerations might seem like a lot and it might seem like the phone would be safe, however, that is simply not true. 10,000 enumerations are very small and powerful password decryption tools can guess 100's of millions of possible enumerations per second. The thing that keeps you safe is how the iPhone will lock itself for increasing number of seconds per failed attempt.
So to answer your question of how many possible enumerations of fourdigit passcodes for your iPhone there are, with a bang:
There Are 10,000 Possible enumerations for Your iPhone 6 FourDigit Passcode
Now let's introduce a new idea.
So far we have discussed enumerations
. Now let's discuss permutations
and combinations
.
Let m
= outcomes
Let n
= sequence
Permutations  Enumerations  Combinations 

m!/(mn)!

m^n

m!/(mn)!n!

Now let's look at some tangible examples:
Say we are playing a lottery and we can draw 6 numbers from a range 0...59. Here is how we will be able to figure out all the possible enumerations
, permutations
, and combinations
:
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