The next concept I want to talk about is conditional probability.
This is the one we're going to need to use to prove that the one-time pad is perfectly secure.
Here's the definition of conditional probability.
If we have two events--we'll call them A and B--
and they're in the same probability space,
the conditional probability of B given that A occured
is written like this: it's the probability of B given A--so we use the bar
to indicate conditional probability--and it's defined by this formula.
It's the probability of A intersected with B
divided by the probability of A.
To get an intuition for that, let's look at these sets.
Here's our universe omega. That's all events.
Our question is given that we know A occurred,
what's the probability of A.
The fact that A occurred means the rest of our probability set doesn't matter.
We're only considering the outcomes where A occurred.
That's the set for A.
We want to know what was the probability that B occurred.
Those are the elements in this intersection--the times when B did occur
divided by the size of the sets A,
which is the probability of A. That's the intuition behind this formula.
Let's see if it makes sense for our example as well.
The question is given that we have a valid coin toss,
what's the probability that it's heads.
I'll remind you the model we had.
It said the probability of heads is 0.49999 with four 9s.
The probability of tails is 0.49999.
The probabiilty of edge is 1 minus the sum of those,
which is 0.00002.
We define valid as the outcomes where it lands on heads or tails not on the edge.
Given that you know a coin toss is valid,
what's the probability that the outcome is heads?
