Applications to Probability

Here is a python script that can help you get a feel for the questions above. You can switch between the two questions by commenting and uncommenting out the appropriate lines (lines that start with a # are comments). See how lucky you are!

If you know some python, you might want to modify the script to run the experiment 1000 times and see how many of those are wins.

We can get a feel for probability empirically by observing how frequently events occur when an experiment is repeated many times. It often happens, as it did with the Chevalier de Mere, that our intuition about probability is not quite right. Using the counting techniques we have studied, we can explain why our intuition is off and what the true probabilities are.

Most of the questions about counting we have considered in this chapter can also be asked as a question about probability. For example: How many passwords of length 8 can you make using just lower-case letters? What is the probability that randomly selecting 8 lower-case letters will give you your password?

While the subject of probability is vast and complex, the basics of discrete probability are little more than counting. So here we will take a brief look at how our study of counting can help us understand probability.

Think about how we use the language of probability in our everyday lives. We might say that tossing a coin has a 50% chance of coming up heads. Or that when rolling two dice, having the sum of the dice result in a 7 is more likely than having the sum be a 2. Casinos certainly rely on certain pairs of cards being consistently more likely than others when setting payouts for Blackjack. All of this assumes that there is some randomness to events, and that even in this randomness, there is some consistency to what can happen. We will assume this model of reality.

The things we can assign probabilities to are called random experiments. These can have different possible outcomes. We will call the (finite) set of possible outcomes to a random experiment the sample space (we will usually denote this set as \(S\)). By definition, performing a random experiment will always result in exactly one outcome from the sample space.

Throughout this section, we will always assume the uniform probability distribution, which means that we insist that each outcome in the sample space is equally likely. Then the probability of any particular outcome in the sample space \(S\) is exactly \(\frac{1}{\card{S}}\).

Consistent with our intuition, we define the probability of an event as follows.

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

Here are a few basic probability facts that follow easily from our definition of probability and understanding of counting. While we are still under the assumption that the outcomes in the sample space are equally likely (the uniform probability distribution), these rules will hold for all probability distributions.

First, we often are interested in the probability that an event does not occur. We call this the complement of the event. Remember, events are subsets of the sample space, and not being in the event means you are in the complement of that subset. Using the same notation we have for sets, the complement of an event \(E\) will be written \(\bar{E}\). Here is the relationship between the probability of an event and its complement.

Let's illustrate this proof with an example.

Complementary probabilities are very useful when answering historical questions about dice.

The proof of this fact is one of the exercises in this section. However, it should become clear how this works with an example.

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.

The famous probability problem, known as the Monty Hall problem, presents the following conundrum. You are on the game show Let's Make a Deal and will win whatever is behind one of three doors you decide to open. Behind one door is a car; behind the other two are goats. You pick a door, but before opening it, the host (Monty Hall) reveals one of the other doors that has a goat behind it. You then have the opportunity to switch doors. Should you switch? What is the probability of getting the car if you do?

This problem was perhaps one of the first math problems to go viral, although it did so when it appeared in the Sunday newspaper magazine Parade. After its publication, around 10,000 readers (including close to 1000 with PhDs) wrote in complaining that the author, Marilyn vos Savant, was wrong. She wasn't.

You might be tempted to say that the probability of getting the car when you switch is \(\frac{1}{2}\). After all, there are two doors left, and the car is behind one of them. However, we must ask what the probability of getting the car is given that Monty has revealed a goat behind another, unpicked door.

Does this definition agree with our intuition for what conditional probability should mean? Let's think about the sample space. We want to know the chances of \(A\) occurring under the assumption that \(B\) has already occurred. In other words, we only care about the elements of the sample space that belong to \(B\).

If \(B\) becomes the sample space, then the only outcomes from \(A\) that can possibly occur are the outcomes that are in \(A\) and \(B\). So perhaps the definition of conditional probability really should be, \[P(A | B) = \frac{\card{A \cap B}}{\card{B}}\].

Unfortunately, I'm not in charge of probability definitions. It turns out that the standard definition is just as good though. This is because, \[P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{\card{A \cap B}}{\card{S}}}{\frac{\card{B}}{\card{S}}} = \frac{\card{A \cap B}}{\card{B}}\]. Phew. Another crisis averted.

Condensed — the full section is in Levin, Discrete Mathematics: An Open Introduction.