Base Rate Fallacy
September 10th, 2016
Today we will discuss an interesting concept, Base Rate Fallacy. This is a phenomenon resulting from our minds’ tendency to remember specific information and ignore broader information. Let’s dive in.
Let’s say 5% of sober people fail breathalyzers, while a truly drunk person always fails a breathalyzer test. One in a thousand drivers is drunk. Suppose a police officer pulls someone over and forces them to take a breathalyzer test. They fail. What is the probability that they are drunk?
According to Wikipedia, where this problem comes from, many people answers reach .95, or 95%. Since 5% of sober people fail, that seems reasonable, right? Let’s do the math.
I. Conditional Probability
We will start by looking at the basics of conditional probability. Conditional probability is the probability of an even occurring given the outcome of a prior event. For example, let’s say we have an Ace and a King. What’s the probability of drawing the Ace as the second card? Well, in one scenario, we draw the King first and the Ace second, and in the other possible scenario we draw the Ace first then the King. So, the probability of drawing the Ace the second time is .5, or 50%. But let’s say we know that the King is drawn first, then we have a 100% probability that the Ace is drawn second. This is called conditional probability. The conditional probability of A given B is written as
II. Monty Hall Problem
A really good example of this is the Monte Hall Problem. You are on a game show, and there are 3 doors. Behind 2 of the doors is a goat, and behind the 3rd door is a car. You pick a door. What’s the probability that your door has the car? Hopefully it’s pretty clear that the answer is 1/3 or 33.3%.
But, we are going to change the game now. Now, you pick a door, and then the host opens a door with a goat behind it. Now there remains one door with a goat and one with a car. He asks you if you want to switch your door. You would think that there is a 50% chance that your door holds the car, so switching wouldn’t matter. Analyzing the situation yields a different result, however. Let’s say you pick door 1.
Either door 2 or 3 has to hold a goat. There are 3 scenarios. In the first, your door is the car, while in the other scenarios, your door holds a goat. In the first, if you switch, you don’t get the car, but in both of the other scenarios, since your door has a goat, he is forced to show you the only other goat, and switching yields the car.
We can also do this mathematically using the formula for conditional probability. The conditional probability formula is
Again, let’s say you pick the first door. What is the probability that your door holds the car? We will say the game show host opens door 2, but it’s exactly the same if he opens door 3. Our A is that door 1 is the car, and our B is that door 2 is opened.
This is because we have a 1/3 chance that our door holds the car, since we picked between the 3 doors, and then since our door is door 1, there is a 50% chance he opens door 2, and a 50% chance he opens door 3.
Since as stated above, there is a 50/50 chance he opens door 2. Therefore, using the formula,
So, it’s advantageous to switch your door. You are twice as likely to get the car if you switch.
II. The Drunk Driving Problem
We will start by defining A as the event that the person is drunk, and B as the event that they fail a breathalyzer. The probability that someone is drunk and fails a breathalyzer, since 1 in a thousand drivers are drunk and every drunk driver fails a breathalyzer. Now, we need to calculate the probability that any random driver fails a breathalyzer.
This is because the 1 drunk person fails, and 5% of the 999 sober people fail. So, using the formula, we get
So, according to the facts given, less than 2% of people that fail a breathalyzer test will be drunk. This is largely due to the fact that 5% of the 999 sober people is still a lot of people, around 50, and 1/50 is approximately the ratio of drunk/sober people failing, which is 2%.