Heston

You are called Heston. You need to make a bar of chocolate. The process to make this bar of chocolate uses three machines, one after the other. All three need to work for the chocolate bar to be made. Each machine has a success rate of 90%. What is the probability that the whole process is a success and the chocolate bar gets made?

Answer = 73%. (0.9 x 0.9 x 0.9 = 0.73)

People often guess a higher success rate than that. This is an example of the confirmation heuristic, whereby people overestimate the probability of conjunctive events. A 'heuristic' just means a rule of thumb, we use them all the time and mostly that's okay, but sometimes they lead us to make systematic mistakes just like the one above.

The Confirmation Trap

Example taken from Bazerman and Moore (2009) based on Wason (1960):

Imagine the following sequence of numbers follows a rule, and that your task is to diagnose that rule. When you write down other sequences of three numbers, your instructor will tell you whether or not your sequences follow the rule.

2-4-6

What sequences would you write down?

Commonly guessed patterns include "numbers go up by two" and "the difference between the first two two numbers equals the difference between the last two numbers". In fact, the rule was much broader: "any three ascending numbers". But people had only tried sequences that tested their hypothesis by accumulating evidence that confirmed it, but didn't test for wider rules. They fell into the confirmation trap by just seeking to confirm their suspicions by trying sequences like 1-3-5 and 22-24-26 instead of trying, for example, 1-2-3 or 1-2-10.

This is called the confirmation bias. In all walks of life, from politics to business, people seek to confirm their beliefs rather than really test them.

People ask "May I believe it?" rather than "Must I believe it?"

Econitus

The following problem is adapted from Bazerman and Moore (2009) Judgement in Managerial Decision Making:

Lisa is worried about her health. Her doctor tells her not to worry too much as there is only a 1 in 1,000 chance that women of her age has the dreaded Econitus virus. Nevertheless, Lisa remains anxious about this possibility and decides to obtain a test that can detect Econitus. The test is moderately accurate: When someone has Econitus it delivers a positive result 86% of the time. But there is, however, a small 'false positive' rate: 5% of people produce a positive result despite not having Econitus. Lisa takes the test and obtains a positive result. What are the chances that she has Econitus?

0-20 percent chance
21-40 percent chance
41-60 percent chance
61-80 percent chance
81-100 percent chance

The correct answer is 1.7%!

If you answered 86% then you fell into the common trap of ignoring 'base rates'...

If 1,000 women like Lisa take the test, 999 will not have Econitus. But the false positive result means 50 will be told they have Econitus. Therefore there is only 1.7% chance that Lisa has Econitus (trust me!).

What this demonstrates is that people ignore background information in favour of more salient information about a specific case.

Hopefully this will be useful the next time you are given statistics by a doctor...

Boy Or Girl??

The following problem is taken from Bazerman and Moore (2009) Judgement in Managerial Decision Making:

You and your spouse have had three children together, all of them girls. Now that you are expecting your fourth child, you wonder whether the odds favour having a boy this time. What is the best estimate of your probability of having another girl?

6.25% (1 in 16), because the odds of getting four girls in a row is 1 out of 16

50% (1 in 2), because there is roughly an equal chance of getting each gender

A percentage that falls somewhere between the two estimates (6.25-50 percent)

The answer is 50%. (The sperm that determines gender of the child does not know the gender of previous children! In other words, the gender of each child is independent of that their siblings.)

So what explains our potentially wayward intuition here?

According to Kahneman and Tversky (1974) "Chance is commonly viewed as a self-correcting process in which a deviation in one direction induces a deviation in the opposite direction to restore the equilibrium. In fact, deviations are not corrected as a chance process unfolds, they are merely diluted."

So there you go, hope this helps when you're expecting your fourth child...

Russian Roulette

Scenario 1: There is a gun pointed at your head. It has six chambers. There are four bullets inside. The chambers will be spun randomly before the gun is fired once.

Scenario 2: There is a gun pointed at your head. It has six chambers. There is only one bullet inside. The chambers will be spun randomly before the gun is fired once.

In which scenario would you be willing to pay more to have one bullet removed from the chambers?

Interestingly, according to economic theory you should be willing to pay more to reduce the bullets from 4 to 3 than from 1 to 0. Is this the case for you?


For those who are economically inclined, here is a brief sketch of the proof of the theory:

Where u(.) is a von Neumann-Morgenstern utility function, p1 is what you are willing to pay in Scenario 1, p2 is what you are willing to pay in Scenario 2 and Y is your income.

In Scenario 2:

u(alive, Y - p2) = 1/6u(dead) + 5/6u(alive, Y)

In Scenario 1:

1/2u(dead) + 1/2u(alive, Y - p1) = 2/3u(dead) + 1/3u(alive, Y)

Re-write Scenario 1:

u(alive, Y - p1) = 2/6u(dead) + 2/3u(alive, Y)

Comparing scenarios:

u(alive, Y - p2) > u(alive, Y - p1)

Therefore:

p1 > p2

QED

The Monty Hall Problem - Fast Car

This is a probability puzzle named after the American quiz show host Monty Hall. Strictly speaking it's not a behavioural economics puzzle, but I'm interested in what you do...

There are three doors (1, 2 and 3) with one prize hidden behind each one. There is one car and two goats. Obviously, the idea is to get lucky and pick the car. Equally obviously, there is a one third chance of getting the car.

You can pick whichever door you like, for example door 1.

Monty knows where the car is. He has a think. He opens door 3 and reveals a goat.

He then offers you the chance to switch from door 1 to door 2 (or to stick with door 1).

Do you switch from 1 to 2?

When I first faced this problem I said no. I would stick with door 1. I reasoned that now there is a 50% chance of getting it right, and I might as well stick with what I put down first. It turns out that I was wrong.

If you switch to door 2 there is a higher probability of getting the car!

Wikipedia is full of differing explanations for the maths behind it, but I'll explain it in the way I reasoned it out.

At the start of the problem there is a one third chance of picking the car and you pick door 1. Thus there is a two thirds chance that one of doors 2 and 3 have the car. Therefore there is a two thirds chance that when Monty chose the door to open, he had no choice (he could not open the other door as that would have revealed the car).

There is only a one third chance that neither 2 or 3 had the car. In this case Monty could have picked either 2 or 3 to open.

Thus because there is a 2/3 probability that Monty had to open 3 and there is only a 1/3 chance he could have chosen either, we should switch to the one he did not open: door 2.

It took me a long time to work that out. Please let me know whether my explanation is adequate...

Recommended listening:
Fast Car by Tracy Chapman