## Edinburgh Sci Fest 2012 (Part 2)

Welcome back to part 2 of my write up of our exhibit at the Edinburgh International Science Festival. As you may remember, we were running a series of games and activities to test people’s probability skills and to see how people would react to the stats in a courtroom. In this post I will go through the solutions to the various questions we asked, so if you haven’t had a go at them yet then make sure to have a go now!

So, here are the answers!

The Monty Hall problem is always a good excuse for a picture of a goat.

1) The Monty Hall Problem: 3 doors; 2 hiding goats and 1 hiding a car. You pick a door, then the gameshow host shows you another door hiding a goat. He asks you if you want to swap to the other remaining door, or stick with the one you first picked. The vast majority of the public, of whatever age, said that there was no difference between sticking or swapping, and most preferred to stick. (People seem to find it psychologically more distressing to lose a prize they had briefly owned than to not win a prize they could potentially have had.)

In actual fact, they would have been twice as likely to win the car by swapping doors than by sticking with their original choice. How can this be? After the host opens a door to reveal a goat, there are two doors left, so surely the probability is 50-50? The reason this is not true is because the host’s choice of door depends on the player’s initial choice of door. If the player chooses a door hiding a goat the first time (which happens 2 times out of 3) then the host has no choice but to reveal the other goat-hiding door, meaning that he definitely leaves the car behind the other door. To repeat, 2 times out of 3, the car has to be behind the other door. In other words, the player only wins the goat by sticking if they pick correctly the first time, which happens 1 in 3 times.

2) Heads or Tails: I flip two coins and tell you that at least one of them is a head. What is the chance that the other coin is also a head? Again, intuition tells most people that it should be 50-50. How can the first coin influence the second? And yet the knowledge that at least one coin is a head doubles the chance of the other coin being a tail! Consider the different possible combinations of two coin tosses: Tail/Tail, Tail/Head, Head/Tail or Head/Head. The knowledge that at least one coin is a head eliminates the first possibility, leaving 3 remaining, of which two give the other coin as being a tail.

What’s the chance of correctly choosing 4 numbers from 10?

3) Lottery: what is the chance of winning the jackpot in a lottery where 4 numbers are chosen from 10? Younger visitors to the museum tended to vastly overestimate the probability, quoting odds of about 1 in 4 or 1 in 5. Older visitors tended to be extremely skeptical, thinking the odds were more like 1 in a million! The actual odds were somewhere in between, being about 1 in 200. Here’s how to work it out.

The first ball chosen has a 4 in 10 chance of matching with one of the numbers you’ve picked. That’s because there are 10 numbers in total and it could match any one of your 4 chosen numbers. Now we want the next ball to match. There are 9 numbers left to choose from, but the chosen ball can match with any 3 of your remaining numbers, so that’s a 3 in 9 chance. Continuing similarly, we get a 2 in 8 chance for the third ball to match and a 1 in 7 chance for the final fourth ball to match.  To get the final probability, multiply all the chances together:

${\frac{4}{10}\times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} = \frac{1}{210} }$

[As an aside, I analysed the lottery tickets of 480 people who played our lottery and found that 0 was by far the least popular number, followed (surprisingly) by 3. The number 2 was (marginally) the most popular.]

4) The coin flip: if 100 people flip a fair coin 5 times each, how many of them would we expect to flip 5 heads? There is a probability of ${\frac{1}{2}}$ of the coin being a head, and we need this to happen 5 times in a row. Each flip is independent of the last, so the calculation is ${(\frac{1}{2})^5 = \frac{1}{32}}$. So we expect 3 people out of 100 to flip 5 heads. Small children are especially incredulous that such an event could ever happen, claiming that anyone who flipped 5 heads in a row had to be cheating.

Motto: Even unlikely events become probable if the experiment is repeated enough times. There may be a 1 in a million chance of a piece of DNA matching, but this means we expect 60 people in the UK to be potential suspects of a crime. Just because there is a small chance of finding a piece of evidence (such as a DNA sample), this does not mean there is a correspondingly small chance that the suspect is innocent. This argument is known as the prosecutor’s fallacy and I discussed an example of it in my last blog post.

5) The Birthday Problem: How many people need to be in the room for it to become likely that there is a shared birthday? The most common reply to this is about 180 – simply by

If I had a birthday this would be my cake.

dividing the total number of possible birthdays (365) by 2. People are always shocked to be told that the answer is only 23. How is it possible that only 23 people could produce a shared birthday when there are so many possible birth dates? The key to thinking about this is to remember that it could be a shared birthday between any two people, not between any two particular people. As the number of people in the room increases, the number of possible pairings gets exponentially big and the corresponding chance that nobody shares a birthday becomes very low. (For details of the calculations, see the Wikipedia page.) This mathematical result was well demonstrated on the first few days in the museum, when we had our first match at the 17th person (3 days in a row!). The next couple of days we had to wait until about the 30th person, but this was still well within most people’s expectations. And by 100 people we always had a match of 3 people on the same day!

The calculation relies on birthdays being equally distributed throughout the year, which in fact they are not, so the matches become even more likely. A recent article documented the frequency of birthdays on each day of the year, finding that August – October was the most popular time to have a baby. Lots of cuddling up in the winter months it seems!

6) The eyewitness: if an eyewitness to a crime says she saw a person wearing a hat, and if she has a 2/3 chance of correctly spotting hats, and if only 10% of the population wear hats, what is the chance that the suspect was actually wearing a hat? This is a difficult question to get across so we did this by having 30 Playmobil figures of which 3 were wearing hats. The visitors then had to sort the characters into the different categories:

• Wearing a hat and correctly identified: there are (2/3) x 3 = 2 people in this category;
• Wearing a hat and incorrectly identified: there is (1/3) x 3 = 1 person in this category;
• Not wearing a hat and correctly identified: there are (2/3) x 27 = 18 people in this category;
• Not wearing a hat and incorrectly identified: there are (1/3) x 27) = 9 people in this category.

In total then, there are 2+ 9 = 11 people who were thought to be wearing hats, and only 2 of them were actually wearing  a hat. So the chance of the suspect wearing a hat is only 22% – much lower than the 66% most people expect!

Motto: If a situation is particularly rare, then even a test for it which is quite accurate will produce more false positives than true positives. This is a common fallacy in both law and medicine and is one of the reasons we need to have better teaching of Bayesian statistics in these professions.

How did you do on these questions? Did you fall into the common fallacies? Have you heard of them before? Do you disagree with any of the answers? Leave me your comments!