# What are the chances?

In *Why coincidences happen* we saw how the chance of a rare event occurring to someone, somewhere, depended on the number of opportunities for it to occur.

Here we consider some specific situations.

## Throwing a double six

Mabel tells you she's thrown a double six with two dice. That has a 1/36 chance of occurring so it's lucky, but not that remarkable. Then she tells you that she threw the two dice 36 times. Suddenly this event seems even less remarkable. In fact the expected number of double sixes is now 1/36 $\times$ 36 = 1. Looking at the Table of *Why coincidences happen*, we could say there is approximately a 63% chance of getting a double six at least once in 36 throws.

To get the exact probability of obtaining at least one double six, it is best to consider the event that she does *not *get a double six. This will happen if all 36 throws are of something other than a double six, which will occur with probability 35/36, so the chance that they are all something other than a double six is $(35/36)^{36} = 0.36$, since probabilities of independent events are multiplied. So, there is a 36% chance of not getting a double six, which means that there is a 64% chance of getting a double six in 36 throws, which is very close to the 63% reported in the Table of *Why coincidences happen*.

#### There are 23 people in a room, and Bill and George have the same birthday

This is a classic matching problem. Again we can go through the four stages outlined in *Why coincidences happen*

*What is the chance of the specific event?*Assuming that people's birthdays are spread uniformly throughout the year, and ignoring leap-years, the chance that any two people will have the same birthday is 1/365. Bill's birthday can be any date in the year, but George's must be the same, and that occurs with probability 1/365.*How many opportunities were there for such an event to occur?*There are 23 people in the room, and if they all shook hands with each other there would be 23 $\times$ 22 / 2 handshakes, making 253 pairs of people that shook hands.*What is the expected number of events?*This is 253/365 = 0.69, which is fairly high, so we should already be suspecting that this is not such a surprising event. The Table in*Why coincidences happen*suggests the chance of a match will be high.*What is the chance of a match occurring?*Just as in the double six problem, it is easiest to work out the chance of there*not*being a match. In matching problems it is best to think of all the people lined up and announcing their birthdays one by one. The first can be anything; the second must be different from the first which has probability 364/365, the third must be different from the first two which has probability 363/365, and so on, so that the probability that they all have different birthdays = $$\frac{364}{365}\times \frac{363}{365} \times \frac{362}{365} \times.... \times\frac{343}{365} = 0.51.$$

We see that, perhaps rather remarkably, there is a more than 50:50 chance that two people will share the same birthday. A nice way of illustrating this concerns football matches: there are 23 people on the pitch (2 teams of 11 and one referee), so that in around half of football matches there should be two people on the pitch with the same birthday! Robert Matthews checked this and found it to work.

#### 20 people pick numbers between 1 and 100 at random - will they all choose different numbers?

The reasoning follows exactly the same lines as the birthday problem.

*What is the chance of the specific event?*The chance that any two*specific*people will choose the same number is 1/100.*How many opportunities were there for such an event to occur?*There are 20 people, making 20$\times$19/2 = 190 pairs of people.*What is the expected number of events?*This is 190/100= 1.9. The Table in*Why coincidences happen*suggests the chance of a match will be very high, close to 87%.*What is the exact chance of a match occurring?*Just as in the birthday problems, it is best to think of all the people lined up and choosing a number one by one. The first can be anything; the second must be different from the first which has probability 99/100, the third must be different from the first two which has probability 98/100, and so on, so that the probability that they all choose different numbers = $$\frac{99}{100}\times \frac{98}{100} \times \frac{97}{100} \times.... \times\frac{81}{100} = 0.13.$$

In *Maths of coincidence* we extend these results to more general situations.

## Comments

Anonymous (not verified)

Sat, 27/02/2010 - 7:12am

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## 4 same digits picked by two people

Anonymous (not verified)

Fri, 04/02/2011 - 9:56am

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## I'm very inetersted to know