# Same Number!

Imagine you are in a class of thirty students. The teacher asks everyone to secretly write down a whole number between 1 and 225.

Do you think it's likely or unlikely that everyone's number will be different?

You could try this out in your class a few times, or experiment with this simulation:

How often was everyone's number different?

Are you surprised by this?

Let's try to explain what happens:

Imagine the teacher asks students to read out their numbers one at a time.

What is the probability that the second student reads out a number that is different to the one read out by the first student?

If the first two students have read out different numbers, what is the chance that the third student will read out another new number?

What is the probability that the first three students all read out different numbers?

If the first three students have all read out different numbers, what is the chance that the fourth student will read out another new number?

What is the probability that the first four students all read out different numbers?

...

What is the probability that the whole class of thirty students read out different numbers?

What is the probability that at least two students have written the same number?

There is a famous mathematical problem called the Birthday Problem:

*How many people do you need in a room so that the chance that there*

*will be at least two people with the same birthday is greater than 50%?*

One way to solve this is to imagine that people enter the room one at a time, and that each new person doesn't share a birthday with anyone already there...

Take a look at the problem At Least One... which provides an introduction to using tree diagrams and a neat approach to solving problems like this one.

Same Number! poses some really challenging problems and we hope that you enjoyed exploring them.

Here are some of the solutions which have been previously submitted to the team:

Joseph from Wilson's School shared his thoughts about the problem:

I think it is likely that everyone's number will be different, as there are 225 numbers and only 30 people. I expect there will only be a few times when everyone's numbers are different.

Yes, I am surprised by the fact that everyone's number were different only a few times.

Rajeev from Haberdashers' Aske's Boys' School gave this detailed answer:

In a class of 30, if a teacher asks the class to write down a number from 1 to 225, then the probabilitity of the same number being read out is as follows:

The probability that the $2^{nd}$ student reads out a number that is different to the one read out by the $1^{st}$ student is $\frac{224}{225}$

The chance of a $3{rd}$ student reading out a different number is $\frac{223}{225}$

The probability that the first 3 students will all read out different numbers is $\frac{224}{225}\times\frac{223}{225}$

The chance of a $4^{th}$ student reading out a different number is $\frac{222}{225}$

The probability that the first 4 students will all read out different numbers is $\frac{224}{225}\times\frac{223}{225}\times\frac{222}{225}$

The probability that the whole class will all read out different numbers is $\frac{224}{225}\times\frac{223}{225}\times\frac{222}{225}.......\times\frac{196}{225}$ = 0.13 (to 2 dec. places)**This implies the probability that at least two students have written the same number is1 - 13% = 87%**

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In the Birthday Problem we were asked:

*How many people do you need in a room so that the chance that therewill be at least two people with the same birthday is greater than 50%?*

Imagine the people enter into the room one by one...

The probability that the $2^{nd}$ person to enter does not share their birthday with the $1^{st}$ person is $\frac{364}{365}$= 0.997

The probability that the first three people to enter do not share the same birthday is $\frac{364}{365}\times\frac{363}{365}$= 0.992

If you keep multiplying like this you will eventually find an answer which is just less than 0.50 which happens once 23 people have entered the room:

The probability that the first 23 people to enter do not share the same birthday is

$\frac{364}{365}\times\frac{363}{365}.....\times\frac{343}{365}$ = 0.49

So when there are 23 people in the room, the probability that at least two people share the same birthday is $1- 0.49 = 0.51$

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A student presented a very clear explanation of the general calculation process for these types of problem:

The probability P that nobody shares a number/birthday in a given population of x, where n is the maximum number, is:

$$\frac{n(n-1)(n-2)....(n-(x-1))}{n^x}$$

Knowing two of n, x, or P means we can feed in the numbers to find the last variable.

Therefore the probability of two or more people sharing numbers/birthday in a given population of x, where n is the maximum number, is:

$$1-\frac{n(n-1)(n-2)....(n-(x-1))}{n^x}$$

Well done to everyone!

### Why do this problem?

This problem is one of a set of problems about probability and uncertainty. Intuition can often let us down when working on probability; these problems have been designed to provoke discussions that challenge commonly held misconceptions. You can read more about it in this article.

### Possible approach

**without letting anyone see**. (If the class size is $N$, ask them to write down a number between $1$ and $(\frac{N}{2})^2$)

Once everyone has chosen, go round the class and ask each person to read out their number, with everyone listening to hear if their number is read out again. On most occasions, there will be duplicates.

"OK, let's have another go, you have $225$ numbers to choose from so let's see if you can manage to make them all different!" (Don't allow collaboration on which numbers to choose!)

The experiment can be repeated a few times, or the interactivity can be used to generate sets of random numbers for a class of 30 to get a sense of how often everyone manages to pick a unique number.

On one occasion, after the first person has read their number, interrupt and ask "What's the probability that the next person has written a different number?", and highlight that it's very likely!

Ask a similar question a little further along the line: "What's the probability that the next person has a number that has not been read out so far?" Though the fraction will have changed, it will still be very close to 1, so learners' intuition may be that it is very likely that everyone will choose a different number.

Once everyone has had the chance to be surprised that duplicate numbers happen much more often than not, hand out this worksheet (Word, PDF) and ask the class to discuss the questions and work out their answers with their partner. (You may wish to detach the Extension task from the bottom of the
sheet, and hand it out later to those who need it.)

The Extension task is the well-known Birthday Problem; students could work out the probability of shared birthdays in various class sizes and then use school records to see if the tutor groups in the school have shared birthdays with the expected frequency.

### Key questions

What's the probability (at each stage) that the next person has written a number that has not been read out so far?

### Possible support

The problem At Least One... provides an introduction to using tree diagrams and working with mutually exclusive events whose probabilities sum to 1.