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# Same Number!

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Age 14 to 16

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

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 is
1 - 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 there
will 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!