# Flippin' discs

## Problem

*Flippin' Discs printable sheet*

Imagine you have two discs. Each disc is red on one side and green on the other.**You flip the discs, and when they land, you win if both discs show the same colour.**

Click "Run Once" to complete a throw. Did you win?

Click a few more times and look at your results after several throws.

Approximately how often do you think you would win if you completed $100$ throws?

Make a prediction and then check it by doing the experiment.

Now click on the purple cog in the top right corner and change the number of discs to $3$. **You win if all the discs show the same colour.**

Can you predict what the probability of winning will be this time?

Check that your prediction matches the results from the interactivity.

Do the same with $4$ and $5$ discs.

Do you notice a pattern in your results? Can you explain it?

Can you explain how to find the probability of winning for $n$ discs?

## Getting Started

Try listing the possible outcomes.

You could use a tree diagram.

## Student Solutions

Ruby and Emma from Yewdale Primary in the UK estimated the numbers of wins they would get by thinking about winning combinations:

2 discs- out of 1000 we got 490 wins which is 49%- We predicted 45% to 55% as there are 4 possibilities and only 2 of them give you a point.

Mahdi from Mahatma Ghandi International School in India showed the 4 combinations as an image, and calculated a probability:

From the 4 possible outcomes only 2 are valid.

$\text{Probability} = \dfrac{\text{favourable outcomes}}{\text{total number of outcomes}} = \dfrac24 = 0.5 = 50\%$

Atharva from Wilson's School used the same method, and concluded:

So you should win 50% of the time.

This is the rest of Ruby and Emma's answer. They made mistakes finding the numbers of possibilities, but their answer is still correct in principle:

3 discs- out of 1000 we got 311 wins which is 31.1%- We predicted 20% to 30% as there are nine possibilities and again only 2 of them give you a point.

4 discs- out of 1000 we got 134 wins which is 13.4% - we predicted 10% to 20% as there are 16 possibilities and only 2 of them give you a point.

5 discs - out of 1000 we got 54 wins which is 5.4% - we predicted about 4% to 9% as there are 25 possibilities and only 2 give you a point.

The pattern is that whenever the numbers of discs are bigger, the percentage decreases.

We estimated with 6 discs and out of 1000- we predicted about 3% to 4.5% as

there are 36 possibilities and only 2 give you a point.

Sanika P from PSBBMS, OMR in India, Mahdi and Atharva all continued listing the possibilies. This is Sanika's work for 3 discs:

D1 | D2 | D3 |
---|---|---|

R | R | R |

G | G | R |

R | G | R |

G | R | R |

R | R | G |

G | G | G |

R | G | G |

G | R | G |

Among these, the first and the sixth (2) combinations are considered as wins out of the 8 possible outcomes.

Therefore the probability of winning when there are 3 discs is $\frac28$ or 25%

This is Mahdi's work for 4 discs:

From the 16 possible outcomes only 2 are valid.

$\text{Probability} = \dfrac{\text{favourable outcomes}}{\text{total number of outcomes}} = \dfrac2{16} = 0.125 = 12.5\%$

Atharva spotted a pattern:

2 discs = 50%

3 discs = 25%

4 discs = 12.5%

5 discs = 6.25%

Ergo you just half the percentage for the next one, e.g.

6 discs = 3.125%

*Can you see how this follows from Mahdi's image? If you ignore the last disc in each image, the possibilities for 3 discs all appear twice - so the number of possibilities doubles each time you add another disc. How is that related to Atharva's pattern?*

Ahan from Tanglin Trust School in Singapore, Mahdi and Sanika found the same pattern but described it in a different way. Sanika wrote:

By extending this is $n$ discs

$D_1,D_2,...D_n$ Each of the discs can have $2$ possible states - Red or Green. So the total [number of] possible outcomes [is] $2^n$

Among the possible outcomes only $2$ could be win wherein all the discs are of same colour.

Therefore the probability of winning with $n$ discs is equal to $\dfrac2 {2^n} = \dfrac{1}{2^{n-1}}$

## Teachers' Resources

### Why do this problem?

This problem offers an opportunity to explore and discuss two types of probability: experimental and theoretical. The simulation generates lots of experimental data quickly, freeing time to focus on predictions, analysis and justifications.

The interactivity starts with a simple context of just two discs with two equally likely outcomes, together with a visual way of representing successive trials. As students work on the problem, they can increase the number of discs and switch between representations, before moving on to other related probability tasks. For more information about Low Threshold High Ceiling tasks, you may be
interested in this article.

### Possible approach

Introduce the problem, and invite students to make hypotheses about the likelihood of winning with two discs. Encourage them to explain their thinking and try to justify their hypotheses.

Invite students to think for a minute on their own about what will happen with three discs. Ask them to justify their conjectures with a partner and then the whole class. Use the interactivity to demonstrate what happens. Students could then work in pairs, to explain the results.

You may like to stop the group part way through their work in order to talk about useful representations that they have devised. This could be an opportunity to introduce tree diagrams, sample space diagrams, or the systematic listing of outcomes.

Students could then use the different representations to work out the theoretical probability of winning for 4, 5, and $n$ discs.

### Key questions

How can you decide if a game is fair?

How do we know we haven't missed any possibilities?

### Possible support

Teachers can also use this recording tool to gather the results of other similar experiments that their students are carrying out:

### Possible extension

Students could also work on Odds and Evens, Cosy Corner, and Two's Company.