Challenge Level

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%

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 $\dfrac2 {2^n} = \dfrac{1}{2^{n-1}}$