# In a box

*In a Box printable sheet*

Chris and Jo decide to play a game.

They put some red and some blue ribbons in a box.

They each pick a ribbon from the box without looking (and without replacing them).

Jo wins if the two ribbons are the same colour and Chris wins if the two ribbons are a different colour.

**How many ribbons of each colour would you need in the box to make it a fair game?**

**Is there more than one way to make a fair game?**

*This problem is based on one offered by Doug Williams at the 2003 ATM conference in Bath UK. See also http://www.blackdouglas.com.au/taskcentre*

Click here for a poster of this problem.

What about drawing a sample space of the possibilties.

Try different numbers of each ribbon.

What would happen if you had ribbons of three different colours?

You might like to draw your own tree diagrams for these solutions.

Several of you suggested that having the same number of the same colour ribbons would make the game fair but this is not the case. Take, for example, one that many of you chose, which was four red and four blue ribbons. What happens?

\begin{eqnarray}\mbox{P (two the same colour)}&=& \mbox{P (Red and Red)}+\mbox {P (Blue and Blue)}\\ &=& \mbox{P(RR)}+\mbox{P(BB)}\\ &=&4/8 \times3/7 + 4/8 \times3/7 = 24/56 \end{eqnarray}

But this is less than a half so you have less chance of selecting two of the same colour than choosing two ribbons of different colours.

Laura and David of Cannock Chase High School showed that the game is not fair with 4 ribbons of one colour and two of the other

The game is not fair because:

$$2/6 \times1/5 = 2/30$$

$$4/6 \times3/5 =12/30$$

Probability that the two ribbons are the same is $14/30$ $(2/30 + 12/30)$.

The game is not fair as it is not an even chance; if it was even it would be $15/30$.

Nathan H. and Colby B. offered a solution to making the game fair:

There only need to be 4 ribbons (3 same, 1 different) therefore there can only be a 50/50 chance that one will win.

Working:

\begin{eqnarray}\mbox{P (Red and Red)} &=& \mbox{P (R)} \times \mbox{P (R)} \\ &=& 3/4 \times 2/3 \\ &=& 6/12 \; \mbox{OR}\; 1/2 \end{eqnarray}

\begin{eqnarray}\mbox{P (Red and Blue)} &=& \mbox{P (B)} \times \mbox{P (R)} \\ &=& 3/4 \times1/3 \\ &=& 3/12 \; \mbox{OR}\; 1/4 \end{eqnarray}

\begin{eqnarray}\mbox{P (Blue and Red)} &=& \mbox{P (R)} \times \mbox{P (B)} \\

&=& 1/4 \times3/3\\

&=& 3/12 \; \mbox{OR}\; 1/4 \end{eqnarray}

$$\text{P (Same)} = 1/2$$

$$\text{P (Different)} = 1/2$$

There is another way to even the game: Put 2 ribbons of one colour and 2 of another, and they each have to get the same colour (Red or Blue).

Amy of Kingwinford Secondary School suggested the following modification to the game to make it fair. Is she right?

There is more than one way to make a fair game - one way is:

The two players pick 1 ribbon, and say 1 player picks a red ribbon and the other picks a blue ribbon, then 1 point would get added on to you score, because you have both picked 2 different colours, but if they both picked the same colour then they would have points taken off their score.

There is another way of playing the game but differently, you could have 4 different coloured ribbons (red, blue, green and yellow) 2 of each colour then the players have to try and pick 2

ribbons of the same colour, if they do they win!

Andrei of School 205, Bucharest started looking for general solutions to a fair game and found some interesting patterns:

To solve the problem, I used combinations. The formula of combinations, i.e numbers of groups of k objects taken from n objects, where the order does not matter is: $${n\choose{r}} = \frac{n!}{r!(n - r)!} $$ I calculated the number of pairs of ribbons of the same colour. There are two red ribbons and I take two: $${2\choose2}$$ There are four blue ribbons and I take two: $${4\choose2}$$
The total is: $${2\choose2} + {4\choose2}$$ In total, there are 6 ribbons, and I take 2. So, the total number of groups of 2 ribbons that could be formed from 6 is $${6\choose2}$$ The probability that the two ribbons are of the same colour (that Jo wins) is: $$\frac{{2\choose2} + {4\choose2}}{{6\choose2}} = \frac{1 + 6}{15} = \frac{7}{15}$$ Now, the probability that the two ribbons are of
different colours (that Chris wins) is: $$1 - \frac{7}{15} = \frac{8}{15}$$ As the two probabilities are not equal, the game is not fair. Now, I try to find possibilities to make the game fair. I made a small program in which I introduced the two numbers, and it calculated the probability. The numbers I obtained formed an unexpected pattern. In the following table, the numbers in the first column
represent the ribbons of the first colour, the numbers in the second column the ribbons of the second colour, while in the third column their difference is calculated:

$1$ | $3$ | $3-1=2$ |

$3$ | $6$ | $6-3=3$ |

$6$ | $10$ | $10-6=4$ |

$10$ | $15$ | $15-10=5$ |

$15$ | $21$ | $21-15=6$ |

$21$ | $28$ | $28-21=7$ |

Paris from IES Täby in Sweden sent in a full solution which shows algebraically the patterns that Andrei noticed.

Click

### Why do this problem?

This problem offers opportunties to consider different methods of listing systematically. It can be used to introduce or revisit sample space diagrams, and with some students, tree diagrams.

### Possible approach

Play the game a few times for real.

"Is this a fair game? How can we be sure?"

Class work in pairs trying to decide and to develop an argument to justify their conjectures.

After about ten minutes, stop to discuss the merits of different arguments and representations. This may be an appropriate point to highlight the benefits of different systematic methods for listing all possibilities, using sample space diagrams and, if pupils have met them before, tree diagrams. This is an opportunity for everyone's method to be valued, and for the whole class to work
collaboratively. They should agree on some representations or methods for identifying fairness.

Finding a fair game can become a class activity:

Students help to create a class list of all distinct starting points for the game (for example, four ribbons can be either $1R$ and $3B$ or $2R$ and $2B$). These are written on the board for $3, 4, 5, \ldots$ribbons.

Distribute the task of checking which combinations are fair and record them on the board as pairs of pupils decide.

There are not many solutions that work and if pupils are to notice a pattern amongst the combinations that are fair they may need to consider up to a total of $16$ ribbons.

Spend some time conjecturing about more than $16$ ribbons and test.

### Key questions

How can you decide if the game is fair?

How many goes do you think we need to be confident of the likelihood of winning?

Are there efficient systems for recording the different possible combinations?

Can you justify your conclusions?

### Possible support

When the work is being shared out amongst pupils in the class the smaller number of ribbons can be given to pupils who struggle most. They can also use physical objects, such as coloured counters, to check their findings.

### Possible extension

Justifying the general rule