A Bit of a Dicey problem

1077 /www/nrich/html/content/01/10/penta4/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <mdo:image alt="2dice" height="215" src="2%20Dice%20Pic.jpg" width="291"></mdo:image> Look at these two dice. If you add together the numbers of dots on them, the total is $2$. When you roll two ordinary six-faced dice like these and add together the two numbers, what results could you get? Do you have more chance of getting one answer than any other? If so, what is that answer? And why? What if you use a dice with $10$ numbers on, like these? <mdo:image alt="Deca Dice" height="176" src="DecaDicePic.jpg" width="268"></mdo:image> What answers could you get now? Do you have more chance of getting one answer than any other? If so, what is that answer? And why? Now try with more dice or a mixture of dice. What do you notice now? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">William from Churchers College Junior School sent in his good explanations. $1$a You can roll numbers between $2$ and $12$ with two six-sided dice b $7$: Because you can roll combinations which use all the numbers on the dice. $1-6$, $6-1$, $5-2$, $2-5$, $4-3$, $3-4$ $6$ combinations give $7$. There are $36$ possible combinations when you threw two six-sided dice SO $6$ /$36 \times 100 = 16.67$ % chance of rolling $7$ with two six-sided dice. $2$a With two ten-sided dice you can roll numbers between $2$ and $20$. b $11$; because there are ten combinations which give $11$ which use all the numbers on the dice There are are $100$ possibilities when you roll two ten-sided dice. So $10$/$100 \times 100 = 10$ % chance of rolling $11$ with $2$, $10$ sided dice. From Oscar at CCJS (which I think is Cheltenham College Junior School - but forgive me if I'm wrong!) also worked on this and sent in the following; I wrote down every possible throw I could get using a black and white six-faced dice - it came to $36$ possibilites. I counted how many of each total number I could get and found that the commonest total number to get was $7$ Two six-faced dice gave me $36$ possibilities. That is $6$ squared or $6 \times 6$. The two six-faced dice had $7$ as the commonest number that we could get - that is, one greater than the number of sides on each dice. Finally Tom from Fen Ditton C.P. School wrote to tell us what he already knew or what he found out by doing the activity: $6$ numbered dice: the most likely number is $7$ because is has got the most combinations. $10$ numbered dice: $11$ is the most likely because it has the most combinations. The formula for this problem is:n $+1$ n= no. of sides on the die. Well done all of you, three excellent solutions sent in, keep up the good work!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>A Bit of a Dicey Problem</h2> <mdo:image alt="2dice" height="215" src="2%20Dice%20Pic.jpg" width="291"></mdo:image> Look at these two dice. If you add together the numbers of dots on them, the total is $2$. When you roll two ordinary six-faced dice like these and add together the two numbers, what results could you get? Do you have more chance of getting one answer than any other? If so, what is that answer? And why? What if you use a dice with $10$ numbers on, like these? <mdo:image alt="Deca Dice" height="176" src="DecaDicePic.jpg" width="268"></mdo:image> What answers could you get now? Do you have more chance of getting one answer than any other? If so, what is that answer? And why? Now try with more dice or a mixture of dice. What do you notice now? </div> <h3>Why do this problem?</h3> <div>This <a href="http://nrich.maths.org/public/viewer.php?obj_id=1077&part=">problem</a> gives plenty of chance for practising addition in a meaningful context whilst avoiding the monotony of doing a set of written calculations. As the context is an investigative one, the teacher will have plenty of opportunity to observe the children at work and assess their skills, with addition in the first instance, but subtraction and multiplication would also be possible by changing the operation in the question. The game is also an opportunity to model working systematically.</div> <h3>Possible approach</h3> <div>This is a <a href="http://nrich.maths.org/7701">low threshold high ceiling activity</a> that would be suitable for a range of pupils and enable them all to be working on the same activity but using different skills to reach their conclusions. Introduce the task by getting the children to throw two dice in pairs with one throwing the dice and one recording the results. If each pair in the class records their results for 5 throws then collectively they can make a conjecture about what the outcomes may be. After a number of trials they may be willing to make a hypothesis about the most likely result. It is important to press them to offer a rationale for their choice at this stage. They may be willing to justify their choice on the basis of an argument that doesn't involve collecting lots of examples or they may only be prepared to use the evidence that they have gathered. They could record their arguments and justifications by using a video camera, or on paper and then share them with the rest of the class at intervals during the lesson.</div> <h3>Key questions</h3> <div>What results can you make? How can you be sure you have considered all the possibilities? Are you getting one result more often than the others? What do you notice about the number of 1s, 2s, 3s,... that you have?</div> <h3>Possible extension</h3> <div>Consider other operations such as the difference between the numbers or the total on the white die subtract the total on the black die. What possible results would these give? Why? Which would be the most likely now? Consider other polyhedra dice or a mix of different dice. What about dice that have different sets of numbers on them, such as a six sided dice having 2,4,6,8,10,12 on the faces?</div> <h3>Possible support</h3> <div>Some children will need to spend longer convincing themselves about the possible answers to the addition sums and even to use apparatus to help them. Allow them the time to do so. If children work in pairs then this often helps them to articulate their thinking to one another, which will help them both. The child who has 'got it' will clarify their thinking as they explain. The one who hasn't should be encouraged to ask questions until they do understand. It is the repsonsibility of them both to ensure that they are secure in their understanding of what they are doing.</div> <div style="text-align: center;"> </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Perhaps start with small numbers like the tetrahedral dice. <div style="text-align: center;"><mdo:image alt="TetraDice" height="143" src="TetraDice%20Pic.jpg" width="248"></mdo:image></div> Here we read the number that's on three faces that touch the table - in this case $1$ and $2$. If you do not have any then they can be made from this <a href="/content/01/10/penta4/Tetrahedral%20Dice.pdf">net</a> or bought <a href="http://mathartfun.com/shopsite_sc/store/html/PolyDice.html">here</a>. <div style="text-align: center;"> </div> <div style="text-align: left;">Then you can move onto the $6$ sided and the $10$ sided.</div> <div style="text-align: left;"> </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We have had some lovely solutions for this problem. Some of you made long lists like Marion from Tattingstone: <table border="1"> <tbody> <tr> <td> </td> <td>A</td> <td>B</td> <td> </td> </tr> <tr> <td>2</td> <td>1</td> <td>1</td> <td>1 possibility</td> </tr> <tr> <td>3</td> <td>1 2</td> <td>2 1</td> <td>2 possibilities</td> </tr> <tr> <td>4</td> <td>1 2 3</td> <td>3 2 1</td> <td>3 possibilities</td> </tr> <tr> <td>5</td> <td>1 2 3 4</td> <td>4 3 2 1</td> <td>4 possibilities</td> </tr> <tr> <td>6</td> <td>1 2 3 4 5</td> <td>5 4 3 2 1</td> <td>5 possibilities</td> </tr> <tr> <td>7</td> <td>1 2 3 4 5 6</td> <td>6 5 4 3 2 1</td> <td>6 possibilities</td> </tr> <tr> <td>8</td> <td>2 3 4 5 6</td> <td>6 5 4 3 2</td> <td>5 possibilities</td> </tr> <tr> <td>9</td> <td>3 4 5 6</td> <td>6 5 4 3</td> <td>4 possibilities</td> </tr> <tr> <td>10</td> <td>4 5 6</td> <td>6 5 4</td> <td>3 possibilities</td> </tr> <tr> <td>11</td> <td>5 6</td> <td>6 5</td> <td>2 possibilities</td> </tr> <tr> <td>12</td> <td>6</td> <td>6</td> <td>1 possibility</td> </tr> </tbody> </table> <mdo:image alt="Marion's solution." src="marion.gif"></mdo:image> Fiona from Tattingstone drew a table: <table border="1"> <tbody> <tr> <td> </td> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> </tr> <tr> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td> </tr> <tr> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td> <td>8</td> </tr> <tr> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td> <td>8</td> <td>9</td> </tr> <tr> <td>4</td> <td>5</td> <td>6</td> <td>7</td> <td>8</td> <td>9</td> <td>10</td> </tr> <tr> <td>5</td> <td>6</td> <td>7</td> <td>8</td> <td>9</td> <td>10</td> <td>11</td> </tr> <tr> <td>6</td> <td>7</td> <td>8</td> <td>9</td> <td>10</td> <td>11</td> <td>12</td> </tr> </tbody> </table> Therefore; <mdo:image alt="Fiona's solution." src="fiona2.gif"></mdo:image> Here is another solution with some pictures from Martha at Tattingstone: <div><mdo:image alt="Martha's solution." src="martha.gif"></mdo:image></div> <div class="editorial">Emily from Stoke-by-Nayland Middle School drew a table like Fiona's to prove that 11 is the correct solution:</div> <mdo:image alt="" height="200" src="Dice%20problem%20emily.gif" width="400"></mdo:image> Thank you for sending in your solutions and well done! </mdoxml> 2 5 0 1 0 0 0 A Bit of a Dicey problem When you throw two regular, six-faced dice you have more chance of getting one particular result than any other. What result would that be? Why is this? Probability Theoretical probability Mathematics Tools Dice Using, Applying and Reasoning about Mathematics Working systematically Admin Upper primary mapping document