7346
/www/nrich/html/content/id/7346/
1
2011-02-01T00:00:01
<mdoxml version="1.0"><br></br>
<span style="font-style: italic;">This problem follows on
from</span> <a style="font-style: italic;" href="http://nrich.maths.org/7284&part=">Troublesome
Dice</a><span style="font-style: italic;">.</span><br></br>
<br></br>
Jenny and her family often roll a die to decide who gets to eat the
last slice of pie, but when they roll it, the die often seems to
land "edge-up" or "corner-up" on a napkin, so they roll
again. <br></br>
<br></br>
One day, Jenny suggests that instead of rolling again, when the die
lands corner-up or edge-up, they simply add the visible faces
together: <br></br>
<div style="text-align: center;"> </div>
<div style="text-align: center;"><mdo:image width="200" height="164" src="edge-up.jpg" alt="edge-up die"></mdo:image><mdo:image width="184" height="164" src="corner-up.jpg" alt="corner-up die"></mdo:image></div>
<br></br>
What assumptions would you need to make to come up with a fair way
of determining who gets the last slice of pie?<br></br>
<br></br>
<span style="font-weight: bold;">Here are some questions you might
like to consider:</span><br></br>
<br></br>
Suppose the die lands face-up with probability $\frac {1}{2}$,
edge-up with probability $\frac{1}{3}$ and corner-up with
probability $\frac{1}{6}$. How could you allocate totals to three
people to give each of them an equal chance of
getting the last slice of pie?<br></br>
<br></br>
Suppose the die lands face-up with probability $p$, edge-up with
probability $q$, and corner-up with probability $r = 1-(p+q)$. Are
there any family sizes for which you can always allocate totals
fairly, regardless of the values of $p$, $q$ and $r$?<br></br>
<br></br>
The relative frequency of face-up, edge-up and corner-up
scores for the standard 6-sided die are distributed symmetrically.
Can you explain why?<br></br>
<br></br>
<br></br>
<span style="font-style: italic;">This problem is based on an idea
shared by Paul Stephenson.</span><br></br></mdoxml>
<mdoxml version="1.0"><p><span class="editorial">Daniel from Savile Park School sent us a solution </span><span class="editorial">to the first part of the problem.</span></p>
<p>Here's a list of all the possible scores and how many possibilities they are for each score when an <strong>edge</strong> faces upwards:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 250px;">
<tbody>
<tr>
<td style="text-align: center;"><strong>Score</strong></td>
<td style="text-align: center;"><strong>Combination</strong></td>
</tr>
<tr>
<td style="text-align: center;">3</td>
<td style="text-align: center;">(1,2)</td>
</tr>
<tr>
<td style="text-align: center;">4</td>
<td style="text-align: center;">(1,3)</td>
</tr>
<tr>
<td style="text-align: center;">5</td>
<td style="text-align: center;">(1,4), (2,3)</td>
</tr>
<tr>
<td style="text-align: center;">6</td>
<td style="text-align: center;">(1,5), (2,4)</td>
</tr>
<tr>
<td style="text-align: center;">8</td>
<td style="text-align: center;">(2,6), (3,5)</td>
</tr>
<tr>
<td style="text-align: center;">9</td>
<td style="text-align: center;">(3,6), (4,5)</td>
</tr>
<tr>
<td style="text-align: center;">10</td>
<td style="text-align: center;">(4,6)</td>
</tr>
<tr>
<td style="text-align: center;">11</td>
<td style="text-align: center;">(5,6)</td>
</tr>
</tbody>
</table>
<p>and another table for the possible scores and combinations for ending up a <strong>corner</strong> facing upwards:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 250px;">
<tbody>
<tr>
<td style="text-align: center;"><strong>Score</strong></td>
<td style="text-align: center;"><strong>Combination</strong></td>
</tr>
<tr>
<td style="text-align: center;">6</td>
<td style="text-align: center;">(1,2,3)</td>
</tr>
<tr>
<td style="text-align: center;">7</td>
<td style="text-align: center;">(1,2,4)</td>
</tr>
<tr>
<td style="text-align: center;">9</td>
<td style="text-align: center;">(1,3,5)</td>
</tr>
<tr>
<td style="text-align: center;">10</td>
<td style="text-align: center;">(1,4,5)</td>
</tr>
<tr>
<td style="text-align: center;">11</td>
<td style="text-align: center;">(2,3,6)</td>
</tr>
<tr>
<td style="text-align: center;">12</td>
<td style="text-align: center;">(2,4,6)</td>
</tr>
<tr>
<td style="text-align: center;">14</td>
<td style="text-align: center;">(3,5,6)</td>
</tr>
<tr>
<td style="text-align: center;">15</td>
<td style="text-align: center;">(4,5,6)</td>
</tr>
</tbody>
</table>
<p>In the first case where the probabilities of landing with a face, edge or corner facing upwards are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{6}$ respectively, the total probabilities of getting a number are shown in this table:</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 500px;">
<tbody>
<tr>
<td>Score</td>
<td>Probability</td>
<td>144$\times$ probability</td>
</tr>
<tr>
<td>1</td>
<td>1/12</td>
<td>12</td>
</tr>
<tr>
<td>2</td>
<td>1/12</td>
<td>12</td>
</tr>
<tr>
<td>3</td>
<td>1/9</td>
<td>16</td>
</tr>
<tr>
<td>4</td>
<td>1/9</td>
<td>16</td>
</tr>
<tr>
<td>5</td>
<td>5/36</td>
<td>20</td>
</tr>
<tr>
<td>6</td>
<td>23/144</td>
<td>23</td>
</tr>
<tr>
<td>7</td>
<td>1/48</td>
<td>3</td>
</tr>
<tr>
<td>8</td>
<td>1/18</td>
<td>8</td>
</tr>
<tr>
<td>9</td>
<td>11/144</td>
<td>11</td>
</tr>
<tr>
<td>10</td>
<td>7/144</td>
<td>7</td>
</tr>
<tr>
<td>11</td>
<td>7/144</td>
<td>7</td>
</tr>
<tr>
<td>12</td>
<td>1/48</td>
<td>3</td>
</tr>
<tr>
<td>14</td>
<td>1/48</td>
<td>3</td>
</tr>
<tr>
<td>15</td>
<td>1/48</td>
<td>3</td>
</tr>
</tbody>
</table>
<p>We need to arrange the outcomes into sets so that the sum of the probabilities in each set are equal. For example, for 3 people, a possible arrangement is:</p>
<p>Person 1: scores of 1,6,7,10,12</p>
<p>Person 2: scores of 2,3,5</p>
<p>Person 3: scores of 4,8,9,11,14,15</p>
<p>These 3 sets all sum to $\frac{1}{3}$ as required. There are other solutions similar to this example as you can freely exchange scores 10 and 11, scores 3 and 4, scores 1 and 2 and scores 12,14 and 15, without changing the sums of the probabilities. </p>
<p><span class="editorial">There are likely to be more solutions, have you found any others?</span></p>
<p><span class="editorial">Meg, from All Hallows school, wrote to us with a solution for the second part.</span></p>
<p>When the dice lands on a face, edge or corner with probability p, q and r, the total probabilities for getting each score are:<br></br>
</p>
<table border="1" cellpadding="1" cellspacing="1" style="width: 500px;">
<tbody>
<tr>
<td>Score</td>
<td>Probability</td>
<td>24$\times$ probability</td>
</tr>
<tr>
<td>1</td>
<td>p/6</td>
<td>4p</td>
</tr>
<tr>
<td>2</td>
<td>p/6</td>
<td>49</td>
</tr>
<tr>
<td>3</td>
<td>p/6 + q/12</td>
<td>4p+2q</td>
</tr>
<tr>
<td>4</td>
<td>p/6 + q/12</td>
<td>4p+2q</td>
</tr>
<tr>
<td>5</td>
<td>p/6 +2q/12</td>
<td>4p+4q</td>
</tr>
<tr>
<td>6</td>
<td>p/6+ 2q/12 + r/8</td>
<td>4p+4q+3r</td>
</tr>
<tr>
<td>7</td>
<td>r/8</td>
<td>3r</td>
</tr>
<tr>
<td>8</td>
<td>2q/12</td>
<td>4q</td>
</tr>
<tr>
<td>9</td>
<td>2q/12 + r/8</td>
<td>4q+3r</td>
</tr>
<tr>
<td>10</td>
<td>q/12 + r/8</td>
<td>2q+3r</td>
</tr>
<tr>
<td>11</td>
<td>q/12 + r/8</td>
<td>2q+3r</td>
</tr>
<tr>
<td>12</td>
<td>r/8</td>
<td>3r</td>
</tr>
<tr>
<td>14</td>
<td>r/8</td>
<td>3r</td>
</tr>
<tr>
<td>15</td>
<td>r/8</td>
<td>3r</td>
</tr>
</tbody>
</table>
<p>If there are n members of the family, we need to split the outcomes into n sets, such that the sums of the probabilities in each set are constant (equal to 1/n) for <strong>all</strong> p,q and r<strong>.</strong></p>
<p>This means we need to partition these numbers into sets so that the probabilities sum in each set to give the same quantity of p, the same quantity of q and the same quantity of r. These sums of probabilities will therefore be the same, whatever the values p, q and r take.</p>
<p>The sums will only involve integer multiples of 4p, 2q and 3r. We need to evenly split 6 multiples of 4p, 12 multiples of 2q and 8 multiples of 3r among n family members. What restrictions does this place on n? We need n to divide 6, 12 and 8, which implies n=2. </p>
<p>One possible arrangement is:</p>
<p>1) 1,3,5,8,10,12,14</p>
<p>2) 2,4,6,7,9,11,15</p>
<p>These sets both add up to 1/2. Again, there's degeneracy due to being able to interchange scores with the same probability. </p>
<p><strong>Symmetry of probabilities:</strong></p>
<p>The probabilities for the scores are symmetric because a cube, and therefore the set of possible scores, is unchanged if you interchange 1 and 6, 2 and 5, 3 and 4.</p>
<p> </p></mdoxml>
<mdoxml version="1.0"><br></br>
<h3>Why do this problem?</h3>
<div>This problem explores the familiar context of probability with
dice, in an unfamiliar way.</div>
<h3>Possible approach</h3>
<div>This problem is set in the same context as the Stage 3
problem <a href="http://nrich.maths.org/7284&part=">Troublesome
Dice</a>. </div>
<div> </div>
<div>Make sure students understand the standard way of labelling
the faces of a six-sided die. They will need to calculate all the
possible edge and corner totals in order to calculate the relative
frequencies. There are different areas for exploration suggested in
the problem - one possibility is to assign particular values to the
probability of landing face, edge, and corner up. Values of
$\frac{1}{2}$,$\frac{1}{3}$ and $\frac{1}{6}$ are suggested -
students could compute the overall probabilities for each total.
Alternatively, the general case with probabilities $p$, $q$ and $r
= 1 - (p + q)$ could be explored - this would work well in small
groups with students sharing out the work.</div>
<div> </div>
<div>Once students have computed the probability of each total
(either for specific values or in terms of $p$ and $q$), challenge
them to find fair ways of allocating the numbers for different
sizes of family. They could work in small groups and present their
solutions to the rest of the class.</div>
<br></br>
<h3>Key questions</h3>
<div>I can get a 6 as a face-up, edge-up or corner-up score. Which
scores would you want to be allocated to balance this out?</div>
<div>A 6-sided die can be labelled in a "left-handed" or
"right-handed" manner - do the two dice give different
answers?</div>
<div>The probabilities are distributed symmetrically for the face,
edge and corner sums. Is this surprising? Can you explain
why?</div>
<div> </div>
<h3>Possible extension</h3>
<div>Consideration of the probabilities for dodecahedral or
icosahedral dice provide a challenging extension.</div>
<h3>Possible support</h3>
<div>Spend some time working on <a href="http://nrich.maths.org/7284&part=">Troublesome Dice</a>
to make sure students are happy with the relative probabilities of
the edge-up and corner-up scores.</div>
<br></br>
<br></br></mdoxml>
<mdoxml version="1.0">It might help to look at a die (or make one) to refer to when you
are working out the edge and corner totals.<br></br></mdoxml>
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Dicey Decisions
Can you devise a fair scoring system when dice land edge-up or
corner-up?
Probability
Probability
Probability
Combining probabilities
Transformations and their Properties
Symmetry