Odds and Evens made fair

9538 /www/nrich/html/content/id/9538/ 1 2012-10-11T16:42:42 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">In the problem <a href="/4308">Odds and Evens</a>, we introduced the following game and invited you to work out whether the game was fair: Here is a set of numbered balls used for a game. <mdo:image alt="Set of balls: 2, 3, 4, 5, 6" src="odds1.png" style="width: 130px; height: 89px; float: right;"></mdo:image> To play the game, the balls are mixed up and two balls are randomly picked out together. The numbers on the balls are added together. If the total is even, you win. If the total is odd, you lose. Can you find a set of balls where the chance of getting an even total is the same as the chance of getting an odd total? How many sets of balls with this property can you find? What do you notice about the number of odd and even balls in your sets? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This problem follows on from an earlier one. Make sure you have a think about <a href="http://nrich.maths.org/4308">Odds and Evens</a> first! Iris and many others realised that the game given in this problem is not fair. There are more ways of making an odd total than an even one. Finding a game which is fair is a harder challenge! A good thing to notice is that the actual numbers on the balls don't matter - just whether they are even or odd. This is sometimes called a number's parity. Charlotte, Lokhin, Matthew, Jake and Hashim from The Cherwell School sent in a fair game they found by trial and improvement: To start with we tried two even and two odd balls because this is the most logical, but this made you more likely to lose. The table shows ossible outcomes here where the parity of the first ball is along the top, and the second is down the left side. The diagonal cells don't count because you can't choose the same ball twice. <table style="border-spacing: 1px; width: 200px; height: 200px;" border="1"> <tbody> <tr> <td> </td> <td>E</td> <td>E</td> <td>O</td> <td>O</td> </tr> <tr> <td>E</td> <td>-</td> <td>E</td> <td>O</td> <td>O</td> </tr> <tr> <td>E</td> <td>E</td> <td>-</td> <td>O</td> <td>O</td> </tr> <tr> <td>O</td> <td>O</td> <td>O</td> <td>-</td> <td>E</td> </tr> <tr> <td>O</td> <td>O</td> <td>O</td> <td>E</td> <td>-</td> </tr> </tbody> </table> You can see that there are 8 odd outcomes and only 4 even ones - and each outcome is equally likely. This game is unfair, so we tried swapping one of the evens with an odd and the result was a fair game. Great work! Try to construct the table of outcomes for this new game with 1 even and 3 odd balls. Check for yourself that the game is fair. Michael, Benjie, Oliver, Conall gave very similar algebraic solutions. They used different ways of turning the problem into a set of equations to solve, and then found the answers. Fantastic work, see if you can follow the method: Write $a$ for the number of even balls, and $b$ for the number of odd ones. We need to count ways of finding even and odd totals. It's important to be able to count ways of choosing pairs of things. If you choose 2 things from $n$ objects there are $n(n-1)/2$ ways to do this because there are $n$ ways of choosing the first, then $(n-1)$ ways of choosing the second; but then we count each pair twice; choosing it in either order. Even, Even: This makes an even total. There are $a(a-1)/2$ ways of choosing two evens. Odd, Odd: This makes an even total. There are $b(b-1)/2$ ways of choosing two odds. Even, Odd or Odd, Even: This makes an odd total. There are exactly $ab$ ways of choosing one odd and one even number. Sometimes counting pairs can be tricky, make sure you understand why these numbers work. So there are $\frac{a(a-1) + b(b-1)}{2}$ ways of choosing an even total and $ab$ ways of choosing an odd total. Since each individual outcome is equally likely to occur we need these numbers to be equal. This is where we find an equation: $$\frac{a(a-1) + b(b-1)}{2} = ab$$ There are a few good ways of solving this. One extra problem is that $a$ and $b$ are integers. Rearrange the equation to give \begin{align} a(a-1) + b(b-1) &= 2ab\\ a^2-a+b^2-b-2ab &= 0\\ (a-b)^2 -(a+b) &= 0\\ (a-b)^2 &= a+b \end{align} Which we can solve using our knowledge that since $a$ and $b$ are integers; so are $a-b$ and $a+b$. but then $a+b$ is a square number! Let's write $a+b=n^2$. Then we also have $$a-b=\pm n$$ But since $a$ and $b$ play the same role in the equation solutions are valid if we swap $a$ and $b$. Let's just take $a-b=n$ then. The other sign is dealt with by swapping. Now we have $a-b= n$ and $a+b=n^2$. So solve for $a$ and $b$: \begin{align} a &=\frac{n^2+n}{2}\\ b &=\frac{n^2-n}{2} \end{align} Wonderful. The same solutions can be found if you use the quadratic formula on the big equation for $a$ and $b$ too. This is equally good! Now put all this together: So for the game to be fair; we need the number of odd and even balls to have the formulas above for any positive integer $n$. This also means the total number of balls is a square number. For example if $n=2$ we have the solution found by trial and improvement above, with odd and even swapped.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=9538">This problem</a> offers a meaningful context in which algebraic fractions and tree diagrams can help explain a surprising result in a probability problem. Collaborative working makes it possible for students to tackle an otherwise unmanageable task.</div> <h3>Possible approach</h3> This problem follows on from the problem <a href="/4308">Odds and Evens</a>. "Imagine you had a bag containing a set of balls with whole numbers on them. You choose two balls, and find the total. Can you find a set of balls where the chance of getting an odd total is equal to the chance of getting an even total?" Click below for a numerical approach suitable for students with little experience working algebraically with tree diagrams. <div class="toggle">"How could we go about finding out whether there are any sets that would give a fair game?" One way of organising the search is to draw up a table on the board showing different combinations of odds and evens: <mdo:image alt="board divided into different numbers of odds and evens" height="400" src="odds4.jpg" width="500"></mdo:image> Divide the class into groups working on different combinations and ask them to report back. Students could use sample space diagrams or tree diagrams to work out the probability for each case. Then they could record combinations that have been checked on the board with a tick or a cross to show whether they are fair or not. There will be opportunities while the class are working to stop everyone and share students' insights that will make the job easier. For example: "None of the combinations with zero will work because..." "If 3 odds and 2 evens won't work, 2 odds and 3 evens won't either, because..." "You can't have the same number of evens and odds because..." Eventually, there will be a sea of crosses on the board and just a few combinations that work (four, if the class have gone up to 9 balls in total). Ask the class to stop and consider what the fair sets have in common. This may lead to some new conjectures about the total number of balls, so organise the class to test the conjectures. Once there is some confirmation about the total number of balls needed for fair games, conjectures can also be made about how these should be split into odds and evens. Students can test examples with large numbers, using the simplified sample space method detailed in the <a href="/4308/note">Teachers' notes to Odds and Evens</a>. Draw attention to how valuable it is to work collaboratively as part of a mathematical community, and how difficult it would have been to have reached the same insights working alone. Finally, invite students to prove their conjectures algebraically. <a class="pdflink" href="/content/id/9538/Odds%20and%20Evens.pdf">This worksheet</a> outlines one possible proof.</div> Click below for an algebraic approach using tree diagrams. <div class="toggle">"To explore this, imagine there are p balls with odd numbers and q balls with even numbers. Can you work out the probability of an even total or an odd total in terms of p and q?" Give the class time to work on this. Draw attention to clear methods of representation such as tree diagrams or sample space diagrams. Eventually, students should find that p and q satisfy the equation $(p-q)^2 = p+q$ or equivalent. "Can we find some values of p and q that satisfy the conditions?" At this point, encourage students to work systematically, and depending on their form for the equation they could set square-number values for $p-q$ and solve the resulting simultaneous equations, or choose a value for p and see whether there is a corresponding integer value for q. Invite students to write up on the board any pairs of numbers p and q they find that work. "What do you notice about p and q? Can anyone explain why?" Finally, a little algebra can show why p and q belong to the set of triangle numbers. <a class="pdflink" href="/content/id/9538/Odds%20and%20Evens.pdf">This worksheet</a> shows one possible proof method.</div> <h3> Key questions</h3> <div>How can you decide if a game is fair?</div> <div>How can we make this difficult task (of finding a fair game) more manageable?</div> <h3>Possible extension</h3> The problem <a href="http://nrich.maths.org/919&part=" style="font-weight: 400;">In a Box</a> offers another context for exploring exactly the same underlying mathematical structure, and could be used as a follow-up problem a few weeks after working on this one. <h3>Possible support</h3> <div>Take plenty of time to work on the original <a href="/4308">Odds and Evens</a> problem before trying this extension task.</div> <div> </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> If I had 1 odd and 1 even ball would it be a fair game? If I had 2 odds and 1 even ball would it be a fair game? If I had 1 odd and 2 even balls would it be a fair game? If I had 1 odd and 3 even balls would it be a fair game? ... </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Congratulations to Sam, James and Esther who sent in correct solutions to this problem. Several others of you were really close, but made slight errors. Here is Esther's solution: Set D would give the greatest proportion of even numbers. The possibilities for this set are: (1,3) (1,4) (1,5) (1,7) (1,9) (3,4) (3,5) (3,7) (3,9) (4,5) (4,7) (4,9) (5,7) (5,9) (7,9) of which 10 out of 15 are even, a probability of 2/3 . When you work out A in this way, the probability is 2/5, B is 3/5 and C is 7/15. Esther correctly spotted that if you could select any numbers with at least one odd, the best numbers would be if they were all odd - two odds always add to make an even . If assume that picking 3 and 5 is same as picking 5 and 3, then: A - 4 out of possible 10 are even: 3+5, 2+6, 2+4, 6+4 B - 6 out of possible 10 are even: 3+1, 3+5, 3+7, 1+5, 1+7, 5+7 C - 7 out of possible 15 are even: 8+2, 8+6, 8+4, 3+5, 2+6, 2+4, 4+6 D - 10 out of possible 15 are even: 7+9, 7+3, 7+1, 7+5, 3+9, 3+1, 3+5, 9+1, 9+5, 5+1 If all five balls or all six balls were odd, every combination would give an even total. </mdoxml> 2 3 0 0 0 1 1 Odds and Evens made fair In this follow-up to the problem Odds and Evens, we invite you to analyse a probability situation in order to find the general solution for a fair game. Probability Experimental probability ajk44 solution needs editing