Odds and Evens made fair
Odds and Evens Made Fair printable sheet
In the problem Odds and Evens, 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.
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?
You might like to use the Odds and Evens Interactivity to show the experimental probabilities for different sets of up to nine numbers. You can click on the purple cog to change the sets of numbers - just list the numbers you want to use, separated by a space, as in the screenshot
below:
This problem featured in an NRICH video in June 2020.
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?
...
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.
E | E | O | O | |
E | - | E | O | O |
E | E | - | O | O |
O | O | O | - | E |
O | O | O | E | - |
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.
Why do this problem?
This problem 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.
Possible approach
This problem featured in an NRICH video in June 2020.
This problem follows on from the problem Odds and Evens.
"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 without experience working algebraically with tree diagrams.
![Odds and Evens made fair Odds and Evens made fair](/sites/default/files/styles/large/public/thumbnails/content-id-9538-odds4.jpg?itok=yS6EZU17)
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 Teachers' notes to Odds and Evens. 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. This worksheet outlines one possible proof.
Click below for an algebraic approach using tree diagrams.
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. This worksheet shows one possible proof method.
You might like to use the Odds and Evens Interactivity to show the experimental probabilities for different sets of up to nine numbers. You can click on the purple cog to change the sets of numbers - just list the numbers you want to use, separated by a space, as in the screenshot
below:
Key questions
How can you decide if a game is fair?
Possible support
Take plenty of time to work on the original Odds and Evens problem before trying this extension task.
Possible extension
The problem In a Box 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.