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# Odds and Evens

##### Age 11 to 14 Challenge Level:

Jamie from Wilson's School used the interactivity to see which gave the fairest chance of winning:
Set C seems to be the fairest, because out of 100 tries, 50 won and 50 lost.

Alvin, also from Wilson's school, showed one way of calculating if a game is fair:
There are many possible ways to decide whether the game is fair or not. When you look at the set A, you can clearly see 3 odd balls and 2 evens.
Below is a simple way to find if the game is fair:
2+3=5
2+4=6
2+5=7
2+6=8
3+4=7
3+6=9
4+5=9
4+6=10
5+6=11
We need to count the total of odds and the total of the evens. There are 6 odd totals and 4 evens. This can be expressed as a ratio 6:4 which can be cancelled to 3:2 It can be concluded that the game is unfair.
You can also use a probability table to represent it.

Jacob (Wilson's School) realised that he didn't need to work out the answers to each addition, but just whether the answer was odd or even, which he represented with O and E.

James (from Wilson's too!) also used a listing method to get the probability for set A as 0.4, and then compared the theoretical probability with the experimental probability:
When played 100 times the win/lose ratio is 0.42. It evens it self out after more games:
200 times=0.385
300 times=0.407
400 times=0.385
500 times=0.388

Fred, Johannes, and Lok, all from St Barnabas School listed all the ways of making odd and even numbers for the four sets.

Charlie and Shaun (both from Wilson's School) worked out the probabilities for all four sets. You can see Shaun's solution here.

Akeel (Wilson's School) explained what he would do to maximise the chance of winning:
To decide whether the game is fair you can find all the possible results and from that you can find out if the amount of times you can win is the same as the amount of times you can lose - if it is, the game is fair and if it isn't, the game isn't fair.

In the case of set A, the game is not fair - the probability of you winning is $\frac{8}{20}$, which can be simplified to $\frac{4}{10}$ or 40%.
The probability of winning with B is $\frac{12}{20}$ and the probability of winning with D is $\frac{20}{30}$. The probability of winning with set C is $\frac{14}{30}$.
To maximise my chance of winning I would play with either set B or set D.

Paul (Wilson's School) also identified that Set D gave the best chance of winning, and Tim (Wilson's School) suggested WHY set D gave the best chance of winning:
I would select set D to play with. This is because all of the numbers apart from the four are odd. This maximises your chance of winning because when you add two odd numbers together, you get an even and you are going to be adding more odd and odd numbers together than odd and even. I worked out that by playing set D, you have a 2 in 3 chance of winning.

Hannah from Munich International School pointed out the following:
If you had only odd numbers, all the combinations of numbers would be even, meaning you would win every time! And if you only had even numbers, all the combinations of numbers would be even, meaning, again, you would win all the time!

Tom and Hussein (Wilson's School) found a set that produced a fair game:
To produce a fair set of numbers from what we found out you would need 4 balls, one odd and three even, or the other way round, one even and three odd.

Elliot (Wilson's School) also found a fair set:
A fair set can happen. If you have:
1, 3, 5, 2
2, 4, 6, 1
These are fair, as three results are odd, and three even. Any other groups with three of odd or even, then the one of even or odd are also fair.

Philip, Tahmid and Jacques (Wilson's School) worked out the probabilities but also noticed something interesting along the way:
We've also noticed that the total number of possibilities for each set of balls can be worked out by this formula, in which $n$ = number of balls and $t$ = total possibilities:
$t = n(n-1)$

Krystof from Uhelny Trh in Prague sent us this solution.
Finally, Chi from Raynes Park High School sent us this solution.