Cherry’s Lottery Ball Mystery: Can You Solve It?
The Guardian’s Alex Bello published an interesting lottery logic puzzle about common knowledge. Give it a go to see if you can solve it!
Here is the lottery puzzle…
Nine balls – marked 1, 2, 3, 4, 5, 6, 7, 8 and 9 – are in a lottery machine. The machine dispenses one ball each to three people, Apple, Bean and Cherry. Each person knows only the number of their own ball; they do not know the balls that the others were given, nor the ones left in the machine.
Before the game begins, each of them shows their balls to a fourth person, Zog. Zog says: “On one of the balls is a number that is the sum of the numbers on the other two balls.”
At which point the following bantz begins:
- Apple: “There are 8 possibilities for Bean’s ball.”
- Bean: “There are 8 possibilities for Cherry’s ball.”
- Cherry: “There a are 4 possibilities for Apple’s ball.”
- Apple: “I know Bean’s ball!”
- Bean: “I know Cherry’s ball!”
What is Cherry’s ball?
We can assume that all participants have been truthful at all times. Can you figure it out?
IF YOU WANT TO FIGURE IT OUT FOR YOURSELVES, STOP READING! FOR THE SOLUTION, KEEP READING.
The Solution
The first challenge is to figure out why Zog’s statement is relevant.
If the sum of two of the balls is equal to the third, then we can deduce that it is impossible for both the 8 and the 4 to be among the three selected balls. This is because if the 8 and the 4 were selected, the third ball would have to also be 4, so 4 + 4 = 8. However, this is impossible since there is only one 4-ball.
So, if Apple had either the 8, she would know that Bean does not have the 4, and if she had the 4, she would know Bean does not have the 8. In either case, she would deduce that there are only seven possibilities for Bean’s ball, i.e it could be any ball exception the 8-ball and the 4-ball.
In the question, Apple says there are 8 possibilities for Bean’s ball. We know she does not have either the 4 or the 8.
For the same reasons, when Bean says there are 8 possibilities for Cherry’s ball, we can deduce he doesn’t have the 4 or the 8 either.
Now to Cherry. She knows that her two pals don’t have the 4 or the 8. If she has the 4, or the 8, she would state that there are seven possibilities for Apple’s ball, which are all the balls except the 4-ball and the 8-ball. But she states there are four possibilities, so we can rule out the chance she has the 4 or the 8.
So, no one has the 4 or the 8. This means the three pals must have balls from the following selection: 1,2,3,5,6,7 and 9.
From Zog’s statement we can now deduce that these combinations are also impossible: (1,9), (2,6), (5, 9) and (3,7). In each of these cases, it is impossible for there to be a third ball such that the number on a ball is equal to the sum of the numbers on the others.
Therefore, if Cherry has a 1, there are five possibilities for Apple (all those except 1, 4, 8 and 9). If Cherry has a 2, there are also five possibilities for Apple (all those except 2, 4,8 and 6), and so on. It is clear that only if Cherry has the 9-ball, will she know that there are 4 possibilities Apple can have, which are 2, 3, 6 and 7, (i.e all those except 1, 4, 5, 8 and 9).
So Whats Cherry’s Ball?
The question asked us to find Cherry’s ball – which is 9!
Apple knows Bean’s ball since its number is 9 minus the number on her ball. However, neither Cherry nor us puzzlers can deduce Bean’s ball.
