Two presents are to be given to two children at a birthday party.
The children are told to sit in a circle and a die (singular of dice, but you knew that!) is rolled, giving a number between one and six. Counting Eddy as number one, and going clockwise, the number on the die is used to determine the recipient of the first present.
That child stands up and is no longer in the circle, and rolls the die a second time. Starting with the child on the left of the winner, and again going clockwise, the number determines the second child to get a present.
As the children are originally seated, Vicky has the best chance of a present, and Pete the second-best chance. Rachel has the same chance as Eddy, and Paul's chances are the worst. However Kitty has four times as much chance as another child.
How are the named children seated?
The probability of the first six children winning the first prize is 1/6.
If child one wins the first prize then children two to seven have a 1/6 chance of winning the second prize.
If child two wins the first prize then children three to eight have a 1/6 chance of winning the second prize, and so on.
If there are twelve or more children then the chance of children two to twelve are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 divided by 36.
And the chances of each child winning either first or second prize (probability of winning the first added to the probability of winning the second) is 6, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 1 divided by 36.
From this we can see that two children have four times as much chance of winning as another child. So there must be eleven children or fewer.
With eleven children the probability of child twelve winning is added to the probability of child one to give child one's chances. So for child one to child eleven the chances are 7, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2. This fits the data, and with fewer children it does not.
So the eleven children going clockwise from Eddy are 1) Eddy, 2) Rachel, 3) Kitty, 4) not named, 5) Pete, 6) Vicky . . . 11) Paul.