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In the first example Voting Paradox there are 3 candidates for election. The voters have to rank them in order of preference. Consider the case where 3 voters cast the following votes: ABC, BCA and CAB:
A beats B by 2 choices to 1
B beats C by 2 choices to 1
but A cannot be the preferred candidate because A loses to C, again by 2 choices to 1.
This is an example of intransitivity. You can imagine it happening on an interview panel with three interviewers and three candidates. Can you find the probability of an outcome like this with no outright winner where there are 3 candidates A, B and C, and 3 voters, who place the candidates in order of preference? This was a problem from August 1998 and its solution is here.
The second example 'Winning Team'; involves nine runners belonging to three clubs. Runners A, F and H belong to the Cyber Club, runners B, D and I belong to the Champs Club and runners C, E and G belong to the Kings Club. Whenever they run in the same race A always beats B, and B always beats C, and C always beats D etc. so that they always finish in the order A, B, C, D, E, F, G, H, I.
The tournament consists of three races with 6 runners in each race and the winner scores 6 points, the second 5 points, the third 4 points, the fourth 3 points, the fifth 2 points and the last one 1 point, a total of 21 points between the two teams so there is always a winning team. You can easily calculate the points when the Cybers run against the Champs, the Cybers win with 11 points to the Champs 10 points. Now find the results when the Champs run against the Kings. Which is the best team? What happens when the Cybers run against the Kings? This problem, called Winning Team , appeared in October 1998 with subsequent discussion of the solutions.
The third example `Dicey' involves probabilities rather than events which always happen. Dicey is a game for two players, each throws their own die and the highest score wins. Four fair dice (Efron's dice) are marked on their six faces, using the mathematical constants $\pi$, $e$ and $\phi$ (where $\phi$ is the divine proportion or golden ratio) as follows:
|B:||$\pi \;$||$\pi \;$||$\pi \;$||$\pi$||$\pi$||$\pi$||where $\pi$ is approximately 3.142|
|C:||$e$||$e$||$e$||$e$||7||7||where $e$ is approximately 2.718|
|D:||5||5||5||$\phi \;$||$\phi \;$||$\phi \;$||where $\phi$ is approximately 1.618|
You play the game with your friend Jo and you invite Jo to choose first, and to choose ANY one of the dice. Then you can always choose another die so that you will have a better chance of winning than Jo. Now Jo may think this is unfair and want to play with the die you chose thinking it is the best one. In that case you can always chose another die so that you still have a better chance of winning than Jo. Consider all the cases and decide what choice you will make in each case. Is one of the dice the best one?
Does it make any difference if the dice are marked with 3 instead of $\pi$, 2 instead of $e$ and 1 instead of $\phi$?
Dicey features in one of the January 1999 problems.
For more on this subject you might like to read the chapter 'Nontransitive dice and other probability paradoxes' in Martin Gardner's book 'Wheels, Life and other Mathematical Amusements' published by W.H.Freeman in 1983.