Odd or Even...Fair or Unfair?
To determine who starts on offense in an ultimate Frisbee game, two Frisbee discs are flipped and one team's captain calls "odd" or "even." If both discs land face up or face down, the discs are "even." If one disc lands face up and the other face down, the discs are "odd."
Tommy, though unable to find a way to support his claim, insists that this process is fair and ensures a probability of 0.5 for both "odd" and "even." He does not know the probability of a disc landing face up or face down.
Question 1: Is Tommy's reasoning correct or incorrect? Justify.
Question 2: What happens if you use more than two discs (e.g. four)?
Source: TL (a WWU student)
Hint: Do an experiment, simulating the situation. Can you build a tree diagram?
Solution Commentary: Adapted from student TL: The Prob(single disc landing face up) = x where x is a number between 0 and 1. Similarily, Prob(single disc landing face down) = 1x.
The Prob(even) = Prob(both discs face up) + Prob(both discs face down) = (x)(x) + (1x)(1x) = 1  2x + 2x^{2}. Then, Prob(odd) = 1  Prob(even) = 1  (1  2x + 2x^{2}) = 2x  2x^{2}.
Finally, the process is fair if Prob(even) = Prob(odd). But 1  2x + 2x^{2} = 2x  2x^{2} implies x = 0.5. Thus, the process is fair only if the discs have an equal probability of landing face up or down (which is unlikely).
As a final note, you might graph both Prob(even) and Prob(odd) as an interesting confirmation of what you might expect to happen.
This same process can be used for the case of 4, 6, etc. discs. The algebra gets more complicated....and making graphs of both Prob(even) and Prob(odd) is suggested. Also, a little surprise lies waiting for you regarding the results....
