Who Has to Do the Dishes? [Part 1]
The Dent twins have a dilemma....they live together but neither likes to wash the dishes. So, to make their mathematics teacher proud, they came up with a mathematical way to determine who would wash the dishes and who would
dry.
Their decision process: Two red spoons and one yellow spoon were placed in a bag. Polly would then draw two spoons randomly out of the bag. If the two spoons pulled out were the same color, then Polly would wash and Pepso would dry. But, if the two spoons were different colors, then Pepso would wash and Polly would dry.
Question 1: Why is this process unfair? Who will end up washing the dishes more often (and what percent of the time)?
Seeing that the process was not fair, they decided to add one more yellow spoon to the bag, so that the number of each color was equal. Then, they repeated the experiment under the same decisionrules.
Question 2: Why is this process still unfair? Who will end up washing the dishes more often (and what percent of the time)? What is the big surprise?
In frustration, Polly and Pepso reverted to the original bag of two red spoons and one yellow spoon. Then, they tried two different options...adding either another red spoon or adding a blue spoon.
Question 3: Are either of these new options fair? Explain....and what is the surprise now?
Hint: For each question, first perform an experiment simulating the situation (about 20 times). Then, with that data as helpful background, try to use probability notions (e.g. tree diagrams, combinatorics, sample sets) to explore the mathematical side of things.
Solution Commentary: Relative to Question 1: The key is to treat the two red spoons as being unique. Then, the three possible outcomes are R_{1}Y, R_{2}Y, or R_{1}R_{2} [as _{3}C_{2} = 3!/(2!1!) = 3]. Thus, Prob(same color) = 1/3 and Prob(different color) = 2/3, making Pepso unhappy!
Relative to Question 2: The key is to treat both the two red spoons and the two yellow spoons as being unique. Then, the six possible outcomes are R_{1}Y_{1}, R_{1}Y_{2}, R_{2}Y_{1}, R_{1}Y_{2},
R_{1}R_{2}, or
Y_{1}Y_{2} [as _{4}C_{2} = 4!/(2!2!) = 6]. Thus, Prob(same color) = 2/6 = 1/3 and Prob(different color) = 4/6 = 2/3, making Pepso still unhappy! The surprise is that even after adding an additional yellow spoon, the probabilities did not change.
Relative to Question 3: The exploring and resolution to this question is left to you....one of the two options is fair, but which one? You need to know before doing next week's problem.
