Tomorrow is my Probability Final. My Lie Algebra and Commutative Algebra are long behind me. All that is left is Probability. So for the last week I have had my nose in a Probability book nearly all day every day. It's going to be tough.
So as my Tuesday evening studying is winding down I notice the label one of my empty Peanut Butter Jars. It has a great big "#1" on it with the words "Choice of Choosey moms". Now here is how the fried brain of some one who has been staring at theoretical probability thinks about that statement:
Hmm what percentage of the set of all moms is the subset of moms who are also choosy? What is the probability that if I select a mom at random that mom will satisfy the conditions of being choosy? Come to think of it what DOES make a mom a choosy mom? Do there exist moms in the world that would not define them selves choosy? Or is it determined by the kids? For an example there could exist a mom who considers herself rather choosy, but if you were to ask her kids they would say that mom is not very choosy at all. If this was the case would you define her to be choosy, or would she fall into the category of non-choosy moms? Apparently those are the types of moms that Peanut butter company's frown upon.
All right so lets suppose we have figured out the "choosy" part of this sticky situation. Can we in fact precisely determine what the number one peanut butter in the opinion of these so called choosy moms? I'm guessing there might be quite a few of them. Are you polling ALL of them or only a select few? And even if you were to accurately poll the entire earth's population of choosy mom how would you weight the voting? Surely there exists at least one choosy mom who has another Peanut Butter as choice number one. And there clearly exist three or more brands of Peanut Butter in the world. So there may not even be a single peanut butter that holds a pure majority of choosy mom's votes. Would that Peanut Butter still be number one? And would it not be valid to some how also weight the second and third place votes. Like Jif might hold 34 percent of the choosy mom vote while no other Peanut butter holds more than about 25. But then suppose that all 34 percent would take Skippy as a second choice while the non-Jif voting choosy mom DESPISE Jif not even ranking it in the top five. While 90 percent of mom's might put Skippy in the top three. You see in a two Peanut butter system that would never be a problem. But clear our world is a three or more Peanut Butter system. Our choosy problem it kind of in a tailspin here.
There clearly is an easy solution here. Do what any good Mathematician would do, redefine the terms. So I will ignore your reality and substitute my own. Let us define the set of "Choosy" moms to be the set of all moms who are either executives of the Jif Peanut Butter Company or married to an executive at the Jif Peanut Butter Company. This is clearly a reasonably smaller group of moms to survey and I am fairly certain that this group of "Choosy" moms have a very high probability of choosing Jif.”
On that note I think I am heading to bed. I wonder what is the probability that there will be any Peanut Butter questions on my final tomorrow?
