Controversial Facts In Mathematics

Here's the question as posed in Marilyn Vos Savant's column [http://marilynvossavant.com/game-show-problem/]: "Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?"

But what exactly is the procedure followed by the host? I can think of three:

A. If you pick a goat, then the host shows you the other goat. If you pick the car, then the host picks one of the goats at random to show it to you.

B. The host picks one of the other doors at random and shows it to you. So you might pick a goat, then he shows you the car, and gives you a chance to switch, but you already know that the remaining doors have goats, so there's no point.

C. If you pick the car, then the host shows you one of the goats picked at random. If you pick the goat, then the host says "You picked the goat!", and you don't get a chance to switch.

People assume that the procedure is A, but I don't see how the other procedures are excluded by the description.

I did take notice that a sub-set is likely to be smaller than the set it came from: the set of all accountants vs. the sub-set of all accountants who are also plumbers. So therefore P(accountant) > P(accountant and plumber). Unless P(accountant) be zero, or P(plumber|accountant) be one, in which case neither is greater.

Reflecting on my own human biases, I think I ended up reading it as if this were a multiple-choice question, and I had to pick the "best" choice. In other words, it isn't enough that this person is an accountant, but that this person is also a plumber. Or put another way, if I had to wager on whether anyone would pay just any old accountant to go mess with their water heater, I would find it more surprising that the particular accountant in fact hired had no demonstrable plumbing skills. I'd imagine Kahneman could teach me more on this.

My favorite word problem joke goes something like this. If there are three cans on a fence, and you hit one with a thrown rock, how many cans remain on the fence? Answer: Two. Now, if there are three birds on a fence, and you hit one with a thrown rock, how many birds remain on the fence? Answer: None - they'd all start flying off when there's a rock hurtling towards them.

I'm more inclined to believe this is a (poor) illustration of the idea that a sub-set is likely to be smaller than the set it came from: the set of all accountants vs. the sub-set of all accountants who are also plumbers. In which case - duh! - P(accountant) > P(accountant and plumber). As long as one can indulge in the naivety of this pedagogical tale, that the narrator (the "I" who said "I went to a person") wasn't too particular on whom he hired for the job, then sure, why not?, 'twas likely the accountant that fixed it.

In the real world, which has teachers and mechanics in addition to accountants and plumbers and other occupations, and where people may tend to be pickier on who they'll let near their water heater - leave alone pay someone to mess with it - would this conclusion of "accountant" still prevail necessarily?

Here's one way of presenting this formally. Define a 'decimal' to be an ordered sequence of integers (called 'digits') a_1, a_2, a_3, and so on. (By 'and so on', I formally mean that for each positive integer k we have a digit a_k at the kth position in the sequence). Let's say each a_k has to be between 0 and 9 inclusive.

For each positive integer k, define the 'kth partial sum' of the sequence to be the sum from j=1 to j=k of (1/10^j) x a_j.

I'll skip over what it means if we say that the partial sums converge as k->Infinity, because it sounds like you understand what limits are and how they work. If not, I'd be happy to explain.

Now, if the partial sums converge to some value 'd', we say that the decimal has value equal to d. It can be shown that any decimal has at most one such d (which is good, because a decimal shouldn't have two values).

Now, I think you'd be happy to say that when we write 0.9999999..., what we mean is the decimal where a_k=9 for all k. Given this definition, it follows that the value of the decimal is precisely 1, using properties of geometric sequences.

It is up to you how you define '...', but all mathematicians would agree that 0.99999... should be interpreted as above if it is to have any meaning at all.

If you really want a more precise eplanation of '...' at the end of a truncated decimal, I would provide the following: "Writing 0.abcdef... asserts that the digits abcdef of the truncation provided have a pattern which should be obvious to the reader. Assign the first few digits a_1, a_2 etc as per the part of the decimal that is explicitly given; and then assign all subsequent digits values according to said pattern." - it's not a formal notation, as I say, but rather a convenient shorthand that is understood by working mathematicians. It is always possible to be more precise if one has to be.

Mathematical definitions aren't like dictionary definitions. A closed ellipsis:

    1...100

An open-ended ellipsis is accepted as meaning an infinite repetition of what precedes it. So this ellipsis:

    0.9999...

http://en.wikipedia.org/wiki/0.999...

A quote: "In mathematics, the repeating decimal 0.999... (sometimes written with more or fewer 9s before the final ellipsis, or as 0.9, , 0.(9)) denotes a real number that can be shown to be the number one. In other words, the symbols "0.999..." and "1" represent the same number. Proofs of this equality have been formulated with varying degrees of mathematical rigor, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience."

Remember that mathematics, like science, avoids ambiguity where possible. And mathematical notation strives for less ambiguity than mathematics itself.