I attempted to answer this question on Quora, and was told that I am thinking about the problem incorrectly. The question was:
Two distinct real numbers between 0 and 1 are written on two sheets of
paper. You have to select one of the sheets randomly and declare
whether the number you see is the biggest or smallest of the two. How
can one expect to be correct more than half the times you play this
game?
My answer was that it was impossible, as the probability should always be 50% for the following reason:
You can't! Here's why:
The set of real numbers between (0, 1) is known as an Uncountably Infinite Set
(https://en.wikipedia.org/wiki/Uncountable_set). A set that is
uncountable has the following interesting property:Let $\mathbb{S}$ be an uncountably infinite set. Let, $a, b, c, d \in \mathbb{S} (a \neq b, c \neq d)$. If $x$ is an uncountably infinite subset of
$\mathbb{S}$, containing all elements in $\mathbb{S}$ on the interval $(a, b)$; and $y$
is another uncountably infinite subset of $\mathbb{S}$, which contains all
elements of $\mathbb{S}$ on the interval $(c, d),$ $x$ and $y$ have the same
cardinality (size)!So for example, the set of all real numbers between (0, 1) is actually
the exact same size as the set of all real numbers between (0, 2)!
It is also the same size as the set of all real numbers between (0,
0.00001). In fact, if you have an uncountably infinite set on the interval $(a, b)$, and $a<n<b$, then then exactly 50% of the numbers
in the set are greater than $n$, and 50% are less than $n$, no matter
what you choose for $n$. This is important because it tells us
something unintuitive about our probability in this case. Let's say
the first number you picked is 0.03. You might think "Well, 97% of the
other possible numbers are larger than this, so the other number is
probably larger." You would be wrong! There are actually exactly
as many numbers between (0, 0.03) as there are between (0.03, 1). Even
if you picked 0.03, half of the other possible numbers are smaller
than it, and half of the other possible numbers are larger than it.
This means there is still a 50% probability that the other number is larger, and a 50% probability that it is smaller!"But how can that be?" you ask, "why isn't $\frac{a-b}{2}$ the
midpoint?"The real question is, why is it that we believe that
$\frac{a-b}{2}$ is the midpoint to begin with? The reason is probably
the following: it seems to make the most sense for discrete
(finite/countably infinite) sets. For example, if instead of the real
numbers, we took the set of all multiples of $0.001$ on the interval
$[0, 1]$. Now it makes sense to say that 0.5 is the midpoint, as we
know that the number of numbers below 0.5 is equal to the number of
numbers above 0.5. If we were to try to say that the midpoint is 0.4,
we would find that there are now more numbers above 0.4 then there are
below 0.4. This no longer applies when talking about the set of all
real numbers $\mathbb{R}$. Strangely enough, we can no longer talk
about having a midpoint in $\mathbb{R}$, because every number in
$\mathbb{R}$ could be considered a midpoint. For any point in
$\mathbb{R}$, the numbers above it and the numbers below it always
have the same cardinality.See the Wikipedia article on Cardinality of the continuum
(https://en.wikipedia.org/wiki/Cardinality_of_the_continuum).
My question is, from a mathematical point of view, is this correct? The person who told me that this is wrong is fairly well known, and not someone who I would assume to often be wrong, especially for these types of problems.
The reasoning given for my answer being wrong was as follows:
Your conclusion is not correct.
You're right that the set of real
numbers between 0 and 1 is uncountable infinite, and most of what you
said here is correct. But that last part is incorrect. If you picked a
random real number between 0 and 1, the number does have a 97% chance
of being above 0.03. Let's look at this another way. Let K = {all
integers divisible by 125423423}. Let M = {all integers not divisible
by 125423423}. K and M are the same size, right? Does this mean, if
you picked an random integer, it has a 50% chance of being in K and a
50% chance or not? A random integer has a 50% chance of being
divisible by 125423423?
The reason I disagreed with this response was because the last sentence should actually be true. If the set of all numbers that are divisible by 125423423 is the same size as the set of numbers that aren't, there should be a 50% probability of picking a random number from the first set, and a 50% chance that a number would be picked from the second. This is cirtainly the case with finite sets. If there are 2 disjoint, finite sets with equal cardinality, and you choose a random number from the union of the two sets, there should be a 50% chance that the number came from the first set, and a 50% chance that the number came from the second set. Can this idea be generalized for infinite sets of equal cardinality?
Is my answer wrong? If so, am I missing something about how cardinalities of two set relate to the probability of choosing a number from one of them? Where did I go wrong in my logic?
Best Answer
Yes, your answer is fundamentally wrong. Let me point at that it is not even right in the finite case. In particular, you are using the following false axiom:
However, this is wrong even if we have just two events. For a somewhat real life example, consider some random variable $X$ which is $1$ if I will get married exactly a year from today and which is $0$ otherwise. Now, clearly the sets $\{1\}$ and $\{0\}$ are equally large, each having one element. However, $0$ is far more likely than $1$, although they are both possible outcomes.
The point here is probability is not defined from cardinality. It is, in fact, a separate definition. The mathematical definition for probability goes something like this:
One puts various conditions on $\mu$ to make sure it makes sense, but nowhere do we link it to cardinality. As an example, in the previous example with outcomes $0$ and $1$ which are not equally likely, one might have $\mu$ defined something like: $$\mu(\{\})=0$$ $$\mu(\{0\})=\frac{9999}{10000}$$ $$\mu(\{1\})=\frac{1}{10000}$$ $$\mu(\{0,1\})=1$$ which has nothing to do with the portion of the set of outcomes, which would be represented by the function $\mu'(S)=\frac{|S|}2$.
In general, your discussion of cardinality is correct, but it is irrelevant. Moreover, the conclusions you draw are inconsistent. The sets $(0,1)$ and $(0,\frac{1}2]$ and $(\frac{1}2,1)$ are pairwise equally large, so your reasoning says they are equally probable. However, the number was defined to be in $(0,1)$ so we're saying all the probabilities are $1$ - so we're saying that we're certain that the result will be in two disjoint intervals. This never happens, yet your method predicts that it always happens.
On a somewhat different note, but related in the big picture, you talk about "uncountably infinite sets" having the property that any non-trivial interval is also uncountable. This is true of $\mathbb R$, but not all uncountable subsets - like $(-\infty,-1]\cup \{0\} \cup [1,\infty)$ has that the interval $(-1,1)=\{0\}$ which is not uncountably infinite. Worse, not all uncountable sets have an intrinsic notion of ordering - how, for instance, do you order the set of subsets of natural numbers? The problem is not that there's no answer, but that there are many conflicting answers to that.
I think, maybe, the big thing to think about here is that sets really don't have a lot of structure. Mathematicians add more structure to sets, like probability measures $\mu$ or orders, and these fundamentally change their nature. Though bare sets have counterintuitive results with sets containing equally large copies of themselves, these don't necessarily translate when more structure is added.