Using the digits of $\pi$ to generate random numbers. Let's say I've been captured by Russian operatives and am locked in a room with only one object: a book listing the digits of $\pi$.
I'm told to generate a sequence of binary digits. If this sequence is random, they will cut off one of my arms and let me free; if this sequence is not random, however, I will be killed.
My first solution was to take the digits of $\pi \ \text{mod} \ 2$, so that:
$$3.1415926535897...$$
$$\downarrow$$
$$1.1011100111011...$$
And I would read the digits from left-to-right of the second number.
My Question
Is there any way to prove that the bits I generate are random (no discernible pattern)?
Are the digits of $p \ \text{mod} \ 2$ random for any transcendental $p$? How about any irrational $p$?
I feel like this should be a really easy question (with an affirmative answer), but I don't know how to show it.
 A: $\pi$ has an infinite number of digits. 
Thus a book listing all the digits of $\pi$ would be infinitely big. 
So the room containing both you and the book would be infinitely big as well. 
You are in an infinitely big room, hence you are free.
A: You should be precise about what you mean when you say "random", since "no discernible pattern" is still ambiguous.
You should read about pseudorandom number generators, because that may be the kind of randomness you want.
As for $\pi$, it isn't even known whether $\pi$ is a normal number in any base. 
A: There are long lists of digits for $\pi$ online. You could write a computer program that calculates different types of statistics to test your hypothesis.

One interesting statistic could be to measure $$P(X_{i+1}=1|X_{i},\cdots,X_{i-n})$$That is: the conditioned probability of one bit being $1$ given that we know the binary number of length $n$ preceding it. This conditioned probability should converge to the stationary $P(X_{i+1}=1)$ for all such $n$ bit numbers if there is no short term-memory in our source.
A: I suspect you are wondering if you could begin arbitrarily at a point somewhere within the infinite pool of decimal numbers and then start creating One Time Pads (OTP).
Although it would be hard to detect by most people, and it does emulate randomness, it's not truly "pure random" because it does technically have a discernable pattern. It would fool most, but not all. It likely could be used for non-cryptographic applications. You would not want to use Pi(π), Phi(Φ), Euler's Constant(e), the Bible, War and Peace, Moby Dick, or any other known, discernable sources if the intent is for "pure random" crytography.
A: What does "random" mean? In one natural sense (or class of senses, rather) coming from computability theory, your sequence is not random: even though it's hard to notice it, there is a pattern to the bits. 
While it's clear that in some intuitive sense "$\pi$ mod 2" is somewhat random, and "$\pi$" is somewhat less random, I suspect it will be very difficult to pin down exactly what you mean by this (and so very difficult to give a satisfying answer to your question!).
A: Someone could say that there's nothing random in Pi's digits, meaning that those digits are there already, you just don't know what they are! Just don't tell the Russian how you got those binary digits!
