Dr. Stern's Nasty Problem

Ok, let's consider the toss of a coin that has probability p(head) = 1/2, p(tail) = 1/2 with tail := 0 and head := 1.

Now, every point on the unit interval can be located precisely by a sequence of an infinite number of coin tosses. Think of it this way. Every point (number) in the unit interval [0,1) can be converted from our decimal system into the binary system, which is nothing more than an infinite number of 0's and 1's ( coin tosses ). Even the numbers that you may think have only a finite number of digits achieve an infinite number of digits by merely adding on an infinite number of 0's, which represents getting an infinite number of tails after a certain point. Doesn't sound very random, does it.

Ok, ... let's look at the following diagram to get a visual feel for what's going on here. Notice that if we toss our coin 4 times that every possible sequence is shown. Let's say that the coin comes up tail, head, head, head, i.e., 0111.

Each of the horizontal lines above is defined specifically by a binary function. The top line is defined as:

In general the infinite number of lines possible are defined by the binary function:

Now notice that we can associate the sequence 0111 with any number in the interval [7/16,1/2) that we want to, but, specifically, let's associate it with the left endpoint, 7/16. Now, notice what happens when we take the definition of our binary function and apply it as follows:

b_1 (7/16)/2^1 + b_2(7/16)/2^2 + b_3(7/16)/2^3 + b-4(7/16)/2^4 = 0 + 1/4 + 1/8 + 1/16 = 7/16

Wow, that just happens to be the left end point. In fact if we had continued to toss our coin an infinite number of times and got tails every time thereafter, we would have the infinite series 0111000,...,000,...000... which would locate specifically the point 7/16 in the unit interval [0,1).

Actually, this is just a definition of converting a decimal number in the unit interval [0,1) into binary - nothing too conceptually difficult here. Notice that

b_1 (7/16)/2^1 + b_2(7/16)/2^2 + b_3(7/16)/2^3 + b-4(7/16)/2^4 = 0/(2^1) + 1/(2^2) + 1/(2^3) + 1/(2^4)

This is simply the long hand form of the binary number .0111 in which the ( . ) is a binary point, not a decimal point.

If we were in the decimal system, .0111 = 0/(10^1) + 1/(10^2) + 1/(10^3) + 1/(10^4)

Well, at any rate, it is true that t = b_i(t)/2^i , i = 1, infinity

But what if our coin is not fair. What if p can be any value in the unit interval [0,1)? Then we will call our function p_i(t) to indicate that the probability of the coin does not have to be fair.

It is not true that t = p_i(t)/2^i , i = 1, infinity

For example, if probability = 2/3 for a head, 1/3 for a tail, then p_i(1/3)/2^i = 1/(2^1) + 0/(2^2) = 1/2

and we see that 1/3 is not = to 1/2.

SO HERE'S THE PROBLEM.

Since it is true that t = b_i(t)/2^i , i = 1, infinity , what is the generalization for p_i(t) ?

Please know that I still don't know the answer, and that I doubt I will have time to think about it until at least May. Maybe I'll get lucky and never think about it again.